Continuity and Differentiability
204 JEE Maths previous year questions on Continuity and Differentiability — options free on every question; 20 include the answer & explanation free, the rest unlock with PYQ Pass.
\(y=[x]+|x-2|\) if number of discontinuous point are " \(p\) " and number of non-differentiable points are " \(q\) " then find \(p+q\) where \(x \in(-2,3)\).
[JEE Main 2025]
8
\(f(x)= \begin{cases}-x ; & x \in(-2,-1) \\ 1-x ; & x \in[-1,0) \\ 2-x ; & x \in[0,1) \\ 3-x ; & x \in[1,2) \\ x ; & x \in[2,3)\end{cases}\)
As we can see that function is discontinuous at \(-1,0,1\) and \(2\) . Therefore, function is non-differentiable at \(-1,0,1\) and 2.
Therefore, \(\mathrm{p}=4\) and \(\mathrm{q}=4\).
Then, \(\mathrm{p}+\mathrm{q}=8\).
The value of \(\lim _{x\to 0}\frac{{\log }_{e}\left(\sec \left(ex\right)⋅\sec \left({e}^{2}x\right)⋅\ldots ⋅\sec \left({e}^{10}x\right)\right)}{{e}^{2}−{e}^{2\cos x}}\) is equal to
[JEE Main 2026, 28 Jan (Shift 1)]
\(\frac{\left({e}^{20}−1\right)}{2\left({e}^{2}−1\right)}\)
\(L=\lim _{x\to 0}\frac{{\log }_{e}\left(\sec \left(ex\right)⋅\sec \left({e}^{2}x\right)⋅\ldots ⋅\sec \left({e}^{10}x\right)\right)}{{e}^{2}−{e}^{2\cos x}}\)
\(=\lim _{x\to 0}\frac{\ln \left(\sec \left(ex\right)\right)+\ln \left(\sec \left({e}^{2}x\right)\right)+\ldots ..+\ln \left(\sec \left({e}^{10}x\right)\right)}{{e}^{2\cos x}\left(\frac{{e}^{2−2\cos x}−1}{2−2\cos x}\right)\times \frac{2−2\cos x}{{x}^{2}}\times {x}^{2}}\)
\(=\lim _{x\to 0}\frac{\ln \left(\sec \left(ex\right)\right)+\ln \left(\sec \left({e}^{2}x\right)\right)+\ldots \ldots +\ln \left(\sec \left({e}^{10}x\right)\right)}{{e}^{2}{x}^{2}}\)
Using L'H rule
\(=\lim _{x\to 0}\frac{e\tan \left(ex\right)+{e}^{2}\tan \left({e}^{2}x\right)+\ldots ..+{e}^{10}\tan \left({e}^{10}x\right)}{2{e}^{2}x}\)
\(=\frac{1}{2{e}^{2}}\left[{e}^{2}+{e}^{4}+{e}^{6}+\ldots .+{e}^{20}\right]\)
\(=\frac{1}{2}\frac{{e}^{2}\left({\left({e}^{2}\right)}^{10}−1\right)}{{e}^{2}\left({e}^{2}−1\right)}\)
\(=\frac{1}{2}\frac{\left({e}^{20}−1\right)}{\left({e}^{2}−1\right)}\)
The product of all possible values of \(\alpha\), for which \(\lim _{x\to 0}\left(\frac{1-\cos (\alpha x)\cos ((\alpha +1)x)\cos ((\alpha +2)x)}{{\sin }^{2}((\alpha +1)x)}\right)=2\), is:
[JEE Main 2026, 5 Apr (Shift 1)]
\(-1\)
Using \(\cos \theta=1-\frac{\theta^2}{2}+\ldots\)
\(\cos (\alpha x) \cos ((\alpha+1) x) \cos ((\alpha+2) x) =1-\frac{\alpha^2 x^2+(\alpha+1)^2 x^2+(\alpha+2)^2 x^2}{2}+\ldots\)
So, \(1-\cos (\alpha x) \cos ((\alpha+1) x) \cos ((\alpha+2) x) =\frac{\alpha^2+(\alpha+1)^2+(\alpha+2)^2}{2} x^2\)
Also, \(\sin ^2((\alpha+1) x)=(\alpha+1)^2 x^2\)
\( \therefore \frac{1}{2(\alpha+1)^2}\left[\alpha^2+(\alpha+1)^2+(\alpha+2)^2\right]=2\)
\(\alpha^2+(\alpha+1)^2+(\alpha+2)^2=4(\alpha+1)^2 \)
\( \alpha^2+(\alpha+2)^2=3(\alpha+1)^2 \)
\( \Rightarrow \alpha^2+2 \alpha-1=0\)
Product \(=-1\)
Let \(f(x)\) be a continuously differentiable function on the interval \((0,\infty )\) such that \(f(1)=2\) and \(\lim _{t\to x}\frac{{t}^{10}f(x)-{x}^{10}f(t)}{{t}^{9}-{x}^{9}}=1\) for each \(x>0.\) Then, for all \(x>0, f(x)\) is equal to
[JEE Advanced 2024]
\(\frac{9}{11x}+\frac{13}{11}{x}^{10}\)
\(\lim_{t\to x}\frac{10t^9f(x)-f'(t)x^{10}}{9t^8}=1\)
\(\Rightarrow 10x^9f(x)-f'(x)x^{10}=9x^8\)
\(\Rightarrow f'(x)-\frac{10}{x}f(x)=-\frac{9}{x^2}\)
Let \(y=f(x)\).
\(IF=e^{-\int \frac{10}{x}\,dx}\)
\(IF=\frac{1}{x^{10}}\)
\(\frac{y}{x^{10}}=\int -\frac{9}{x^2}\cdot \frac{1}{x^{10}}\,dx\)
\(\frac{y}{x^{10}}=-9\int x^{-12}\,dx\)
\(\frac{y}{x^{10}}=\frac{9}{11}x^{-11}+C\)
\(\because y(1)=2\Rightarrow C=\frac{13}{11}\)
\(\Rightarrow y=\frac{9}{11x}+\frac{13}{11}x^{10}\)
If \(\lim _{\mathrm{x}\to 0}\frac{\cos (2\mathrm{x})+\mathrm{acos}(4\mathrm{x})-\mathrm{b}}{{\mathrm{x}}^{4}}\) is finite, then \((\mathrm{a}+\mathrm{b})\) is equal to :
[JEE Main 2025, 2 Apr (Shift 2)]
\(\frac{1}{2}\)
Use the Taylor expansions around \(x=0\):
\(\cos (2x)=1−2{x}^{2}+\frac{2}{3}{x}^{4}+O({x}^{6})\)
\(\cos (4x)=1−8{x}^{2}+\frac{32}{3}{x}^{4}+O({x}^{6})\)
\(\cos (2x)+a\cos (4x)−b=(1+a−b)+(−2−8a){x}^{2}+(\frac{2}{3}+\frac{32a}{3}){x}^{4}+O({x}^{6})\)
For
\(\underset{x\to 0}{\lim }\frac{\cos (2x)+a\cos (4x)−b}{{x}^{4}}\)
to be finite, the numerator must be \(O({x}^{4})\). Thus the constant and \({x}^{2}\) terms must vanish:
\(1+a−b=0\ \text{(i)}\)
\(−2−8a=0\ \text{(ii)}\)
From (ii): \(a=−\frac{1}{4}\).
Substitute into (i):
\(1−\frac{1}{4}−b=0\Rightarrow b=\frac{3}{4}\)
So
\(a+b=−\frac{1}{4}+\frac{3}{4}=\frac{1}{2}\mathrm{.}\)
\(y=[x]+|x-2|\) if number of discontinuous point are " \(p\) " and number of non-differentiable points are " \(q\) " then find \(p+q\) where \(x \in(-2,3)\).
8
\(f(x)= \begin{cases}-x ; & x \in(-2,-1) \\ 1-x ; & x \in[-1,0) \\ 2-x ; & x \in[0,1) \\ 3-x ; & x \in[1,2) \\ x ; & x \in[2,3)\end{cases}\)
As we can see that function is discontinuous at \(-1,0,1\) and \(2\) . Therefore, function is non-differentiable at \(-1,0,1\) and 2.
Therefore, \(\mathrm{p}=4\) and \(\mathrm{q}=4\).
Then, \(\mathrm{p}+\mathrm{q}=8\).
\(\lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}} \text { is equal to: }\)
[JEE Main 2025, 23 Jan (Shift 2)]
\(\frac{2}{3 \sqrt{e}}\)
\(\lim _{x\to \infty }\frac{\left(2-\frac{3}{x}+\frac{5}{{x}^{2}}\right){\left(1-\frac{1}{3x}\right)}^{x/2}}{\left(3+\frac{5}{x}+\frac{4}{{x}^{2}}\right){\left(1+\frac{2}{3x}\right)}^{x/2}}\\ \lim _{x\to \infty }\frac{\left(2-\frac{3}{x}+\frac{5}{{x}^{2}}\right)}{\left(3+\frac{5}{x}+\frac{4}{{x}^{2}}\right)}\lim _{x\to \infty }\frac{{\left(1-\frac{1}{3x}\right)}^{x/2}}{{\left(1+\frac{2}{3x}\right)}^{x/2}}\\ =\frac{2}{3}\lim _{x\to \infty }{\left(\frac{\left(1-\frac{1}{3x}\right)}{\left(1+\frac{2}{3x}\right)}\right)}^{x/2}\\ =\frac{2}{3}\cdot \frac{{e}^{\frac{x}{2}\left(1-\frac{1}{3x}-1\right)}}{{e}^{\frac{x}{2}\left(1+\frac{2}{3x}-1\right)}}\\ =\frac{2}{3}\cdot \frac{{\mathrm{e}}^{-\frac{1}{6}}}{{\mathrm{e}}^{1/3}}=\frac{2}{3}{\mathrm{e}}^{-\frac{1}{2}}\)
\(\lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}}\) is equal to
[JEE Main 2025, 23 Jan (Shift 2)]
\(\frac{2}{3 \sqrt{e}}\)
\(\lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}}\)
\(=\lim _{x\to \infty }\frac{\left(2-\frac{3}{x}+\frac{5}{{x}^{2}}\right){\left(1-\frac{1}{3x}\right)}^{x/2}}{\left(3+\frac{5}{x}+\frac{4}{{x}^{2}}\right){\left(1+\frac{2}{3x}\right)}^{x/2}}\)
\(=\lim _{x\to \infty }\frac{\left(2-\frac{3}{x}+\frac{5}{{x}^{2}}\right)}{\left(3+\frac{5}{x}+\frac{4}{{x}^{2}}\right)}\lim _{x\to \infty }\frac{{\left(1-\frac{1}{3x}\right)}^{x/2}}{{\left(1+\frac{2}{3x}\right)}^{x/2}}\)
\(=\frac{2}{3}\lim _{x\to \infty }{\left(\frac{\left(1-\frac{1}{3x}\right)}{\left(1+\frac{2}{3x}\right)}\right)}^{x/2}\)
\(=\frac{2}{3}\cdot \frac{{e}^{\frac{x}{2}\left(1-\frac{1}{3x}-1\right)}}{{e}^{\frac{x}{2}\left(1+\frac{2}{3x}-1\right)}}\)
\(=\frac{2}{3}\cdot \frac{{\mathrm{e}}^{-\frac{1}{6}}}{{\mathrm{e}}^{1/3}}=\frac{2}{3}{\mathrm{e}}^{-\frac{1}{2}}\)
Consider the following statements in respect of a function \(f(x)\)
I. \(f(x)\) is continuous at \(x=a\), if \(\lim _{x\to a}f(x)\) exists.
II. If \(f(x)\) is continuous at a point, then \(\frac{1}{f(x)}\) is also continuous at the point.
Which of the above statement(s) is/are correct?
Neither I nor II
I. Wrong by definition.
II. It's not necessary.
For example:
\(f(x)=x\) is continuous at \(x=0\)
But \(\frac{1}{f(x)}=\frac{1}{x}\) is not continuous at \(x=0\).
Let \(f:[-1,2]\to R\) be given by \(f(x)=2{x}^{2}+x+\left[{x}^{2}\right]-[x]\) , where \([t]\) denotes the greatest integer less than or equal to \(t\). The number of points, where \(f\) is not continuous, is :
[JEE Main 2024, 5 Apr (Shift 2)]
4
Doubtful points: \(-1,0,1,\sqrt{2},\sqrt{3},2\)
At \(x=\sqrt{2},\sqrt{3}\),
\(f(x)=(2x^2+x-[x])+[x^2]\)
Here \(2x^2+x-[x]\) is continuous, but \([x^2]\) is discontinuous.
So, \(f(x)\) is discontinuous at \(x=\sqrt{2},\sqrt{3}\).
At \(x=-1\):
R.H.L. \(\Rightarrow f(-1^+)=2-1-(-1)+0=2\)
\(f(-1)=2-1-(-1)+1=3\)
So, \(f(x)\) is discontinuous at \(x=-1\).
At \(x=2\):
L.H.L. \(\Rightarrow f(2^-)=8+2-1+3=12\)
\(f(2)=8+2-2+4=12\)
R.H.L. \(\Rightarrow f(2^+)=8+2-2+4=12\)
So, \(f(x)\) is continuous at \(x=2\).
At \(x=0\):
L.H.L. \(\Rightarrow f(0^-)=0+0-(-1)+0=1\)
\(f(0)=0\)
So, \(f(x)\) is discontinuous at \(x=0\).
At \(x=1\):
L.H.L. \(\Rightarrow f(1^-)=2+1-0+0=3\)
\(f(1)=2+1-1+1=3\)
R.H.L. \(\Rightarrow f(1^+)=2+1-1+1=3\)
So, \(f(x)\) is continuous at \(x=1\).
If \(f\left(x\right)=\left\{\begin{matrix}{x}^{3}\sin \left(\frac{1}{x}\right), & x\neq 0 \\ 0 & ,x=0\end{matrix}\right.\), then
[JEE Main 2024, 6 Apr (Shift 1)]
\(f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}\)
Given \(f'(x)=3x^2\sin\left(\dfrac{1}{x}\right)-x\cos\left(\dfrac{1}{x}\right)\).
Differentiating again,
\(f''(x)=6x\sin\left(\dfrac{1}{x}\right)-3\cos\left(\dfrac{1}{x}\right)-\cos\left(\dfrac{1}{x}\right)-\dfrac{\sin\left(\dfrac{1}{x}\right)}{x}\)
\(\Rightarrow f''(x)=6x\sin\left(\dfrac{1}{x}\right)-4\cos\left(\dfrac{1}{x}\right)-\dfrac{\sin\left(\dfrac{1}{x}\right)}{x}\)
At \(x=\dfrac{2}{\pi}\),
\(f''\left(\dfrac{2}{\pi}\right)=6\cdot\dfrac{2}{\pi}\sin\dfrac{\pi}{2}-4\cos\dfrac{\pi}{2}-\dfrac{\sin\dfrac{\pi}{2}}{\dfrac{2}{\pi}}\)
\(\Rightarrow f''\left(\dfrac{2}{\pi}\right)=\dfrac{12}{\pi}-0-\dfrac{\pi}{2}\)
\(\Rightarrow f''\left(\dfrac{2}{\pi}\right)=\dfrac{24-\pi^2}{2\pi}\)
Let \(f(x)=a x^3+b x^2+c x+41\) be such that \(f(1)=40, f^{\prime}(1)=2\) and \(f^{\prime \prime}(1)=4\). Then \(a ^2+ b ^2+ c ^2\) is equal to :
[JEE Main 2024, 9 Apr (Shift 1)]
51
Given function \(f(x) = ax^3 + bx^2 + cx + 41\)
\({f}^{'}(x)=3a{x}^{2}+2bx+c\)
\(\Rightarrow {f}^{'}(1)=3a+2b+c=2\ldots (i)\)
\({f}^{"}(x)=6ax+2b\)
\(\Rightarrow f^{\prime \prime}(1)=6 a+2 b=4\)
\(3 a+b=2 \ldots(i i)\)
\((i)-(i i) \Rightarrow b+c=0 \ldots(i i i)\)
\(f(1)=40 \Rightarrow a+b+c+41=40\)
use (iii) use (iii) \(\Rightarrow a+41=40 \Rightarrow a=-1\)
by (ii) \(\Rightarrow-3+b=2 \Rightarrow b=5\) and \( c=-5\)
\(a^2 + b^2 + c^2 = 1 + 25 + 25 = 51\)
Let \([t]\) denote the greatest integer less than or equal to \(t\). If the function
\(f\left(x\right)=\left\{\begin{matrix}{b}^{2}\text{sin}\left(\frac{\pi }{2}\left[\frac{\pi }{2}\left(\text{cos}x+\text{sin}x\right)\text{cos}x\right]\right),\text{ }x<0 \\ \text{ }\frac{\sin x−\frac{1}{2}\sin 2x}{{x}^{3}}\text{ },\text{ }x>0 \\ \text{ }a\text{ },\text{ }x=0\end{matrix}\right.\)
is continuous at \(x = 0\), then \(a^2+b^2\) is equal to
[JEE Main 2026, 24 Jan (Shift 2)]
\(\frac{3}{4}\)
Given \(f(x)\) is continuous at \(x=0\)
\(f(0)=a\)
\(\mathrm{RHL}=\lim _{x\to {0}^{+}}\frac{\sin x(1−\cos x)}{{x}^{3}}\\ =\lim _{x\to {0}^{+}}\left(\frac{\sin x}{x}\right)\left(\frac{1-\cos x}{{x}^{2}}\right)\\ =\frac{1}{2}\)
\(\mathrm{LHL}=\lim _{x\to {0}^{−}}\left({b}^{2}\sin \left(\frac{\pi }{2}\left[\frac{\pi }{2}(\sin x+\cos x)\cos x\right]\right)\right)\\ ={b}^{2}\)
\(∴a=\frac{1}{2}\)
\({b}^{2}=\frac{1}{2}\)
So, \(\left({a}^{2}+{b}^{2}\right)=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)
Let f(x) be a continuously differentiable function on the interval (0, \(\infty\)) such that f(1) = 2
and \(\lim _{t\to x}\frac{{t}^{10}f(x)-{x}^{10}f(t)}{{t}^{9}-{x}^{9}}=1\)for each x > 0. Then, for all x > 0, f(x) is equal to
\(\frac{9}{11x}+\frac{13}{11}{x}^{10}\)
(B)
Let \(g(x)\) be a linear function and \(f(x)=\left\{\begin{array}{cl}g(x) & , x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & , x>0\end{array}\right.\), is continuous at \(x=0\). If \(f^{\prime}(1)=f(-1)\), then the value \(g(3)\) is
[JEE Main 2024, 31 Jan (Shift 1)]
\(\log _e\left(\frac{4}{9 e^{1 / 3}}\right)\)
Let \(g(x)=ax+b\).
Now function \(f(x)\) is continuous at \(x=0\).
\(\therefore \lim_{x\to 0^+}f(x)=f(0)\)
\(\Rightarrow \lim_{x\to 0^+}\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}=b\)
\(\Rightarrow b=0\)
\(\therefore g(x)=ax\)
Now, for \(x>0\),
\(f(x)=\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}\)
Taking log on both sides,
\(\ln f(x)=\frac{1}{x}\ln\left(\frac{1+x}{2+x}\right)\)
Differentiating both sides,
\(\frac{f'(x)}{f(x)}=\frac{1}{x}\cdot\frac{1}{(1+x)(2+x)}-\frac{1}{x^2}\ln\left(\frac{1+x}{2+x}\right)\)
\(\Rightarrow f'(x)=\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}\left[\frac{1}{x(1+x)(2+x)}-\frac{1}{x^2}\ln\left(\frac{1+x}{2+x}\right)\right]\)
\(\therefore f'(1)=\frac{2}{3}\left[\frac{1}{6}-\ln\left(\frac{2}{3}\right)\right]\)
\(\Rightarrow f'(1)=\frac{1}{9}-\frac{2}{3}\ln\left(\frac{2}{3}\right)\)
And \(f(-1)=g(-1)=-a\).
\(\therefore -a=\frac{1}{9}-\frac{2}{3}\ln\left(\frac{2}{3}\right)\)
\(\Rightarrow a=\frac{2}{3}\ln\left(\frac{2}{3}\right)-\frac{1}{9}\)
\(\therefore g(3)=3a\)
\(=2\ln\left(\frac{2}{3}\right)-\frac{1}{3}\)
\(=\ln\left(\frac{4}{9}\right)-\ln e^{\frac{1}{3}}\)
\(=\ln\left(\frac{4}{9e^{\frac{1}{3}}}\right)\)
\(\text{ If }y(\theta )=\frac{2\cos \theta +\cos 2\theta }{\cos 3\theta +4\cos 2\theta +5\cos \theta +2}\text{, then at }\theta =\frac{\pi }{2},{y}^{''}+{y}^{'}+y\text{ is equal to : }\)
2
Given: \(y(\theta )=\frac{2\cos \theta +\cos 2\theta }{\cos 3\theta +4\cos 2\theta +5\cos \theta +2}\)
We know that, \(\cos 2\theta =2{\cos }^{2}\theta -1\)
\(\cos 3\theta =4{\cos }^{3}\theta -3\cos \theta\)
\(y(\theta )=\frac{2\cos \theta +2{\cos }^{2}\theta -1}{4{\cos }^{3}\theta -3\cos \theta +8{\cos }^{2}\theta -4+5\cos \theta +2}\)
\(y(\theta )=\frac{2{\cos }^{2}\theta +2\cos \theta -1}{4{\cos }^{3}\theta +8{\cos }^{2}\theta +2\cos \theta -2}\)
\(y(\theta )=\frac{2{\cos }^{2}\theta +2\cos \theta -1}{(2\cos \theta +2)(2{\cos }^{2}\theta +2\cos \theta -1)}\)
\(y(\theta )=\frac{1}{2(\cos \theta +1)}\)
\(y(\theta )=\frac{1}{2\cdot 2{\cos }^{2}\frac{\theta }{2}}\)
\(y(\theta )=\frac{1}{4}{\sec }^{2}\frac{\theta }{2}\)
\(y'(\theta )=\frac{1}{4}\cdot 2\sec \frac{\theta }{2}\left(\sec \frac{\theta }{2}\tan \frac{\theta }{2}\right)\frac{1}{2}\)
\(y'(\theta )=\frac{1}{4}{\sec }^{2}\frac{\theta }{2}\tan \frac{\theta }{2}\)
\(y''(\theta )=\frac{1}{4}\left({\sec }^{4}\frac{\theta }{2}\times \frac{1}{2}+{\sec }^{2}\frac{\theta }{2}{\tan }^{2}\frac{\theta }{2}\right)\)
\(y''(\theta )=\frac{1}{8}{\sec }^{4}\frac{\theta }{2}+\frac{1}{4}{\sec }^{2}\frac{\theta }{2}{\tan }^{2}\frac{\theta }{2}\)
\(y''(\theta )+y'(\theta )+y(\theta )=\frac{1}{8}{\sec }^{4}\frac{\theta }{2}+\frac{1}{4}{\sec }^{2}\\ \frac{\theta }{2}{\tan }^{2}\frac{\theta }{2}+\frac{1}{4}{\sec }^{2}\frac{\theta }{2}\tan \frac{\theta }{2}+\frac{1}{4}{\sec }^{2}\frac{\theta }{2}\)
\(\text{at }\theta =\frac{\pi }{2},\)
\(y''\left(\frac{\pi }{2}\right)+y'\left(\frac{\pi }{2}\right)+y'\left(\frac{\pi }{2}\right)=\frac{1}{8}\times 4+\frac{1}{4}\times 2\times 1+\frac{1}{4}\times 2\times 1+\frac{1}{4}\times 2\)
\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
If \(f(x)=\left|\begin{array}{ccc}a+\frac{\sin x}{x} & 1 & b \\ a & 1+\frac{\sin x}{x} & b \\ a & 1 & b+\frac{\sin x}{x}\end{array}\right|\)
if \(\lim _{x \rightarrow 0^{+}} f(x)=\mu+\alpha a+\beta b\) then find the value of \((\mu+\alpha+\beta)^2\)
16
\(\begin{aligned}&& \lim _{x \rightarrow 0^{+}} f(x)=\left|\begin{array}{ccc}a+1 & 1 & b \\a & 1+1 & b \\a & 1 & b+1\end{array}\right|=\mu+\alpha a+\beta b \\& R_1 \rightarrow R_1-R_2 \\& R_2 \rightarrow R_2-R_3 \\& \left|\begin{array}{ccc}1 & -1 & 0 \\0 & 1 & -1 \\a & 1 & b+1\end{array}\right|=\mu+\alpha a+\beta b \\& C_2 \rightarrow C_2+C_1 \\& \left|\begin{array}{ccc}1 & 0 & 0 \\0 & 1 & -1 \\a & a+1 & b+1\end{array}\right|=\mu+\alpha a+\beta b \\& a+b+2=\mu+\alpha a+\beta b \\& \quad \text { So } \alpha=1, \mu=2, \beta=1&\text { Now, } \quad(\mu+\alpha+\beta)^2=16\end{aligned}\)
\(\lim _{x\to 0}\csc x\left(\sqrt{2{\cos }^{2}x+3\cos x}-\sqrt{{\cos }^{2}x+\sin x+4}\right)\) is
[JEE Main 2025, 24 Jan (Shift 1)]
\(-\frac{1}{2\sqrt{5}}\)
\(\lim _{x\to 0}\csc x\left(\sqrt{2{\cos }^{2}x+3\cos x}-\sqrt{{\cos }^{2}x+\sin x+4}\right)\)
\(=\lim _{x\to 0}\frac{2{\cos }^{2}x+3\cos x-{\cos }^{2}x-\sin x-4}{\sin x\left(\sqrt{2{\cos }^{2}x+3\cos x}+\sqrt{{\cos }^{2}x+\sin x+4}\right)}\)
\(=\lim _{x\to 0}\frac{{\cos }^{2}x+3\cos x-\sin x-4}{\sin x(\sqrt{5}+\sqrt{5})}\)
on applying, L'Hospitals rule,
\(=\lim _{x\to 0}\frac{-2\cos x\sin x-3\sin x-\cos x}{\cos x\cdot 2\sqrt{5}}=-\frac{1}{2\sqrt{5}}\)
Let \(\mathbb{R}\) denote the set of all real numbers. Define the function \(f:\mathrm{ℝ}\to \mathrm{ℝ}\) by
\(f\left(x\right)=\left\{\begin{matrix}2-2{x}^{2}-{x}^{2}\sin \frac{1}{x}, & \text{ if }x\neq 0 \\ 2, & \text{ if }x=0\end{matrix}\right.\)
Then which one of the following statements is TRUE?
[JEE Advanced 2025]
For any positive real number \(\delta\), the function \(f\) is NOT an increasing function on the interval \((-\delta ,0)\)
(A) RHD at \(x=0: \lim _{h \rightarrow 0} \frac{\left(2-2 h^2-h^2 \sin \frac{1}{h}\right)-2}{h}=0\)
Similarly LHD at \(\mathrm{x}=0\) is also equal to 0 .
∴ Differentiable at \(\mathrm{x}=0\)
(D)
\(\mathrm{f}(0)=2,\\ \mathrm{f}(0+\mathrm{h})<2\\ \mathrm{f}(0-\mathrm{h})<2\)
therefore \(x=0\) is a point of local maxima
(B) Now,
\({f}^{'}(x)=-\left(4x+2x\sin \frac{1}{x}\right)+\cos \frac{1}{x}\)
for any \(\delta>0\), choose \(x=\frac{1}{2 \pi n} \in(0, \delta)\) with \(n\) large.
Then \(\sin \frac{1}{x}=0, \cos \frac{1}{x}=1\), so
\(f^{\prime}(x)=1-4 x>0\)
A decreasing function cannot have \(f^{\prime}(x)>0\) at a point where derivative exists.
(C) for any \(\delta>0\), choose \(x=-\frac{1}{(2 n+1) \pi} \in(-\delta, 0)\)
with \(n\) large. Then \(\cos \frac{1}{x}=-1\) and
\(f^{\prime}(x)=-4 x-2 x \sin \frac{1}{x}-1 \leq 6|x|-1<0\)
for sufficiently small \(|x|\)
Hence \(f\) cannot be increasing on ( \(-\delta, 0\) )
Let \(f\) be a real polynomial of degree \(n\), \(f(\mathrm{x})={f}^{'}(\mathrm{x}){f}^{''}(\mathrm{x}),\) for all \(x\in R\), If \(f(0)=0\), then \( 36\left(f^{\prime}(2)+f^{\prime \prime}(2)+\int_0^2 f(x) d x\right)\) is equal to:
[JEE Main 2026, 4 Apr (Shift 1)]
\(56\)
Let degree of \(f(x)\) be \(n\)
Since \(f(x)=f^{\prime}(x) f^{\prime \prime}(x)\),
comparing degrees gives \(n=(n-1)+(n-2)\), so \(n=3\)
Let \(f(x)=a x^3+b x^2+c x+d\)
Since \(f(0)=0, d=0\), so \(f(x)=a x^3+b x^2+c x\)
Now \(f^{\prime}(x)=3 a x^2+2 b x+c\) and \(f^{\prime \prime}(x)=6 a x+2 b\)
Using \(f(x)=f^{\prime}(x) f^{\prime \prime}(x)\):
\(a x^3+b x^2+c x=\left(3 a x^2+2 b x+c\right)(6 a x+2 b)\)
So \(a x^3+b x^2+c x=18 a^2 x^3+18 a b x^2+\left(6 a c+4 b^2\right) x+2 b c\)
Comparing coefficients:
\(a=18 a^2\)
Hence \(a=\frac{1}{18}\)
Also \(b c=0\) and \(c=6 a c+4 b^2\)
Using \(a=\frac{1}{18}, c=\frac{c}{3}+4 b^2\), so \(c=6 b^2\)
Since \(b c=0\), we get \(b=0\), hence \(c=0\)
Therefore \(f(x)=\frac{x^3}{18}\)
So \(f^{\prime}(x)=\frac{x^2}{6}\) and \(f^{\prime \prime}(x)=\frac{x}{3}\)
Hence \(f^{\prime}(2)=\frac{2}{3}, f^{\prime \prime}(2)=\frac{2}{3}\),
and \(\int_0^2 f(x) d x=\int_0^2 \frac{x^3}{18} d x=\left.\frac{x^4}{72}\right|_0 ^2=\frac{2}{9}\)
Therefore \(36\left(f^{\prime}(2)+f^{\prime \prime}(2)+\int_0^2 f(x) d x\right)=36\left(\frac{2}{3}+\frac{2}{3}+\frac{2}{9}\right)=56\)
If \(y\left(x\right)=\left|\begin{matrix}\mathrm{sinx} & \mathrm{cosx} & \mathrm{sinx}+\mathrm{cosx}+1 \\ 27 & 28 & 27 \\ 1 & 1 & 1\end{matrix}\right|,\mathrm{x}\in \mathrm{ℝ}\), then \(\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}+\mathrm{y}\) is equal to
[JEE Main 2025, 3 Apr (Shift 1)]
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Let \(f(x)={x}^{5}+2{x}^{3}+3x+1,x\in R\), and \(g(x)\) be a function such that \(g(f(x))=x\) for all \(x\in R\). Then\(\frac{g(7)}{{g}^{'}(7)}\) is equal to:
[JEE Main 2024, 5 Apr (Shift 1)]
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At \(x=\frac{\pi^2}{4}, \frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(e^x\right)\right)=\)
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If \(\lim _{\mathrm{x}\to 0}\frac{\cos (2\mathrm{x})+\mathrm{acos}(4\mathrm{x})-\mathrm{b}}{{\mathrm{x}}^{4}}\) is finite, then \((\mathrm{a}+\mathrm{b})\) is equal to :
[JEE Main 2025, 2 Apr (Shift 2)]
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If \(y=\tan ^{-1}\left(\frac{\sqrt{x}-x}{1+x^{3 / 2}}\right)\), then \(y^{\prime}(1)\) is equal to:
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If \(\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha\), then the value of \(\frac{\log _e \alpha}{1+\log _e \alpha}\) equals :
[JEE Main 2025, 22 Jan (Shift 2)]
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Let \([t]\) denote the greatest integer less than or equal to \(t\). If the function
\(f\left(x\right)=\left\{\begin{matrix}{b}^{2}\text{sin}\left(\frac{\pi }{2}\left[\frac{\pi }{2}\left(\text{cos}x+\text{sin}x\right)\text{cos}x\right]\right),\text{ }x<0 \\ \text{ }\frac{\sin x−\frac{1}{2}\sin 2x}{{x}^{3}}\text{ },\text{ }x>0 \\ \text{ }a\text{ },\text{ }x=0\end{matrix}\right.\)
is continuous at \(x = 0\), then \(a^2+b^2\) is equal to
[JEE Main 2026, 24 Jan (Shift 2)]
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Let \(y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1 [JEE Main 2024, 29 Jan (Shift 2)]
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Let \(y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1 [JEE Main 2024, 29 Jan (Shift 2)]
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For \(a , b >0\), let \(f(x)= \begin{cases}\frac{\tan (( a +1) x)+ b \tan x}{x}, & x<0 \\ 3, & x=0 \\ \frac{\sqrt{ a x+ b ^2 x^2}-\sqrt{ a x}}{ b \sqrt{ a } x \sqrt{x}}, & x>0\end{cases}\) be a continous function at \(x=0\). Then \(\frac{ b }{ a }\) is equal to:
[JEE Main 2024, 08 Apr (Shift 2)]
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For \(a , b >0\), let \(f(x)= \begin{cases}\frac{\tan (( a +1) x)+ b \tan x}{x}, & x<0 \\ 3, & x=0 \\ \frac{\sqrt{ a x+ b ^2 x^2}-\sqrt{ a x}}{ b \sqrt{ a } x \sqrt{x}}, & x>0\end{cases}\) be a continous function at \(x=0\). Then \(\frac{ b }{ a }\) is equal to:
[JEE Main 2024, 08 Apr (Shift 2)]
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If \(\lim _{\mathrm{x}\to {1}^{+}}\frac{(\mathrm{x}-1)(6+\mathrm{λcos}(\mathrm{x}-1))+\mathrm{μsin}(1-\mathrm{x})}{(\mathrm{x}-1{)}^{3}}=-1\), where \(\lambda ,\mu \in R\), then \(\lambda +\mu\) is equal to
[JEE Main 2025, 4 Apr (Shift 1)]
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If \(f(x)=\left\{\begin{array}{cc}\frac{a|x|+x^2-2(\sin |x|)(\cos |x|)}{x} & , x \neq 0 \\ b & , x=0\end{array}\right.\)
is continuous at \(x = 0\), then \(a+b\) is equal to
[JEE Main 2026, 23 Jan (Shift 2)]
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Let \(x_0\) be the real number such that \(e^{x_0}+x_0=0\). For a given real number \(\alpha\), define \(g\left(x\right)=\frac{3x{e}^{x}+3x-\alpha {e}^{x}-\alpha x}{3\left({e}^{x}+1\right)}\) for all real numbers \(x\). Then which one of the following statements is TRUE?
[JEE Advanced 2025]
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Let \(f: \mathrm{R} \rightarrow \mathrm{R}\) be a twice differentiable function such that \((\operatorname{sinxcos} \mathrm{y})(f(2 \mathrm{x}+2 \mathrm{y})-f(2 \mathrm{x}-2 \mathrm{y})) =(\cos x \sin y)(f(2 x+2 y)+f(2 x-2 y))\), for all \(x, y \in R\). If \(f^{\prime}(0)=\frac{1}{2}\), then the value of \(24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)\) is:
[JEE Main 2025, 2 Apr (Shift 1)]
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If \(\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha\), then the value of \(\frac{\log _e \alpha}{1+\log _e \alpha}\) equals :
[JEE Main 2025, 22 Jan (Shift 2)]
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Let the function \(f(x)=\left(x^2-3\right)\left|x^2-a x+2\right|+\cos |x|\) be not differentiable at two points \(x=\alpha=2\) and \(x=\beta\). Then the distance of the point \((\alpha, \beta)\) from the line \(12 x+5 y+10=0\) is equal to :
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For the function \(f(x)={e}^{\sin |x|}-|x|,x\in R\), consider the following statements:
Statement I: \(f\) is differentiable for all \(x\in R\)
Statement II: \(f\) is increasing in \(\left(-\pi ,-\frac{\pi }{2}\right)\)
In the light of the above statements, choose the correct answer from the options given below:
[JEE Main 2026, 8 Apr (Shift 2)]
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If \(y={\tan }^{-1}\left(\frac{3\cos x-4\sin x}{4\cos x+3\sin x}\right)+2{\tan }^{-1}\left(\frac{x}{1+\sqrt{1-{x}^{2}}}\right)\)then \(\frac{dy}{dx}\) at \(x=\frac{\sqrt{3}}{2}\) is equal to:
[JEE Main 2026, 4 Apr (Shift 1)]
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Let \(f(x)={x}^{5}+2{e}^{x/4}\) for all \(x\in R\). Consider a function \(g(x)\) such that \((g∘f)(x)=x\) for all \(x\in R\). Then the value of \(8{g}^{'}(2)\) is :
[JEE Main 2024, 4 Apr (Shift 1)]
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Let \(f:R\to R\) be a twice differentiable function such that
\((\mathrm{sinxcosy})(f(2\mathrm{x}+2\mathrm{y})-f(2\mathrm{x}-2\mathrm{y}))\)\(=(\mathrm{cosxsiny})(f(2\mathrm{x}+2\mathrm{y})+f(2\mathrm{x}-2\mathrm{y}))\),
for all \(\mathrm{x},\mathrm{y}\in R\).If \({f}^{'}(0)=\frac{1}{2}\), then the value of \(24{f}^{''}\left(\frac{5\pi }{3}\right)\)is:
[JEE Main 2025, 2 Apr (Shift 1)]
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Let \(f(x)=a x^3+b x^2+c x+41\) be such that \(f(1)=40, f^{\prime}(1)=2\) and \(f^{\prime \prime}(1)=4\). Then \(a ^2+ b ^2+ c ^2\) is equal to :
[JEE Main 2024, 9 Apr (Shift 1)]
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Let \(f: R \rightarrow R\) be defined as : \(f(x)= \begin{cases}\frac{ a - b \cos 2 x}{x^2} ; & x<0 \\ x^2+ c x+2 ; & 0 \leq x \leq 1 \\ 2 x+1 & ; x>1\end{cases}\)
If \(f\) is continuous everywhere in \(R\) and \(m\) is the number of points where \(f\) is NOT differentiable then \(m + a + b + c\) equals :
[JEE Main 2024, 1 Feb (Shift 1)]
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The value of \(\lim _{x\to 0}\frac{{\log }_{e}\left(\sec \left(ex\right)⋅\sec \left({e}^{2}x\right)⋅\ldots ⋅\sec \left({e}^{10}x\right)\right)}{{e}^{2}−{e}^{2\cos x}}\) is equal to
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Suppose for a differentiable function \(h, h(0)=0, h(1)=1\) and \(h^{\prime}(0)=h^{\prime}(1)=2\). If \(g (x)=h\left( e ^x\right) e ^{h(x)}\), then \(g^{\prime}(0)\) is equal to :
[JEE Main 2024, 6 Apr (Shift 2)]
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Suppose for a differentiable function \(h, h(0)=0, h(1)=1\) and \(h^{\prime}(0)=h^{\prime}(1)=2\). If \(g (x)=h\left( e ^x\right) e ^{h(x)}\), then \(g^{\prime}(0)\) is equal to :
[JEE Main 2024, 6 Apr (Shift 2)]
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If \(y=\tan ^{-1}\left(\frac{\sqrt{x}-x}{1+x^{3 / 2}}\right)\), then \(y^{\prime}(1)\) is equal to:
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If \(f(x)=\left\{\begin{array}{cc}\frac{a|x|+x^2-2(\sin |x|)(\cos |x|)}{x} & , x \neq 0 \\ b & , x=0\end{array}\right.\)
is continuous at \(x = 0\), then \(a+b\) is equal to
[JEE Main 2026, 23 Jan (Shift 2)]
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\(\lim _{x\to 0}\csc x(\sqrt{2{\cos }^{2}x+3\cos x}\)\(-\sqrt{{\cos }^{2}x+\sin x+4})\) is
[JEE Main 2025, 24 Jan (Shift 1)]
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\(\lim _{x\to {0}^{+}}\frac{\tan \left(5(x{)}^{\frac{1}{3}}\right){\log }_{e}\left(1+3{x}^{2}\right)}{{\left({\tan }^{-1}3\sqrt{x}\right)}^{2}\left({e}^{5(x{)}^{\frac{4}{3}}}-1\right)}\) is equal to
[JEE Main 2025, 7 Apr (Shift 1)]
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Let \(f(x)= \begin{cases}\frac{\mathrm{a} x^2+2 \mathrm{a} x+3}{4 x^2+4 x-3} & , x \neq-\frac{3}{2}, \frac{1}{2} \\ b & , x=-\frac{3}{2}, \frac{1}{2}\end{cases}\)
be continuous at \(x=-\frac{3}{2}\text{. If }f∘f(x)=\frac{7}{5}\), then \(x\) is equal to:
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If \(\log _e y=3 \sin ^{-1} x\), then \(\left(1-x^2\right) y^{\prime \prime}-x y^{\prime}\) at \(x=\frac{1}{2}\) is equal to
[JEE Main 2024, 9 Apr (Shift 2)]
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Given below are two statements :
Statement I : \(\lim _{x\to 0}\left(\frac{{\tan }^{-1}x+{\log }_{e}\sqrt{\frac{1+x}{1-x}}-2x}{{x}^{5}}\right)=\frac{2}{5}\)
Statement II : \(\lim _{x\to 1}\left({x}^{\frac{2}{1-x}}\right)=\frac{1}{{e}^{2}}\)
In the light of the above statements, choose the correct answer from the options given below :
[JEE Main 2025, 8 Apr (Shift 1)]
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Let [.] denote the greatest integer function, and let \(f(x)=\min \left\{\sqrt{2} x, x^2\right\}\). Let \(S=\{x \in(-2,2)\) : the function \(g(x)=|x|\left[x^2\right]\) is discontinuous at \(\left.x\right\}\). Then \(\sum_{x \in S} f(x)\) equals
[JEE Main 2026, 22 Jan (Shift 2)]
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Let \(f(x)= \begin{cases}\frac{\mathrm{a} x^2+2 \mathrm{a} x+3}{4 x^2+4 x-3} & , x \neq-\frac{3}{2}, \frac{1}{2} \\ b & , x=-\frac{3}{2}, \frac{1}{2}\end{cases}\)
be continuous at \(x=-\frac{3}{2}\text{. If }f∘f(x)=\frac{7}{5}\), then \(x\) is equal to:
[JEE Main 2026, 23 Jan (Shift 1)]
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If \(\lim _{x\to 2}\frac{\sin \left({x}^{3}-5{x}^{2}+ax+b\right)}{(\sqrt{x-1}-1){\log }_{e}(x-1)}=m\), then \(a+b+m\) is equal to:
[JEE Main 2026, 2 Apr (Shift 1)]
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If the function
\(f\left(x\right)=\left\{\begin{matrix}\frac{{72}^{x}-{9}^{x}-{8}^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}},x\neq 0 \\ a{\log }_{e}2{\log }_{e}3,x=0\end{matrix}\right.\)
is continuous at \(x=0\), then the value of \({a}^{2}\) is equal to
[JEE Main 2024, 4 Apr (Shift 2)]
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If \(f(x)=\left|\begin{array}{ccc}a+\frac{\sin x}{x} & 1 & b \\ a & 1+\frac{\sin x}{x} & b \\ a & 1 & b+\frac{\sin x}{x}\end{array}\right|\)
if \(\lim _{x \rightarrow 0^{+}} f(x)=\mu+\alpha a+\beta b\) then find the value of \((\mu+\alpha+\beta)^2\)
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Let \(f:\mathrm{ℝ}\to \mathrm{ℝ}\) be a function defined by \(f(x)=\left\{\begin{array}{cc}x^2 \sin \left(\frac{\pi}{x^2}\right), & \text { if } x \neq 0, \\0, & \text { if } x=0\end{array}\right.\)Then which of the following statements is TRUE?
[JEE Advanced 2024]
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Given below are two statements :
Statement I : \(\lim _{x\to 0}\left(\frac{{\tan }^{-1}x+{\log }_{e}\sqrt{\frac{1+x}{1-x}}-2x}{{x}^{5}}\right)=\frac{2}{5}\)
Statement II : \(\lim _{x\to 1}\left({x}^{\frac{2}{1-x}}\right)=\frac{1}{{e}^{2}}\)
In the light of the above statements, choose the correct answer from the options given below :
[JEE Main 2025, 8 Apr (Shift 1)]
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If \(\lim _{\mathrm{x}\to {1}^{+}}\frac{(\mathrm{x}-1)(6+\mathrm{λcos}(\mathrm{x}-1))+\mathrm{μsin}(1-\mathrm{x})}{(\mathrm{x}-1{)}^{3}}=-1\), where \(\lambda ,\mu \in \mathrm{ℝ}\), then \(\lambda +\mu\) is equal to
[JEE Main 2025, 4 Apr (Shift 1)]
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For \(\alpha ,\beta ,\gamma \in \mathrm{R}\), if \(\lim _{\mathrm{x}\to 0}\frac{{\mathrm{x}}^{2}\mathrm{sinαx}+(\gamma -1){\mathrm{e}}^{{\mathrm{x}}^{2}}}{\sin 2\mathrm{x}-\mathrm{βx}}=3\), then \(\beta +\gamma -\alpha\) is equal to:
[JEE Main 2025, 2 Apr (Shift 1)]
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At \(x=\frac{\pi^2}{4}, \frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(e^x\right)\right)=\)
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Let \(f\left(x\right)=\text{ }\lim _{\theta \text{ }\to \text{ }0}\text{ }\left(\frac{\cos \text{ }\pi \text{ }x−\text{ }{x}^{\left(\frac{2}{\theta }\right)}\text{ }\sin \left(x−1\right)}{1+{x}^{\left(\frac{2}{\theta }\right)}\text{ }\left(x−1\right)}\right),\text{ }x\text{ }\in \text{ }R.\)
Consider the following two statements:
(I) \(f\left(x\right)\) is discontinuous at \(x = 1\).
(II) \(f\left(x\right)\) is continuous at \(x = – 1\).
Then,
[JEE Main 2026, 28 Jan (Shift 2)]
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\(\lim _{n\to \infty }\frac{\left({1}^{2}-1\right)(n-1)+\left({2}^{2}-2\right)(n-2)+⋯+\left((n-1{)}^{2}-(n-1)\right)\cdot 1}{\left({1}^{3}+{2}^{3}+⋯⋯+{n}^{3}\right)-\left({1}^{2}+{2}^{2}+⋯⋯+{n}^{2}\right)}\) is equal to :
[JEE Main 2024, 6 Apr (Shift 2)]
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Suppose \(f(x)=\frac{\left(2^{x}+2^{-x}\right) \tan x \sqrt{\tan ^{-1}\left(x^{2}-x+1\right)}}{\left(7 x^{2}+3 x+1\right)^{3}}\). Then the value of \(f^{\prime}(0)\) is equal to
[JEE Main 2024, 29 Jan (Shift 1)]
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Suppose \(f(x)=\frac{\left(2^{x}+2^{-x}\right) \tan x \sqrt{\tan ^{-1}\left(x^{2}-x+1\right)}}{\left(7 x^{2}+3 x+1\right)^{3}}\). Then the value of \(f^{\prime}(0)\) is equal to
[JEE Main 2024, 29 Jan (Shift 1)]
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Let \(k\in \mathrm{ℝ}\). If \(\lim _{x \rightarrow 0+}(\sin (\sin k x)+\cos x+x)^{\frac{2}{x}}=e^6\), then the value of \(k\) is
[JEE Advanced 2024]
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Let \(f:(-\infty ,\infty )-{0}\to \mathrm{ℝ}\) be a differentiable function such that \(f^{\prime}(1)=\lim _{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right)\). Then \(\lim _{a\to \infty }\frac{a(a+1)}{2}{\tan }^{-1}\left(\frac{1}{a}\right)+{a}^{2}-2{\log }_{e}a\) is equal to
[JEE Main 2024, 6 Apr (Shift 1)]
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If the function \(f(x)=\) \(\left\{\begin{matrix}\frac{2}{x}\left\{\sin \left({k}_{1}+1\right)x+\sin \left({k}_{2}-1\right)x\right\}, & x<0 \\ 4 & x=0 \\ \frac{2}{x}{\log }_{e}\left(\frac{2+{\mathrm{k}}_{1}\mathrm{x}}{2+{\mathrm{k}}_{2}\mathrm{x}}\right), & x>0\end{matrix}\right.\)is continuous at \(x=0,\) then \({\mathrm{k}}_{1}^{2}+{\mathrm{k}}_{2}^{2}\) is equal to
[JEE Main 2025, 23 Jan (Shift 1)]
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If \(y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}\), then at \(\theta=\frac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y\) is equal to :
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Let \(f:R\to R\) be a polynomial function of degree four having extreme values at \(x=4\) and \(x=5\). If \(\lim _{x\to 0}\frac{f(\mathrm{x})}{{\mathrm{x}}^{2}}=5\), then \(f(2)\) is equal to :
[JEE Main 2025, 7 Apr (Shift 2)]
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If \(f(x)\) is defined as follows: \(f(x)=\left\{\begin{matrix}4,-\infty If \(k\) is the number of points where \(f(x)\) is not differentiable, then \(k-2=\)
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Let \(f\left(x\right)=\lim _{\theta \to 0}\left(\frac{\cos \pi x−{x}^{\left(\frac{2}{\theta }\right)}\sin \left(x−1\right)}{1+{x}^{\left(\frac{2}{\theta }\right)}\left(x−1\right)}\right),x\in R.\)
Consider the following two statements:
(I) \(f\left(x\right)\) is discontinuous at \(x = 1\).
(II) \(f\left(x\right)\) is continuous at \(x = – 1\).
Then,
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Let \(f\left(\mathrm{x}\right)=\left\{\begin{matrix}(1+\mathrm{ax}{)}^{1/\mathrm{x}} & , & \mathrm{x}<0 \\ 1+\mathrm{b} & , & \mathrm{x}=0 \\ \frac{(\mathrm{x}+4{)}^{1/2}-2}{(\mathrm{x}+\mathrm{c}{)}^{1/3}-2} & , & \mathrm{x}>0\end{matrix}\right.\) be continuous at \(\mathrm{x}=0\). Then \({\mathrm{e}}^{\mathrm{a}}\mathrm{bc}\) is equal to
[JEE Main 2025, 3 Apr (Shift 1)]
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For \(\alpha ,\beta ,\gamma ,\in R\), if \(\lim _{\mathrm{x}\to 0}\frac{{\mathrm{x}}^{2}\mathrm{sinαx}+(\gamma -1){\mathrm{e}}^{{\mathrm{x}}^{2}}}{\sin 2\mathrm{x}-\mathrm{βx}}=3\), then \(\beta +\gamma -\alpha\)is equal to:
[JEE Main 2025, 2 Apr (Shift 1)]
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If \(y\left(x\right)=\left|\begin{matrix}\mathrm{sinx} & \mathrm{cosx} & \mathrm{sinx}+\mathrm{cosx}+1 \\ 27 & 28 & 27 \\ 1 & 1 & 1\end{matrix}\right|,\mathrm{x}\in \mathrm{ℝ}\), then \(\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}+\mathrm{y}\) is equal to
[JEE Main 2025, 3 Apr (Shift 1)]
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Let [.] denote the greatest integer function, and let \(f(x)=\min \left\{\sqrt{2} x, x^2\right\}\). Let \(S=\{x \in(-2,2)\) : the function \(g(x)=|x|\left[x^2\right]\) is discontinuous at \(\left.x\right\}\). Then \(\sum_{x \in S} f(x)\) equals
[JEE Main 2026, 22 Jan (Shift 2)]
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If the function \(f(x)=\) \(\left\{\begin{matrix}\frac{2}{x}\left\{\sin \left({k}_{1}+1\right)x+\sin \left({k}_{2}-1\right)x\right\}, & x<0 \\ 4 & x=0 \\ \frac{2}{x}{\log }_{e}\left(\frac{2+{\mathrm{k}}_{1}\mathrm{x}}{2+{\mathrm{k}}_{2}\mathrm{x}}\right), & x>0\end{matrix}\right.\)
is continuous at \(x=0,\) then \({\mathrm{k}}_{1}^{2}+{\mathrm{k}}_{2}^{2}\) is equal to
[JEE Main 2025, 23 Jan (Shift 1)]
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Let \(f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, \quad x \text { is odd, }\end{array} x \in N\right.\). If for some \(a \in N , f(f(f( a )))=21\), then \(\lim _{x \rightarrow a ^{-}}\left\{\frac{|x|^3}{ a }-\left[\frac{x}{ a }\right]\right\}\), where \([t]\) denotes the greatest integer less than or equal to \(t\), is equal to :EndFragment
[JEE Main 2024, 01 Feb (Shift 2)]
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Let \(f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, \quad x \text { is odd, }\end{array} x \in N\right.\). If for some \(a \in N , f(f(f( a )))=21\), then \(\lim _{x \rightarrow a ^{-}}\left\{\frac{|x|^3}{ a }-\left[\frac{x}{ a }\right]\right\}\), where \([t]\) denotes the greatest integer less than or equal to \(t\), is equal to :EndFragment
[JEE Main 2024, 01 Feb (Shift 2)]
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Let the function \(f(x)=\left(x^2-3\right)\left|x^2-a x+2\right|+\cos |x|\) be not differentiable at two points \(x=\alpha=2\) and \(x=\beta\). Then the distance of the point \((\alpha, \beta)\) from the line \(12 x+5 y+10=0\) is equal to :
[JEE Main 2025, 29 Jan (Shift 2)]
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\(\lim _{x \rightarrow 0^{+}} \frac{\tan \left(5(x)^{\frac{1}{3}}\right) \log _e\left(1+3 x^2\right)}{\left(\tan ^{-1} 3 \sqrt{x}\right)^2\left(e^{5(x)^{\frac{4}{3}}}-1\right)}\) is equal to
[JEE Main 2025, 7 Apr (Shift 1)]
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If the function \(f\left(x\right)=\frac{\sin 3x+\alpha \sin x-\beta \cos 3x}{{x}^{3}},x\in R\), is continuous at \(x=0\), then \(f(0)\) is equal to:
[JEE Main 2024, 5 Apr (Shift 1)]
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Consider the function.
\(f(x)=\left\{\begin{array}{cc}\frac{ a \left(7 x-12-x^2\right)}{ b \left|x^2-7 x+12\right|}, & x<3 \\ 2^{\frac{\sin (x-3)}{x-[x]}} & , x>3 \\ b & , x=3,\end{array}\right.\)
where \([x]\) denotes the greatest integer less than or equal to \(x\). If \(S\) denotes the set of all ordered pairs \((a, b)\) such that \(f(x)\) is continuous at \(x=3\), then the number of elements in \(S\) is:
[JEE Main 2024, 27 Jan (Shift 1)]
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Consider the function.
\(f(x)=\left\{\begin{array}{cc}\frac{ a \left(7 x-12-x^2\right)}{ b \left|x^2-7 x+12\right|}, & x<3 \\ 2^{\frac{\sin (x-3)}{x-[x]}} & , x>3 \\ b & , x=3,\end{array}\right.\)
where \([x]\) denotes the greatest integer less than or equal to \(x\). If \(S\) denotes the set of all ordered pairs \((a, b)\) such that \(f(x)\) is continuous at \(x=3\), then the number of elements in \(S\) is:
[JEE Main 2024, 27 Jan (Shift 1)]
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Let \(f: R -\{0\} \rightarrow R\) be a function satisfying \(f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}\) for all \(x, y, f(y) \neq 0\). If \(f^{\prime}(1)=2024\), then
[JEE Main 2024, 30 Jan (Shift 2)]
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Let \(f: R -\{0\} \rightarrow R\) be a function satisfying \(f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}\) for all \(x, y, f(y) \neq 0\). If \(f^{\prime}(1)=2024\), then
[JEE Main 2024, 30 Jan (Shift 2)]
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Let \(\alpha, \beta \in \mathbb{R}\) be such that the function \(f(\mathrm{x})= \begin{cases}2 \alpha\left(\mathrm{x}^2-2\right)+2 \beta \mathrm{x} & , \mathrm{x}<1 \\ (\alpha+3) \mathrm{x}+(\alpha-\beta) & , \mathrm{x} \geq 1\end{cases}\) be differentiable at all \(x \in \mathbb{R}\). Then \(34(\alpha+\beta)\) is equal to
[JEE Main 2026, 24 Jan (Shift 1)]
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Let \(f:R\to R\) be a function given by
\(f\left(x\right)=\left\{\begin{matrix}\frac{1-\cos 2x}{{x}^{2}},x<0\\ \alpha x=0,\\ \frac{\beta \sqrt{1-\cos x}}{x},x>0 \\ \end{matrix}\right.\)
where \(\alpha ,\beta \in R\). If \(f\) is continuous at \(x=0\), then \({\alpha }^{2}+{\beta }^{2}\) is equal to :
[JEE Main 2024, 04 Apr (Shift 1)]
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Let \(f:R\to R\) be a function given by
\(f\left(x\right)=\left\{\begin{matrix}\frac{1-\cos 2x}{{x}^{2}},x<0\\ \alpha x=0,\\ \frac{\beta \sqrt{1-\cos x}}{x},x>0 \\ \end{matrix}\right.\)
where \(\alpha ,\beta \in R\). If \(f\) is continuous at \(x=0\), then \({\alpha }^{2}+{\beta }^{2}\) is equal to :
[JEE Main 2024, 04 Apr (Shift 1)]
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Let \(f\left(\mathrm{x}\right)=\left\{\begin{matrix}(1+\mathrm{ax}{)}^{1/\mathrm{x}} & , & \mathrm{x}<0 \\ 1+\mathrm{b} & , & \mathrm{x}=0 \\ \frac{(\mathrm{x}+4{)}^{1/2}-2}{(\mathrm{x}+\mathrm{c}{)}^{1/3}-2} & , & \mathrm{x}>0\end{matrix}\right.\)
be continuous at \(\mathrm{x}=0\). Then \({\mathrm{e}}^{\mathrm{a}}\mathrm{bc}\) is equal to
[JEE Main 2025, 3 Apr (Shift 1)]
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Let \(\mathbb{R}\) denote the set of all real numbers. Define the function \(f:\mathrm{ℝ}\to \mathrm{ℝ}\) by
\(f\left(x\right)=\left\{\begin{matrix}2-2{x}^{2}-{x}^{2}\sin \frac{1}{x}, & \text{ if }x\neq 0 \\ 2, & \text{ if }x=0\end{matrix}\right.\)
Then which one of the following statements is TRUE?
[JEE Advanced 2025]
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If \(f(x)\) is defined as follows: \(f(x)=\left\{\begin{matrix}4,-\infty If \(k\) is the number of points where \(f(x)\) is not differentiable, then \(k-2=\)
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Let \( \mathrm{f}(\mathrm{x}) \) be a differentiable function at \( x=a \) with \( f^{\prime}(a)=2 \) and \( f(a)=4 \). Then \( \lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a} \)equals:
[JEE Main 2021, 26 Feb (Shift 2)]
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Let \(f\) be any continuous function on \([0,2]\) and twice differentiable on \((0,2)\). If \(f(0)=0, f(1)=1\) and \(f(2)=2\), then
[JEE Main 2021, 31 Aug (Shift 2)]
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Let \(f\) be a twice differentiable function defined on \(R\) such that \(f(0)=1, f^{\prime}(0)=2\) and \(f(x) \neq 0\) for all \(x \in R\). If \(\left|\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0\), for all \(x \in R\), then the value of \(f(1)\) lies in the interval.
[JEE Main 2021, 24 Feb (Shift 2)]
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Let \( f: R \rightarrow R \) be defined as \( f(x)=\left\{\begin{array}{cc}2 \sin \left(-\frac{\pi x}{2}\right), & \text { if } x<-1 \\ \left|a x^{2}+x+b\right|, & \text { if }-1 \leq x \leq 1 \\ \sin (\pi x) & \text { if } x>1\end{array}\right. \)
If \( f(x) \) is continuous on \( R \), then \( a+b \) equals
[JEE Main 2021, 26 Feb (Shift 2)]
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The function \( f(x)=\left|x^{2}-2 x-3\right| \cdot e^{\left|9 x^{2}-12 x+4\right|} \) is not differentiable at exactly
[JEE Main 2021, 31 Aug (Shift 1)]
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If \(\alpha = \lim_{x \rightarrow \frac{\pi}{4}}\frac{\tan^{3}x - \tan x}{\cos(x + \frac{\pi}{4})}\) and \(\beta = \lim_{x \rightarrow 0}(\cos x)^{\cot x}\) are the roots of the equation \(ax^{2} + bx -4 = 0\), the ordered pair \((a,b)\) is:
[JEE Main 2021, 31 Aug (Shift 2)]
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Let \(f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}, x \in[0, \pi]-\left\{\frac{\pi}{4}\right\}\). Then \(f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right)\) is equal to
[JEE Main 2023, 8 Apr (Shift 1)]
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If \(x y+y^2=\tan x+y\), then find \(\frac{d y}{d x}\) is
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\(\lim _{x \rightarrow \frac{\pi}{2}}\left(\tan ^2 x\left(\left(2 \sin ^2 x+3 \sin x+4\right)^{\frac{1}{2}}-\left(\sin ^2 x+6 \sin x+2\right)^{\frac{1}{2}}\right)\right)\) is equal to
[JEE Main 2022, 25 Jun (Shift 2)]
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If \(\ 2^x+2^y=2^{x+y} \), then \(\ \frac{d y}{d x}= \)
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Let \( x^{k}+y^{k}=a^{k}(a, k>0) \) and \( \frac{d y}{d x}+\left(\frac{y}{x}\right)^{\frac{1}{3}}=0 \), then \( k \) is
[JEE Main 2020, 7 Jan (Shift 1)]
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Let \( f: R \rightarrow R\) be defined as
\(f(x)=\left\{\begin{array}{cc}\frac{\lambda | x^{2}-5 x+6|}{\mu\left(5 x-x^{2}-6\right)} & , x<2 \\e^{\frac{\tan (x-2)}{x-[x]}} & , x>2 \\\mu & , x=2\end{array}\right.\)
Where \( [x] \) is the greatest integer less than or equal to \( x \). If \( f \) is continuous at \( x=2 \), then \( \lambda+\mu \) is equal to
[JEE Main 2021, 25 Jul (Shift 1)]
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Let \( [t] \) denote the greatest integer \( \leq t \) and \( \lim _{x \rightarrow 0} x\left[\frac{4}{x}\right]=A \). Then the function, \( f(x)=\left[x^{2}\right] \sin (\pi x) \) is discontinuous, when \( x \) is equal to :
[JEE Main 2020, 9 Jan (Shift 2)]
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\( \lim _{x \rightarrow 0}\left(\frac{3 x^{2}+2}{7 x^{2}+2}\right)^{1 / x^{2}} \) is equal to
[JEE Main 2020, 8 Jan (Shift 1)]
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Let \( f: S \rightarrow S \) where \( S=(0, \infty) \) be a twice differentiable function such that \( f(x+1)=x f(x) \). If \( g: S \rightarrow R \) be defined as \( g(x)= \) \( \log _{e} f(x) \), then the value of \( \left|g^{\prime \prime}(5)-g^{\prime \prime}(1)\right| \) is equl to
[JEE Main 2021, 16 Mar (Shift 2)]
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If the function \(f(x)=\left\{\begin{array}{cc}\frac{1}{x} \log _e\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right) & , x<0 \\k & , x=0 \\\frac{\cos ^2 x-\sin ^2 x-1}{\sqrt{x^2+1}-1} & , x>0\end{array}\right.\)
is continuous at \(x =0\), then \(\frac{1}{a}+\frac{1}{b}+\frac{4}{k}\) s equal to:
[JEE Main 2021, 31 Aug (Shift 1)]
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If \(y=\sqrt{\left(\frac{1+\cos 2 \theta}{1-\cos 2 \theta}\right)}\), then \(\frac{d y}{d \theta}\) at \(\boldsymbol{\theta}=\frac{3 \pi}{4}\) is :
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Let \( \mathrm{S} \) be the set of all function \( f:[0,1] \rightarrow \mathrm{R} \), which are continuous on \( [0,1] \) and differentiable on \( (0,1) \). Then for every \( f \) in \( S \), there exists a \( c \in(0,1) \), depending on \( f \), such that
[JEE Main 2020, 8 Jan (Shift 2)]
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Let \( y(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)\left(1+x^{16}\right) \) Then \( y^{\prime}-y^{\prime \prime} \) at \( x=-1 \) is equal to
[JEE Main 2023, 25 Jan (Shift 1)]
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Let \( f \) and \( g \) be differentiable functions on \( \mathrm{R} \) such that \(fog\) is the identity function. If for some \( a, b \in R, g^{\prime}(a)=5 \) and \( g(a)=b \), then \( f^{\prime}(b) \) is equal to :
[JEE Main 2020, 9 Jan (Shift 2)]
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If \( y(x)=x^{x}, x>0 \), then \( y^{\prime \prime}(2)-2 y^{\prime}(2) \) is equal to
[JEE Main 2023, 1 Feb (Shift 2)]
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The value of
\(\lim _{n\to \infty }\frac{1+2-3+4+5-6+\ldots +(3n-2)+(3n-1)-3n}{\sqrt{2{n}^{4}+4n+3-}\sqrt{{n}^{4}+5n+4}}\) is:
[JEE Main 2023, 25 Jan (Shift 1)]
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The derivation of \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) with respect to \(\tan ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right)\) at \(x=\frac{1}{2}\) is:
[JEE Main 2020, 5 Sep (Shift 2)]
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For all twice differentiable functions \(f: R \rightarrow R\), with \(f(0)\) \(=f(1)=f^{\prime}(0)=0 \)
[JEE Main 2020, 6 Sep (Shift 2)]
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Let \( f: R \rightarrow R \) be a function defined as
\(f(x)=\left\{\begin{array}{cc}\frac{\sin (a+1) x+\sin 2 x}{2 x}, & \text { if } x<0 \\b & , \text { if } x=0 \\\frac{\sqrt{x+b x^{3}}-\sqrt{x}}{b x^{5 / 2}} & , \text { if } x>0\end{array}\right.\)
If \( f \) is continuous at \( x=0 \), then the value of \( a+b \) is equal to
[JEE Main 2021, 18 Mar (Shift 2)]
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If \(\ \mathbf{x}=\frac{1-t^2}{1+t^2} \) and \(\mathbf{y}=\frac{2 t}{1+t^2} \), then \(\ \frac{d y}{d x} \) is equal to :
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Let
\[ y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)\right) \text {. } \]
Then, at \( x=1 \),
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Let \(y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)\right) \text {. } \)Then, at \( x=1 \),
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If \( \mathrm{f}(\mathrm{x})=\left\{\begin{array}{cl}\frac{\sin (\mathrm{a}+2) \mathrm{x}+\sin \mathrm{x}}{\mathrm{x}} ; & \mathrm{x}<0 \\ \mathrm{~b} & ; \mathrm{x}=0 \\ \frac{\left(\mathrm{x}+3 \mathrm{x}^{2}\right)^{1 / 3}-\mathrm{x}^{1 / 3}}{\mathrm{x}^{4 / 3}} ; & \mathrm{x}>0\end{array}\right. \)
is continuous at \( x=0 \), then \( a+2 b \) is equal to :
[JEE Main 2020, 9 Jan (Shift 1)]
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If \( \lim _{n \rightarrow \infty}\left(\sqrt{n^{2}-n-1}+n \alpha+\beta\right)=0 \), then \( 8(\alpha+\beta) \) is equal to
[JEE Main 2022, 25 Jul (Shift 1)]
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\(\ \left(\frac{d}{d x} \tan ^{-1} \sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right) \) is equal to
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The set of all values of a for which \(\lim _{x\to a}([x-5]-[2x+2])=0\), where\([\mu ]\) denotes the greater integer less than or equal to \(\mu\) is equal to
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If \(\alpha = \lim_{x \rightarrow \frac{\pi}{4}}\frac{\tan^{3}x - \tan x}{\cos(x + \frac{\pi}{4})}\) and \(\beta = \lim_{x \rightarrow 0}(\cos x)^{\cot x}\) are the roots of the equation \(ax^{2} + bx -4 = 0\), the ordered pair \((a,b)\) is:
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Let \(f\) be the function defined by
\(f(x)=\left\{\begin{matrix}\frac{{x}^{2}−1}{{x}^{2}−2|x−1|−1}, & x\neq 1 \\ 1/2, & x=1\end{matrix}\right.\)
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Let \(f\left(x\right)=\cos \left(2{\tan }^{-1}\sin \left({\cot }^{-1}\sqrt{\frac{1-x}{x}}\right)\right),0 [JEE Main 2021, 26 Aug (Shift 1)]
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Let \(f(x)=\left[x^2-x\right]+|-x+[x]|\), where \(x \in R\) and \([t]\) denotes the greatest integer less than or equal to \(t\). Then, \(f\) is
[JEE Main 2023, 11 Apr (Shift 1)]
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If a function \(f(x)\) defined by
\(f(x)= \begin{cases}\mathrm{ae}^x+\mathrm{be}^{-x}, & -1 \leq x<1 \\ \mathrm{cx}^2 & , 1 \leq x \leq 3 \\ \mathrm{a} x^2+2 \mathrm{cx}, & 3
[JEE Main 2020, 2 Sep (Shift 1)]
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\(\text { If the function }\)
\(f(x)= \begin{cases}(1+|\cos x|)^ \frac{\lambda}{|\cos x|} & , 0 [JEE Main 2023, 25 Jan (Shift 2)]
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Let \(f\) and \(g\) be twice differentiable functions on \(R\) such that
\(\begin{aligned}& f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x \\& f^{\prime}(1)=4 g^{\prime}(1)-3=9 \\& f(2)=3 g(2)=12\end{aligned}\)
Then which of the following is NOT true?
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Let [x] denotes the greatest integer function and \(\text{f}(\text{x})=\max {1+\text{x}+[\text{x}],2+\text{x},\text{x}+2[\text{x}]},\).\(0\leq \text{x}\leq 2\) Let m be the number of points in [0, 2], where f is not continuous and n be the number of points in (0, 2), where f is not differentiable. Then \({(m+n)}^{2}+2\) is equal to
[JEE Main 2023, 15 Apr (Shift 1)]
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If the function is equal to
\(f(x)= \begin{cases}(1+|\cos x|) \frac{\lambda}{|\cos x|}, & 0
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If \( \lim _{x \rightarrow 0} \frac{\sin ^{-1} x-\tan ^{-1} x}{3 x^{3}} \) is equal to \( l \), then the value of \( (6 l+1) \) is
[JEE Main 2021, 18 Mar (Shift 1)]
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If \(x=\frac{1−{t}^{2}}{1+{t}^{2}}\) and \(y=\frac{2t}{1+{t}^{2}},\) then \(\frac{dy}{dx}\) is equal to:
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Let \( f \) be any function defined on \( R \) and let it satisfy the condition: \( |f(x)-f(y)| \leq\left|(x-y)^{2}\right|, \forall(x, y) \in R \), if \( f(0)=1 \), then
[JEE Main 2021, 26 Feb (Shift 1)]
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Let \( f \) and \( g \) be two functions defined by
\( f(x)=\left\{\begin{array}{cc}x+1, & x<0 \\ |x-1|, & x \geq 0\end{array} \quad\right. \) and \( g(x)=\left\{\begin{array}{cc}x+1, & x<0 \\ 1, & x \geq 0\end{array}\right. \)
Then (gof) \( (x) \) is
[JEE Main 2023, 11 Apr (Shift 2)]
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If \( f(x)=\left\{\begin{array}{cl}\frac{1}{|x|} & ; \quad|x| \geq 1 \\ a x^{2}+b & ; \quad|x|<1\end{array}\right. \) is differentiable at every point of the domain, then the values of \( a \) and \( b \) are respectively
[JEE Main 2021, 18 Mar (Shift 1)]
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Let \([x]\) denote the greatest integer function and \(f(x)=\max \{1+x+[x], 2+x, x+2[x]\}, 0 \leq x \leq 2\). Let \(m\) be the number of points in \([0,2]\), where \(f\) is not continuous and \(n\) be the number of points in \((0,2)\), where \(f\) is not differentiable. Then \((m+n)^2+2\) is equal to
[JEE Main 2023, 15 Apr (Shift 1)]
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If \(f(x)=\left\{\begin{matrix}{e}^{x}+ax, & x<0 \\ b{(x−1)}^{2}, & x\geq 0\end{matrix}\right.\)is differentiable at \(x=0\), then \((a,b)\) is
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\(\lim _{n\to \infty }\left\{\left({2}^{\frac{1}{2}}-{2}^{\frac{1}{3}}\right)\left({2}^{\frac{1}{2}}-{2}^{\frac{1}{5}}\right)⋯⋯\left({2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}}\right)\right\}\) is equal to
[JEE Main 2023, 6 Apr (Shift 2)]
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If \(\alpha\) is the positive root of the equation, \(p(x)={x}^{2}-x-2=0\) then \(\lim _{x\to {\alpha }^{+}}\frac{\sqrt{1-\cos (p(x))}}{x+\alpha -4}\) is equal to.
[JEE Main 2020, 5 Sep (Shift 1)]
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\(\text { If the function } f(x)=\left\{\begin{array}{cc}\frac{1}{x} \log _e\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right) & , x<0 \\k & , x=0 \\\frac{\cos ^2 x-\sin ^2 x-1}{\sqrt{x^2+1}-1} & , x>0\end{array}\right.\)
\(\text { is continuous at } x =0 \text {, then } \frac{1}{a}+\frac{1}{b}+\frac{4}{k} \text { is equal to: }\)
[JEE Main 2021, 31 Aug (Shift 1)]
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Let \(f\) and \(g\) be two functions defined by
\(f(x)=\left\{\begin{array}{cc}x+1, & x<0 \\|x-1|, & x \geq 0\end{array} \text { and } g(x)=\left\{\begin{array}{cc}x+1, & x<0 \\1, & x \geq 0\end{array}\right. \text {. }\right.\)
Then \((g \circ f)(x)\) is
[JEE Main 2023, 11 Apr (Shift 2)]
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Let the functions \(f: R \rightarrow R\) and \(g: R \rightarrow R\) be defined as:
\(f(x)=\left\{\begin{array}{ll}x+2 & x<0 \\ x^2, & x \geq 0\end{array}\right.\) and \(g(x)= \begin{cases}x^3 & x<1 \\ 3 x-2, & x \geq 1\end{cases}\)
Then, the number of points in \(R\) where \((fog)(x)\) is NOT differentiable is equal to :
[JEE Main 2021, 16 Mar (Shift 1)]
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If \(f(x)=\left\{\begin{array}{cl}\frac{1}{|x|} & ;|x| \geq 1 \\ a x^2+b & ;|x|<1\end{array}\right.\) is differentiable at every point of the domain, then the values of \(a\) and \(b\) are respectively ;
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Let \( f:\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \rightarrow R \) be defined as
\(f(x)=\left\{\begin{array}{ccc}(1+|\sin x|)^{\frac{3 a} {|\sin x|}} & , & -\frac{\pi}{4} If \( f \) is continuous at \( x=0 \), then the value of \( 6 a+b^{2} \) is equal to [JEE Main 2021, 27 Jul (Shift 1)]
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For the differentiable function \(f: R -\{0\} \rightarrow R\), let \(3 f(x)+2 f\left(\frac{1}{x}\right)=\frac{1}{x}-10\), then \(\left|f(3)+f^{\prime}\left(\frac{1}{4}\right)\right|\) is equal to
[JEE Main 2023, 13 Apr (Shift 1)]
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Let the functions \(f: R \rightarrow R\) and \(g: R \rightarrow R\) be defined as:
\[f(x)=\left\{\begin{array}{cc}x+2, & x<0 \\x^2, & x \geq 0\end{array} \text { and } g(x)=\left\{\begin{array}{cc}x^3, & x<1 \\3 x-2, & x \geq 1\end{array}\right.\right.\]
Then, the number of points in \(R\) where \((fog)(x)\) is NOT differentiable is equal to :
[JEE Main 2021, 16 Mar (Shift 1)]
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Let \(f:R\to R\) is given by \(f(x)=x+1\) then the value of \(\lim _{n\to \infty }\frac{1}{n}\left[f(0)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+....+f\left(\frac{5(n-1)}{n}\right)\right]\) is
[JEE Main 2021, 20 Jul (Shift 2)]
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Let \( f: R \rightarrow R \) be defined as
\(f(x)=\left\{\begin{array}{cc}\frac{x^{3}}{(1-\cos 2 x)^{2}} \log _{e}\left(\frac{1+2 x e^{-2 x}}{\left(1-x e^{-x}\right)^{2}}\right), & x \neq 0 \\\alpha & , x=0\end{array}\right.\)
If \( f \) is continuous at \( x=0 \), then \( \alpha \) is equal to
[JEE Main 2021, 22 Jul (Shift 2)]
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If \(y(x)=x^x, x>0\), then \(y^{\prime \prime}(2)-2 y^{\prime}(2)\) is equal to:
[JEE Main 2023, 1 Feb (Shift 2)]
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If \((a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=2 a^2-3 b^2\), where \(a>b>0\), then \(\frac{d x}{d y}\) at \(\left(\frac{\pi}{4}, \frac{\pi}{4}\right)\) is:
[JEE Main 2020, 4 Sep (Shift 1)]
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Let f, g and h be the real valued functions defined on R as \(f\left(x\right)=\left\{\begin{matrix}\frac{x}{|x|}, & x\neq 0 \\ 1, & x=0\end{matrix},g\left(x\right)=\left\{\begin{matrix}\frac{\sin (x+1)}{(x+1)}, & x\neq -1 \\ 1, & x=-1\end{matrix}\right.\right.\)and \(h(x)=2[x]-f(x)\), where [x] is the greatest integer\(\leq x\). Then the value of \(\lim _{x\to 1}g(h(x-1))\) is
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If \(f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R\), then
[JEE Main 2023, 24 Jan (Shift 2)]
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$$\begin{aligned}\text { Let } \mathrm{f} \text { be the function defined by }\\f(x)= \begin{cases}\frac{x^2-1}{x^2-2|x-1|-1}, & x \neq 1 \\ 1 / 2, & x=1\end{cases}\end{aligned}$$
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\(\text { If } f(x)=\left\{\begin{array}{cc}\int_0^x(5+|1-t|) d t, & x>2 \\5 x+1, & x \leq 2\end{array}\right. \text {, then }\)
[JEE Main 2021, 25 Jul (Shift 2)]
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\(\text{ Let }f:\left(-\frac{\pi }{4},\frac{\pi }{4}\right)\to R\text{ be defined as }\\ f(x)=\left\{\begin{matrix}(1+|\sin x|{)}^{\frac{3a}{|\sin x|}} & , & -\frac{\pi }{4} \(\text{ If }f\text{ is continuous at }x=0\text{, then the value of }6a+{b}^{2}\text{ is equal to: }\) [JEE Main 2021, 27 Jul (Shift 1)]
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Let \(f\) and \(g\) be two functions defined by
\[f(x)=\left\{\begin{array}{cc}x+1, & x<0 \\|x-1|, & x \geq 0\end{array} \text { and } g(x)=\left\{\begin{array}{cc}x+1, & x<0 \\1, & x \geq 0\end{array}\right. \text {. }\right.\]
Then \((g \circ f)(x)\) is
[JEE Main 2023, 11 Apr (Shift 2)]
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Let \(f:\left[0,\text{ }\infty \right)\text{ }\to \text{ }\left[0,\text{ }3\right]\) be a function defined by
\(f(x)=\left\{\begin{array}{cc}\max \{\sin t: 0 \leq t \leq x\}, & 0 \leq x \leq \pi \\2+\cos x, & x>\pi\end{array}\right.\)
Then which of the following is true?
[JEE Main 2021, 27 Jul (Shift 2)]
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Let \(f:R\to R\) be a function defined as
\(f(x)=\left\{\begin{matrix}\frac{\sin (a+1)x+\sin 2x}{2x},\text{ if }x<0 \\ b,\text{ if }x=0 \\ \frac{\sqrt{x+b{x}^{3}}-\sqrt{x}}{b{x}^{5/2}},\text{ if }x>0\end{matrix}\right.\)
If \(f\) is continuous at \(x=0\), then the value of \(a+b\) is equal to:
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Let \( y=y(x) \) be a function of \( x \) satisfying \( y \sqrt{1-x^{2}}=k -x \sqrt{1-y^{2}} \) where \( k \) is a constant and \( y\left(\frac{1}{2}\right)=-\frac{1}{4} \). Then \( \frac{\mathrm{dy}}{\mathrm{dx}} \) at \( x=\frac{1}{2} \), is equal to :
[JEE Main 2020, 7 Jan (Shift 2)]
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If \( y=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) \) then \( \frac{d^{2} y}{d x^{2}} \) is equal to
[JEE Main 2021, 27 Aug (Shift 2)]
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If the function \( f(x)=\left\{\begin{array}{cc}k_{1}(x-\pi)^{2}-1, & x \leq \pi \\ k_{2} \cos x, & x>\pi\end{array}\right. \)
is twice differentiable, then the ordered pair \( \left(\mathrm{k}_{1}, \mathrm{k}_{2}\right) \) is equal to :
[JEE Main 2020, 5 Sep (Shift 1)]
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If \( y(\alpha)=\sqrt{2\left(\frac{\tan \alpha+\cot \alpha}{1+\tan ^{2} \alpha}\right)+\frac{1}{\sin ^{2} \alpha}}, \alpha \in\left(\frac{3 \pi}{4}, \pi\right) \); then \( \frac{\mathrm{dy}}{\mathrm{d} \alpha} \) at \( \alpha=\frac{5 \pi}{6} \) is
[JEE Main 2020, 7 Jan (Shift 1)]
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The function \(f\left(x\right)=\left\{\begin{matrix}\frac{\pi }{4}+{\tan }^{-1}x,\left|x\right|\leq 1 \\ \frac{1}{2}\left(\left|x\right|-1\right),\left|x\right|>1\end{matrix}\right.\)
[JEE Main 2020, 4 Sep (Shift 2)]
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Let a function \(f:R\to R\) be defined as
\(f(x)=\left\{\begin{matrix}\sin x-{e}^{x}\text{ if }x\leq 0 \\ a+[-x]\text{ if }0 where \([x]\) is the greatest integer less than or equal to x. If f is continuous on R, then ( a + b ) is equal to: [JEE Main 2021, 20 Jul (Shift 1)]
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Let a function \( f: R \rightarrow R \) be defined as
\(f(x)=\left\{\begin{array}{ccc}\sin x-e^{x} & , \quad \text { if } x \leq 0 \\a+[-x], & \text { if } 0 where \( [x] \) is the greatest integer less than or equal to \( x \). If \( f \) is continuous on \( R \), then \( (a+b) \) is equal to [JEE Main 2021, 20 Jul (Shift 1)]
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\( \lim _{x \rightarrow 1}\left(\frac{\int_{0}^{(x-1)^{2}} t \cos \left(t^{2}\right) d t}{(x-1) \sin (x-1)}\right) \)
[JEE Main 2020, 6 Sep (Shift 1)]
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\( \text { If } 2 x^y+3 y^x=20 \text {, then } \frac{d y}{d x} \text { at }(2,2) \text { is equal to } \)
[JEE Main 2023, 6 Apr (Shift 1)]
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If \( f: R \rightarrow R \) is given by \( f(x)=x+1 \), then the value of \( \lim _{n \rightarrow \infty} \frac{1}{n}\left[f(0)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+\ldots+f\left(\frac{5(n-1)}{n}\right)\right] \) is:
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If the function
\(f(x)=\left\{\begin{matrix}\frac{1}{x}{\log }_{e}\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right), & ,x<0 \\ k & ,x=0 \\ \frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{{x}^{2}+1}-1}, & x>0\end{matrix}\right.\)
is
continuous at \( x=0 \), then \( \frac{1}{a}+\frac{1}{b}+\frac{4}{k} \) is equal to:
[JEE Main 2021, 31 Aug (Shift 1)]
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If \(y^2+\log _e\left(\cos ^2 x\right)=y, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\),Then:
[JEE Main 2020, 3 Sep (Shift 1)]
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Let \(f(x)=\left[{x}^{2}-x\right]+|-x+[x]|\), where \(x\in R\) and [t] denotes the greatest integer less than or equal to t. Then, f is
[JEE Main 2023, 11 Apr (Shift 1)]
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\(\text { If } y(x)=x^x, x>0 \text {, then } y^{\prime \prime}(2)-2 y^{\prime}(2) \text { is equal to: }\)
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The set of all values of a for which \( \lim _{x \rightarrow a}([x-5]- \) \( [2 x+2])=0 \) where \( [\alpha] \) denotes the greatest integer less than or equal to \( \alpha \) is equal to
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If \( c \) is a point at which Rolles theorem holds for the function \( f(x)=\log _{e}\left(\frac{x^{2}+\alpha}{7 x}\right) \) in the interval \( [3,4] \), where \( \alpha \in R \), then \( f^{\prime \prime}(c) \) is equal to :
[JEE Main 2020, 8 Jan (Shift 1)]
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\(\lim _{x\to a}\frac{{\left(a+2x\right)}^{1/3}-{\left(3x\right)}^{1/3}}{{\left(3a+x\right)}^{1/3}-{\left(4x\right)}^{1/3}}\left(a\neq 0\right)\) is equal to
[JEE Main 2020, 3 Sep (Shift 2)]
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Let \(\alpha \in R\) be such that the function
\(f\left(x\right)=\left\{\begin{matrix}\frac{{\text{cos}}^{−1}\left(1−{{x}}^{2}\right){\text{sin}}^{−1}\left(1−\left\{x\right\}\right)}{\left\{x\right\}−{{x}}^{3}},x\neq 0 \\ \text{ }\alpha ,\text{ }x=0\end{matrix}\right.\)
is continuous at \(x=0\), where \(\{x\}=x-[x],[x]\) is the greatest integer less than or equal to \(x\).
[JEE Main 2021, 16 Mar (Shift 2)]
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Let \(x=2\) be a root of the equation \({x}^{2}+px+q=0\) and
\(f\left(x\right)=\left\{\begin{matrix}\frac{1-\cos \left({x}^{2}-4px+{q}^{2}+8q+16\right)}{(x-2p{)}^{4}}, & x\neq 2p \\ 0 & x=2p\end{matrix}\right.\)
Then \(\lim _{x \rightarrow 2 p^{+}}[f(x)]\) where [.] denotes greatest integer function, is
[JEE Main 2023, 29 Jan (Shift 1)]
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If \(2 x^y+3 y^x=20\), then \(\frac{d y}{d x}\) at \((2,2)\) is equal to
[JEE Main 2023, 6 Apr (Shift 1)]
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Let x=2 be a root of the equation \({x}^{2}+px+q=0\) and
\(f\left(x\right)=\left\{\begin{matrix}\frac{1-\cos \left({x}^{2}-4px+{q}^{2}+8q+16\right)}{(x-2p{)}^{4}}, & x\neq 2p \\ 0 & x=2p\end{matrix}\right.\)
Then \(\lim _{x\to 2{p}^{+}}[f(x)]\) where [.] denotes greatest integer function, is
[JEE Main 2023, 29 Jan (Shift 1)]
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Let \(f\) and \(g\) be twice differentiable functions on \(R\) such that
\[\begin{aligned}& f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x \\& f^{\prime}(1)=4 g^{\prime}(1)-3=9 \\& f(2)=3 g(2)=12\end{aligned}\]
Then which of the following is NOT true?
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If 2xy + 3yx = 20, then \(\frac{dy}{dx}\) at (2, 2) is equal to:
[JEE Main 2023, 6 Apr (Shift 1)]
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The set of all values of a for which \(\lim _{x\to a}([x-5]-[2x+2])=0\), where\([\mu ]\) denotes the greater integer less than or equal to \(\mu\) is equal to
[JEE Main 2023, 24 Jan (Shift 2)]
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If \(x=2 \sin \theta-\sin 2 \theta\) and \(y=2 \cos \theta-\cos 2 \theta, \theta \in[0,2 \pi]\), then \((\frac{d^2 y}{d x^2}) _{\theta=\pi}\) is :
[JEE Main 2020, 9 Jan (Shift 2)]
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Let \( \mathrm{f}:(0, \infty) \rightarrow(0, \infty) \) be a differentiable function such that \( f(1)=e \) and \( \lim _{t \rightarrow x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}=0 \). If \( f(x)=1 \), then \( x \) is equal to :
[JEE Main 2020, 4 Sep (Shift 2)]
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Let [x] denote the greatest integer function and \(f(x)=\max \{1+x+[x], 2+x, x+2[x]\}, 0 \leq x \leq 2\).
Let m be the number of points in [0,2], where f is not continuous and n be the number of points in (0,2), where f is not differentiable. Then \((m+n)^2+2\) is equal to
[JEE Main 2023, 15 Apr (Shift 1)]
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The value of \( \lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(x-[x]^{2}\right) \cdot \sin ^{-1}\left(x-[x]^{2}\right)}{x-x^{3}} \), where \( [x] \) denotes the greatest integer \( \leq x \) is
[JEE Main 2021, 17 Mar (Shift 1)]
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The value of \( \lim _{n \rightarrow \infty} \frac{[r]+[2 r]+\ldots+[n r]}{n^{2}} \), where \( r \) is non-zero real number and \( [r] \) denotes the greatest integer less than or equal to \( r \), is equal to
[JEE Main 2021, 17 Mar (Shift 2)]
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Let f : \(R\to R\) be defined as
\(f(x)=\left\{\begin{matrix}\frac{{x}^{3}}{{(1−\cos 2x)}^{2}}{\log }_{\text{e}}\left(\frac{1+2x{\text{e}}^{−2x}}{{\left(1−x{\text{e}}^{−x}\right)}^{2}}\right) & ,x\neq 0 \\ \text{α} & ,x=0\end{matrix}\right.\)
If f is continuous at x = 0, then α is equal to
[JEE Main 2021, 22 Jul (Shift 2)]
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Let \(y(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right) \text {. }\)Then \(y^{\prime}-y^{\prime \prime}\) at \(x=-1\) is equal to
[JEE Main 2023, 25 Jan (Shift 1)]
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\(\text{ If }f(x)=\left\{\begin{matrix}{\int }_{0}^{x}(5+|1-t|)dt, & x>2 \\ 5x+1, & x\leq 2\end{matrix}\right.\text{, then }\)
[JEE Main 2021, 25 Jul (Shift 2)]
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If the function \(f(x)= \begin{cases}(1+|\cos x|)^ \frac{\lambda}{|\cos x|} & , 0 is continuous at \(x=\frac{\pi}{2}\), then \(9 \lambda+6 \log _e \mu+\mu^6-e^{6 \lambda}\) is equal to [JEE Main 2023, 25 Jan (Shift 2)]
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If \(f(x)=\left\{\begin{array}{cc}\int_0^x(5+|1-t|) d t, & x>2 \\ 5 x+1, & x \leq 2\end{array}\right.\), then
[JEE Main 2021, 25 Jul (Shift 2)]
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Let \(y(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right) \text {. }\)
Then \(y^{\prime}-y^{\prime \prime}\) at \(x=-1\) is equal to
[JEE Main 2023, 25 Jan (Shift 1)]
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\(\text { If } f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R \text {, then }\)
[JEE Main 2023, 24 Jan (Shift 2)]
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Let \(f, g\) and \(h\) be the real valued functions defined on \(R\) as\(f\left(x\right)=\left\{\begin{matrix}\frac{x}{|x|}, & x\neq 0 \\ 1, & x=0\end{matrix},g\left(x\right)=\left\{\begin{matrix}\frac{\sin (x+1)}{(x+1)}, & x\neq -1 \\ 1, & x=-1\end{matrix}\right.\right.\)
and \(h(x)=2[x]-f(x)\), where \([x]\) is the greatest integer\(\leq x\). Then the value of \(\lim _{x \rightarrow 1} g(h(x-1))\) is
[JEE Main 2023, 30 Jan (Shift 2)]
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If \( f: R \rightarrow R \) is a function define by \(f(x)=[x-1] \cos \left(\frac{2 x-1}{2}\right) \pi\) where [·] denotes the greatest integer function, then \( f \) is:
[JEE Main 2021, 24 Feb (Shift 1)]
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\(\text { Let } f(x)=\left\{\begin{array}{cc}x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\0 & , x=0\end{array} ; \text { Then at } x=0\right.\)
[JEE Main 2023, 24 Jan (Shift 1)]
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Let f, g and h be the real valued function defined on R as \(f(x)=\left\{\begin{matrix}\frac{x}{|x|}, & x\neq 0 \\ 1, & x=0\end{matrix}\right.\\ g(x)=\left\{\begin{matrix}\frac{\sin (x+1)}{(x+1)}, & x\neq −1 \\ 1, & x=−1\end{matrix}\right.\\\)and h(x) = 2[x] – f(x), where [x] is the greatest integer \(\leq\) x. Then the value of \(\lim _{x\to 1}g(h(x−1))\) is
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Let \(f(x)=\left\{\begin{array}{cc}x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0\end{array} \right.\) .
Then at \(x=0\)
[JEE Main 2023, 24 Jan (Shift 1)]
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