Continuity and Differentiability

Explanation

$\frac{d}{dx}\left(x{e}^y\right)=\frac{d}{dx}\left(1\right)$

$e^y+x{e}^y\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-e^y}{x{e}^y}$

$\frac{dy}{dx}=-\frac{1}{x}$

${\left.\frac{dy}{dx}\right|}_{x=1}=-\frac{1}{1}=-1$

ExplanationStep 1: Check for Points of Discontinuity

A function is discontinuous at points where:

1. It is not defined.
2. The left-hand limit (LHL) and right-hand limit (RHL) do not match or are not finite.

The given function is undefined at $x = 5$ as the denominator becomes zero, leading to an infinite discontinuity.

Step 2: Verify if There Are Other Discontinuities

The function is well-defined and continuous everywhere else because it is a simple rational function(except $x=5$)

Step 3: Conclusion

Since there is only one point where the function is not defined ($x = 5$x=5), the number of points of discontinuity is 1.

Explanation

Here, $f\left(x\right)$ will be continuous at $x=-\frac{\mathrm{\pi }}{3}$ if,

$\underset{x\to -\frac{\mathrm{\pi }}{3}}{\lim }f\left(x\right)=f\left(-\frac{\mathrm{\pi }}{3}\right)$

Now, $\underset{x\to -\frac{\mathrm{\pi }}{3}}{\lim }f\left(x\right)=\underset{x\to -\frac{\pi }{3}}{\lim }\frac{\sqrt{3}\cos x+\sin x}{3x+\frac{\pi }{2}}$

$=\frac{\sqrt{3}\cos \left(-\frac{\pi }{3}\right)+\sin \left(-\frac{\pi }{3}\right)}{3\left(-\frac{\pi }{3}\right)+\frac{\pi }{2}}$

$=\frac{\sqrt{3}·\frac{1}{2}-\frac{\sqrt{3}}{2}}{-\frac{\pi }{2}}$

$=\frac{0}{-\frac{\pi }{2}}=0$

It is given that $f\left(-\frac{\mathrm{\pi }}{3}\right)=k$.

Since $\underset{x\to -\frac{\mathrm{\pi }}{3}}{\lim }f\left(x\right)=f\left(-\frac{\mathrm{\pi }}{3}\right)$, therefore $k=0$.

Explanation

For \(x \neq 0\) :

$f\left(x\right)=x^2+3$

continuous on \(\mathbb{R}\)

$f^{\text{'}}\left(x\right)=2x$ exists.

At \(x=0\) :

$\underset{x\to 0}{\lim }\left(x^2+3\right)=3$

$f\left(0\right)=1$

\(3 \neq 1\)

\( \Rightarrow\) f is not differentiable at 0

Therefore \(f\) is continuous and differentiable on \(\mathbb{R} - \{0\}\).

Explanation

\(y=\tan ^{-1}\left(x^2\right)\)

\(\frac{d y}{d x}=\frac{1}{1+\left(x^2\right)^2} \cdot \frac{d}{d x}\left(x^2\right)\)

\(\frac{d y}{d x}=\frac{2 x}{1+x^4}\)

Explanation

For \(x \leq-3: f(x)=|x|+3\)

⇒ continuous on \((-\infty,-3]\)

For \(-3<x<3\) : \(f(x)=-2 x\)

⇒ continuous on \((-3,3)\)

For \(x \geq 3: f(x)=6 x+2\)

⇒ continuous on \([3, \infty)\)

At \(x=-3: \lim _{x \rightarrow(-3)^{-}}(|x|+3)=|-3|+3=6\)

\(\lim _{x \rightarrow(-3)^{+}}(-2 x)=-2(-3)=6\)

⇒ continuous at \(x=-3\)

At \(x=3: \lim _{x \rightarrow 3^{-}}(-2 x)=-2(3)=-6\)

\(\lim _{x \rightarrow 3^{+}}(6 x+2)=6(3)+2=20\)

⇒ discontinuous at \(x=3\)

Explanation

Let $\mathrm{u}=e^{{\sin }^{2}x}$ and $v=\cos x$.

Then , $\frac{du}{dx}=e^{{\sin }^{2}x}\left(2\sin x\right)\cos x$

And, $\frac{dv}{dx}=-\sin x$

Now,

$$\begin{gathered} \frac{du}{dv}=\frac{du/dx}{dv/dx} \\ =\frac{e^{{\sin }^{2}x}·2\sin x·\cos x}{-\sin x} \\ =-2\cos x{e}^{{\sin }^{2}x} \end{gathered}$$

Explanation

Let \(y=\sin \left(x^2\right)\).

\(\frac{d y}{d x}=\cos \left(x^2\right) \cdot 2 x\)

\(\text { At } x=\sqrt{\pi}\):

\(\frac{d y}{d x}=2 \sqrt{\pi} \cos (\pi)\)
\(\Rightarrow \frac{d y}{d x}=-2 \sqrt{\pi}\) \([\because\cos (\pi)=-1]\)

Explanation

Here, $f\left(x\right)$ will be continuous at $x=-\frac{\mathrm{\pi }}{3}$ if,

$\underset{x\to -\frac{\mathrm{\pi }}{3}}{\lim }f\left(x\right)=f\left(-\frac{\mathrm{\pi }}{3}\right)$

Calculate the limit at $x=-\frac{\mathrm{\pi }}{3}$.

$\underset{x\to -\frac{\mathrm{\pi }}{3}}{\lim }f\left(x\right)=\underset{x\to -\frac{\pi }{3}}{\lim }\frac{\sqrt{3}\cos x+\sin x}{3x+\frac{\pi }{2}}$

$=\frac{\sqrt{3}\cos \left(-\frac{\pi }{3}\right)+\sin \left(-\frac{\pi }{3}\right)}{3\left(-\frac{\pi }{3}\right)+\frac{\pi }{2}}$

$=\frac{\sqrt{3}·\frac{1}{2}-\frac{\sqrt{3}}{2}}{-\frac{\pi }{2}}$

$=\frac{0}{-\frac{\pi }{2}}=0$

It is given that $f\left(-\frac{\mathrm{\pi }}{3}\right)=k$.

Since $\underset{x\to -\frac{\mathrm{\pi }}{3}}{\lim }f\left(x\right)=f\left(-\frac{\mathrm{\pi }}{3}\right)$, therefore $k=0$.

Explanation

Let \(y=\sin \left(x^2\right)\).

\(\frac{d y}{d x}=\cos \left(x^2\right) \cdot 2 x\)

\(\text { At } x=\sqrt{\pi}: \frac{d y}{d x}=2 \sqrt{\pi} \cos (\pi)\)
\(\cos (\pi)=-1\)
\(\Rightarrow \frac{d y}{d x}=-2 \sqrt{\pi}\)

Explanation

For \(x \leq-3: f(x)=|x|+3\)

⇒ continuous on \((-\infty,-3]\)

For \(-3<x<3\) : \(f(x)=-2 x\)

⇒ continuous on \((-3,3)\)

For \(x \geq 3: f(x)=6 x+2\)

⇒ continuous on \([3, \infty)\)

At \(x=-3\):

L.H.L\(=\lim _{x \rightarrow(-3)^{-}}(|x|+3)=|-3|+3=6\)

R.H.L.\(=\lim _{x \rightarrow(-3)^{+}}(-2 x)=-2(-3)=6\)

\(f(-3)=6\)

\(\because\)L.H.L.\(=\)R.H.L.\(=f(-3)\)

⇒ continuous at \(x=-3\)

At \(x=3\):

L.H.L.\(=\lim _{x \rightarrow 3^{-}}(-2 x)=-2(3)=-6\)

R.H.L.\(=\lim _{x \rightarrow 3^{+}}(6 x+2)=6(3)+2=20\)

\(f(3)=20\)

\(\because\)L.H.L \(\neq\) R.H.L.\(=f(3)\)

⇒ discontinuous at \(x=3\)

Explanation

Given: ${\mathrm{xe}}^{\mathrm{y}}=1$

Differentiate w.r.t. \(x\)

$\frac{d}{dx}\left(x{e}^y\right)=\frac{d}{dx}\left(1\right)$

$\Rightarrow e^y+x{e}^y\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{-e^y}{x{e}^y}$

$\Rightarrow \frac{dy}{dx}=-\frac{1}{x}$

Hence, ${\left.\frac{dy}{dx}\right|}_{x=1}=-\frac{1}{1}=-1$

Explanation

Given: \(f(x)=\left\{\begin{array}{cc}x^2+3, & x \neq 0 \\ 1, & x=0\end{array}\right.\)

Now, \(f(0)=1 \text {. }\)
\(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(x^2+3\right)=3\)
Since \(\lim _{x \rightarrow 0} f(x) \neq f(0), f(x) \) is not continuous at \(x=0 \).

For \(x \neq 0, f(x)=x^2+3\) which is a polynomial and hence continuous everywhere except at \(x=0\).

For \(x \neq 0, f(x)=x^2+3\) is differentiable since it is a polynomial.
At \(x=0\), since \(f(x)\) is not continuous, it cannot be differentiable.

Therefore \(f\) is continuous and differentiable on \(\mathbb{R} - \{0\}\).

Explanation

Let $\mathrm{u}=e^{{\sin }^{2}x}$ and $v=\cos x$.

$\frac{du}{dx}=e^{{\sin }^{2}x}\left(2\sin x\right)\cos x$

$\frac{dv}{dx}=-\sin x$

$\begin{array}{rcl}\frac{du}{dv}& =& \frac{du/dx}{dv/dx}\\ & =& \frac{e^{{\sin }^{2}x}·2\sin x·\cos x}{-\sin x}\\ & =& -2\cos x{e}^{{\sin }^{2}x}\end{array}$

ExplanationStep 1: Check for Points of Discontinuity

A function is discontinuous at points where:

1. It is not defined.
2. The left-hand limit (LHL) and right-hand limit (RHL) do not match or are not finite.

The given function is undefined at $x = 5$ as the denominator becomes zero, leading to an infinite discontinuity.

Step 2: Verify if There Are Other Discontinuities

The function is well-defined and continuous everywhere else because it is a simple rational function(except $x=5$)

Step 3: Conclusion

Since there is only one point where the function is not defined ($x = 5$x=5), the number of points of discontinuity is 1.

Explanation

Given: \(y=\tan ^{-1}\left(x^2\right)\)

Differentiate w.r.t. \(x\)

\(\frac{d y}{d x}=\frac{1}{1+\left(x^2\right)^2} \cdot \frac{d}{d x}\left(x^2\right)\)

\(\Rightarrow \frac{d y}{d x}=\frac{2 x}{1+x^4}\)