Application of Derivatives

$$\begin{gathered} \text{Given,} \\ \mathrm{f}\left(\mathrm{x}\right)=2{\log }_{\mathrm{e}}\left(\mathrm{x}-2\right)-{\mathrm{x}}^2+\mathrm{ax}+1 \\ \text{On differentiating both sides, we get} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{2}{\mathrm{x}-2}-2\mathrm{x}+\mathrm{a} \ge 0 \left[\text{for increasing} f\left(x\right)\right] \\ \text{Again,differentiating both sides, we get} \\ \mathrm{f}"\left(\mathrm{x}\right)=\frac{-2}{(\mathrm{x}-2{)}^2}-2<0 \\ f\text{'}\left(x\right)\text{ is decreasing function } \\ {\mathrm{f}}^{\text{'}}\left(3\right)\ge 0 \\ 2-6+\mathrm{a}\ge 0 \\ \mathrm{a}\ge 4 \\ {\mathrm{a}}_{\min }=4 \\ \text{Now, } \\ \mathrm{g}\left(\mathrm{x}\right)={\left(\mathrm{x}-1\right)}^3{\left(\mathrm{x}+2-\mathrm{a}\right)}^2 \\ \mathrm{g}\left(\mathrm{x}\right)={\left(\mathrm{x}-1\right)}^3{\left(\mathrm{x}-2\right)}^2 \\ \text{On differentiating, we get} \\ {\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)={\left(x-1\right)}^{3}2\left(x-2\right)+{\left(\mathrm{x}-2\right)}^{2}3{\left(x-1\right)}^2 \\ ={\left(x-1\right)}^2\left(\mathrm{x}-2\right)\left(2\mathrm{x}-2+3\mathrm{x}-6\right) \\ ={\left(x-1\right)}^2\left(\mathrm{x}-2\right)\left(5\mathrm{x}-8\right)<0 \\ \mathrm{x}\in \left(\frac{8}{5},2\right) \\ \text{Hence, } \\ 100\left(\mathrm{a}+\mathrm{b}-\mathrm{c}\right)=100\left(4+\frac{8}{5}-2\right)=360 \end{gathered}$$

\(f\) is increasing function.

\(x<5x<7x\)

\(f(x)<f(5x)<f(7x)\)

\(\frac{f(x)}{f(x)}<\frac{f(5x)}{f(x)}<\frac{f(7x)}{f(x)}\)

\(\lim_{x\to\infty}\frac{f(x)}{f(x)}<\lim_{x\to\infty}\frac{f(5x)}{f(x)}<\lim_{x\to\infty}\frac{f(7x)}{f(x)}\)

\(1<\lim_{x\to\infty}\frac{f(5x)}{f(x)}<1\Rightarrow \lim_{x\to\infty}\frac{f(5x)}{f(x)}=1\)

\(\lim_{x\to\infty}\left(\frac{f(5x)}{f(x)}-1\right)=0\)

\(
g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \text { and } f^{\prime \prime}(x)>0 \forall x \in(0,3)
\)

\(\Rightarrow f^{\prime}(x)\) is increasing function

\(
\begin{aligned}
& g^{\prime}(x)=3 \times \frac{1}{3} \cdot f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \\
& =\mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})
\end{aligned}
\)

If g is decreasing in \((0, \alpha)\)

\(
\begin{aligned}
& \mathrm{g}^{\prime}(\mathrm{x})<0 \\
& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})<0 \\
& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)<\mathrm{f}^{\prime}(3-\mathrm{x}) \\
& \Rightarrow \frac{\mathrm{x}}{3}<3-\mathrm{x} \\
& \Rightarrow \mathrm{x}<\frac{9}{4}
\end{aligned}
\)

$$\begin{gathered} f\left(x\right)={\int }_0^{x^2}\frac{t^2-8t+15}{e^t}dt,x\in \mathbf{R} \\ {f}^{\text{'}}\left(x\right)=\left(\frac{x^4-8x^2+15}{e^{x^2}}\right)\left(2x\right) \\ =\frac{\left(x^2-3\right)\left(x^2-5\right)\left(2x\right)}{{\mathrm{e}}^{{\mathrm{x}}^2}} \\ =\frac{(x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5})2x}{e^{x^2}} \end{gathered}$$

using first derivative approach :-

$$\begin{gathered} \text{Maxima at }x\in \{-\sqrt{3},\sqrt{3}\} \\ \text{Minima at }\mathrm{x}\in \{-\sqrt{5},0,\sqrt{5}\} \end{gathered}$$

number of local maxima points=$2$

number of local minima points=$3$

Complete the square for the circle equation:

$$\begin{gathered} x^2+y^2+12y+35=0 \\ ⟹x^2+(y^2+12y)+35=0 \end{gathered}$$

Complete the square for $y$:

$$y^2 + 12y = (y + 6)^2 - 36$$

Substitute back:

$$\begin{gathered} x^2+(y+6{)}^2-36+35=0 \\ ⟹x^2+(y+6{)}^2=1 \end{gathered}$$

The circle has center $(0, -6)$ and radius $r = 1$.

The parabola $y^2 = 8x$ can be parameterized as $\left( \frac{t^2}{8}, t \right)$ (where $t$ is a parameter).

The distance $D$ from the circle’s center $(0, -6)$ to a point $\left( \frac{t^2}{8}, t \right)$ on the parabola is:

$$\begin{gathered} D=\sqrt{{(\frac{t^2}{8}-0)}^2+(t-(-6){)}^2} \\ =\sqrt{\frac{t^4}{64}+(t+6{)}^2} \end{gathered}$$

To minimize $D$, minimize $D^2$ (since the square root is monotonic):

Take the derivative of $f(t)$ and set it to zero:

$$f'(t) = \frac{4t^3}{64} + 2(t + 6) = \frac{t^3}{16} + 2t + 12$$

Set $f'(t) = 0$:

$$\frac{t^3}{16} + 2t + 12 = 0 \implies t^3 + 32t + 192 = 0$$

By the Rational Root Theorem, $t = -4$ is a root (verify: $(-4{)}^3+32(-4)+192$$=-64-128+192=0$). Factor the cubic:

$$t^3 + 32t + 192 = (t + 4)(t^2 - 4t + 48)$$

The quadratic $t^2 - 4t + 48$ has no real roots (discriminant $-176 < 0$), so the only real critical point is $t = -4$.

Evaluate $f(t)$ at $t = -4$:

$f(-4)=\frac{(-4{)}^4}{64}+(-4+6{)}^2$$=\frac{256}{64}+2^2=4+4=8$

Thus, $D = \sqrt{8} = 2\sqrt{2}$.

The shortest distance between the circle and the parabola is the minimum distance from the center to the parabola minus the circle’s radius:

$$\text{Shortest distance} = 2\sqrt{2} - 1$$

\(g(x) = 3f \left( \frac{x}{3} \right) + f(3 - x)\) and \(f''(x) > 0 \forall x \in (0,3)\)
\(\Rightarrow f'(x)\) is increasing function
\(g'(x) = 3 \times \frac{1}{3} f' \left( \frac{x}{3} \right) - f'(3 - x) = f' \left( \frac{x}{3} \right) - f'(3 - x)\)

If g is decreasing in \((0, \alpha)\)
\(g'(x) < 0\)
\(= f' \left( \frac{x}{3} \right) - f'(3-x) < 0\)
\(\Rightarrow f' \left( \frac{x}{3} \right) < f'(3-x)\)
\(\Rightarrow \frac{x}{3} < 3-x \Rightarrow x < \frac{9}{4}\)
Therefore \(\alpha = \frac{9}{4}\)
Then \(8\alpha = 8 \times \frac{9}{4} = 18\)

\(g(x) = 3f \left( \frac{x}{3} \right) + f(3 - x)\) and \(f''(x) > 0 \forall x \in (0,3)\)
\(\Rightarrow f'(x)\) is increasing function
\(g'(x) = 3 \times \frac{1}{3} f' \left( \frac{x}{3} \right) - f'(3 - x) = f' \left( \frac{x}{3} \right) - f'(3 - x)\)

If g is decreasing in \((0, \alpha)\)
\(g'(x) < 0\)
\(= f' \left( \frac{x}{3} \right) - f'(3-x) < 0\)
\(\Rightarrow f' \left( \frac{x}{3} \right) < f'(3-x)\)
\(\Rightarrow \frac{x}{3} < 3-x \Rightarrow x < \frac{9}{4}\)
Therefore \(\alpha = \frac{9}{4}\)
Then \(8\alpha = 8 \times \frac{9}{4} = 18\)

The first derivative of $f(x)$ is:

$$f'(x) = 6x^2 - 18ax + 12a^2$$

Factor the derivative:

$$f'(x) = 6(x^2 - 3ax + 2a^2) = 6(x - a)(x - 2a)$$

Setting $f'(x) = 0$ gives critical points $x = a$ and $x = 2a$.

Analyze the sign of $f'(x)$:

• For $x < a$, $f'(x) > 0$ (function increasing).
• For $a < x < 2a$, $f'(x) < 0$ (function decreasing).
• For $x > 2a$, $f'(x) > 0$ (function increasing).

Thus, $x = a$ is a local maximum ( $p = a$ ) and $x = 2a$ is a local minimum ( $q = 2a$ ).

Given $p^2 = q$, substitute $p = a$ and $q = 2a$:

$$a^2 = 2a$$

Since $a > 0$, divide both sides by $a$:

$$a = 2$$

Substitute $a = 2$ into $f(x)$:

$$f(x) = 2x^3 - 9(2)x^2 + 12(2)^2x + 1 = 2x^3 - 18x^2 + 48x + 1$$

Now, evaluate $f(3)$:

$$f(3) = 2(3)^3 - 18(3)^2 + 48(3) + 1$$

Thus, $f(3)=37$

Sol:

$\begin{array}{l}f\text{'}\left(x\right)=\frac{1}{3}-\frac{3}{x^2}\\ =\frac{x^2-9}{3x^2}=\frac{(x+3)(x-3)}{3x^2}\end{array}$

For strictly increasing ${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)>0 \Rightarrow \mathrm{x}\in (-\infty ,-3)\cup (3,\infty )$

For strictly decreasing ${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)<0 \Rightarrow \mathrm{x}\in (-3,0)\cup (0,3)$

$$\therefore \alpha_1 = -3, \alpha_2 = 3 \quad \alpha_3 = -3, \alpha_4 = 0 \quad \alpha_5 = 3$$

$$\sum _{i=1}^5{\alpha }_i^2=36$$

\(f(x)=\frac{x}{3}+\frac{3}{x}+3,\ x\neq 0\)

\(f'(x)=\frac{1}{3}-\frac{3}{x^2}=0\Rightarrow x=\pm 3\)

\(f'(x)=\frac{x^2-3}{3x^2}\)

\(f'(x)>0\ \forall(-\infty,-3)\cup(3,\infty)\rightarrow\) increasing

\(f'(x)<0\ \forall(-3,0)\cup(0,3)\rightarrow\) decreasing

\(\sum_{i=1}^{5}\alpha_i^2=(-3)^2+(3)^2+(-3)^2+(0)^2+(3)^2=36\)

$$\begin{gathered} f\left(x\right)={\int }_0^{x^2}\frac{t^2-8t+15}{e^t}dt,x\in \mathbf{R} \\ \text{using lebnitz rule:} \\ {f}^{\text{'}}\left(x\right)=\left(\frac{x^4-8x^2+15}{e^{x^2}}\right)\left(2x\right) \\ =\frac{\left(x^2-3\right)\left(x^2-5\right)\left(2x\right)}{{\mathrm{e}}^{{\mathrm{x}}^2}} \\ =\frac{(x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5})2x}{e^{x^2}} \end{gathered}$$

using first derivative approach :-

$$\begin{gathered} \text{ Maxima at }x\in \{-\sqrt{3},\sqrt{3}\} \\ \text{ Minima at }\mathrm{x}\in \{-\sqrt{5},0,\sqrt{5}\} \end{gathered}$$

number of local maxima points=$2$

number of local minima points=$3$

$f\left(x\right)=\left|\mathcal{l}nx\right|-|x-1|$

$=\left\{\begin{array}{cc}\mathcal{l}nx-(x-1)& x\ge 1\\ -\mathcal{l}nx+(x-1)& 0<x<1\end{array}\right.$

$=\left\{\begin{array}{cc}\mathcal{l}nx-x+1& x\ge 1\\ -\mathcal{l}nx+x-1& 0<x<1\end{array}\right.$

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{cc}\frac{1}{x}-1& x\ge 1\\ -\frac{1}{x}+1& 0<x<1\end{array}\right.$

$f^{\text{'}}\left(1^{+}\right)=f^{\text{'}}\left(1^{-}\right)=0$

$\Rightarrow f\left(x\right)$ is differentiable $\forall x > 0$.

$$\begin{gathered} f\text{'} \left(x\right)<0\forall x>1 \\ f\text{'}\left(x\right)<0\forall 0<x<1 \end{gathered}$$

$\Rightarrow f\left(x\right)$ is decreasing $\forall x\in (0, \infty )$.

Given \(f(x)=\sqrt{x}\ln x-x+1\), where \(x>0\).

\(f'(x)=\dfrac{1}{2\sqrt{x}}\ln x+\sqrt{x}\cdot\dfrac1x-1\)

\(f'(x)=\dfrac{\ln x}{2\sqrt{x}}+\dfrac1{\sqrt{x}}-1=\dfrac{\ln x+2}{2\sqrt{x}}-1\)

\(f'(x)=\dfrac{\ln x+2-2\sqrt{x}}{2\sqrt{x}}\)

Put \(t=\sqrt{x}\), so \(x=t^2\) and \(\ln x=2\ln t\).

\(f'(x)=\dfrac{2\ln t+2-2t}{2t}=\dfrac{\ln t+1-t}{t}\)

Now \(\ln t\le t-1\) for \(t>0\).

So \(\ln t+1-t\le0\), hence \(f'(x)\le0\).

Equality occurs only when \(t=1\), i.e. \(x=1\).

Therefore \(f\) is decreasing on \((0,\infty)\), but derivative becomes zero only at one point.

Now check option \(A\):

\(f''(x)=\dfrac{-\ln x}{4x^{3/2}}\)

For \(0<x<1\), \(\ln x<0\), so \(f''(x)>0\).

Thus \(f'\) is increasing on \((0,1)\), not decreasing. So option \(A\) is false.

Since \(f'(x)\le0\) on \((0,\infty)\), \(f\) cannot have a local maximum or local minimum. So options \(B\) and \(C\) are false.

Hence option \(D\) is true.

\(\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{y}}{3}\right)=\frac{\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})}{3}\)

Put \(\mathrm{x}=\mathrm{y}=0\)

\( f(0)=\frac{2 f(0)}{3}\)
\(\Rightarrow f(0)=0 \ldots(1) \)
\(\mathrm{f}^{\prime}\left(\frac{\mathrm{x}+\mathrm{y}}{3}\right) \cdot \frac{1}{3}=\frac{1}{3} \mathrm{f}^{\prime}(\mathrm{x}) \)
Put \( \mathrm{x}=0 \)
\( f^{\prime}\left(\frac{\mathrm{y}}{3}\right) \frac{1}{3}=\frac{1}{3} \times 3\)
\( \mathrm{f}^{\prime}\left(\frac{\mathrm{y}}{3}\right)=3 \)
Put \( \mathrm{y}=3 \mathrm{x}\)
\( \mathrm{f}^{\prime}(\mathrm{x})=3\)

Integrate both sides:

\(f(x)=3 x+C\), and from \(f(0)=0, C=0\), so \(f(x)=3 x\)

Now, \( \mathrm{g}(\mathrm{x})=3+\mathrm{e}^{\mathrm{x}} \cdot 3 \mathrm{x}\)

\( \mathrm{g}^{\prime}(\mathrm{x})=3\left[\mathrm{e}^{\mathrm{x}}+\mathrm{x} \cdot \mathrm{e}^{\mathrm{x}}\right]\)

\(=3 \mathrm{e}^{\mathrm{x}}(\mathrm{x}+1)\)

\( \mathrm{g}^{\prime}(\mathrm{x})=0\), at \( \mathrm{x}=-1\)

Now, \(g^{\prime \prime}(x)=3 e^x(2+x)\)

\(g^{\prime \prime}(-1)=3 e^{-1}(1)=\frac{3}{e}>0 \) (Local Minimum)

\((\mathrm{~g}(\mathrm{x}))_{\min }=3+\mathrm{e}^{-1}(-3)\)

\(=3\left[1-\frac{1}{\mathrm{e}}\right]=\frac{3(\mathrm{e}-1)}{\mathrm{e}}\)

\(f(x)=(x-2)^{\frac{2}{3}}(2x+1)\)

\(f'(x)=\frac{2}{3}(x-2)^{-\frac{1}{3}}(2x+1)+(x-2)^{\frac{2}{3}}\)

\(f'(x)=2\times \frac{(2x+1)+(x-2)}{3(x-2)^{\frac{1}{3}}}\)

\(\frac{3x-1}{(x-2)^{\frac{1}{3}}}=0\)

Critical points \(x=\frac{1}{3}\) and \(x=2\)

\(f(x)=x^x;\ x>0\)

\(\ln y=x\ln x\)

\(\frac{1}{y}\frac{dy}{dx}=\frac{x}{x}+\ln x\)

\(\frac{dy}{dx}=x^x(1+\ln x)\)

For strictly increasing

\(\frac{dy}{dx}\geq 0\Rightarrow x^x(1+\ln x)\geq 0\)

\(\Rightarrow \ln x\geq -1\)

\(x\geq e^{-1}\)

\(x\geq \frac{1}{e}\)

\(x\in\left[\frac{1}{e},\infty\right)\)

$f^{\text{'}}\left(x\right)=3x^2+2ax+\frac{b}{x}$

$f^{\text{'}}\left(-1\right)=3-2a-b=0$

$f^{\text{'}}\left(-2\right)=12+4a-\frac{b}{2}=0$

$a=\frac{-9}{2}, b=12$

$f^{\text{'}}\left(x\right)=3x^2-9x+\frac{12}{x}$

$=\frac{3\left(x+1\right){\left(x+2\right)}^2}{x}$

Max at n = – 1

$f\left(x\right)=x^2-\frac{9}{2}{x}^2+12\ln \left|x\right|+1$

$f\left(-1\right)=-1-\frac{9}{2}$

$M=-4.5$

Min. value at $x = -2$

$f\left(-2\right)=-8-18+12\ln 2+1$

$m=-25+12\ln 2=-16.6$

$\left|M+m\right|=21.1$

1. The triangle is isosceles with base \(2x\) and height \(y\).

2. Area \(A(x) = \frac{1}{2}(2x)(y) = x(-2x^2 + 54) = 54x - 2x^3\).

3. \(A'(x) = 54 - 6x^2 = 0 \Rightarrow x^2 = 9 \Rightarrow x = 3\).

4. Max Area \(A(3) = 54(3) - 2(27) = 162 - 54 = 108\).

1. The triangle is isosceles with base \(2x\) and height \(y\).

2. Area \(A(x) = \frac{1}{2}(2x)(y) = x(-2x^2 + 54) = 54x - 2x^3\).

3. \(A'(x) = 54 - 6x^2 = 0 \Rightarrow x^2 = 9 \Rightarrow x = 3\).

4. Max Area \(A(3) = 54(3) - 2(27) = 162 - 54 = 108\).

\(f(x)=x^3+a x^2+b \log _e|x|+1\)

$f^{\text{'}}\left(x\right)=3x^2+2ax+\frac{b}{x}$

$f^{\text{'}}\left(-1\right)=3-2a-b=0 ...\left(1\right)$

$f^{\text{'}}\left(-2\right)=12+4a-\frac{b}{2}=0 ...\left(2\right)$

on solving equation (1) and (2) we get

$a=\frac{-9}{2}, b=12$

$f^{\text{'}}\left(x\right)=3x^2-9x+\frac{12}{x}$

\(=\frac{3 x^3-9 x^2+12}{x}\)

\(=\frac{(x+1)\left(3 x^2-12 x+12\right)}{x}\)

$=\frac{3\left(x+1\right){\left(x+2\right)}^2}{x}$

since \(x \in\left[-2,-\frac{1}{2}\right]\)

$f\left(x\right)=x^2-\frac{9}{2}{x}^2+12\ln \left|x\right|+1$

$f\left(-1\right)=-1-\frac{9}{2}+1$

$=-4.5$

\(f\left(-\frac{1}{2}\right)=-\frac{1}{8}-\frac{9}{8}-12 \times 0.7+1=-1.25\)

$f\left(-2\right)=-8-18+12\ln 2+1$

$=-25+12\ln 2=-16.6$

Hence \(M=-4.5, m=-16.6\)

$\left|M+m\right|=21.1$

\(f(x)=6x^2-18ax+12a^2=0\)

\(\alpha+\alpha^2=3a\) and \(\alpha\times\alpha^2=2a^2\)

\(\left(\alpha+\alpha^2\right)^3=27a^3\)

\(\Rightarrow 2a^2+4a^4+3(3a)(2a^2)=27a^3\)

\(\Rightarrow 2+4a^2+18a=27a\)

\(\Rightarrow 4a^2-9a+2=0\)

\(\Rightarrow 4a^2-8a-a+2=0\)

\(\Rightarrow (4a-1)(a-2)=0\Rightarrow a=2\)

So \(6x^2-36x+48=0\)

\(\Rightarrow x^2-6x+8=0\)

If we take \(a=\frac{1}{4}\), then \(\alpha=\frac{1}{2}\), which is not possible.

Since \(\frac{x^2}{f\left(a^2+7 a+3\right)}+\frac{y^2}{f(3 a+15)}=1\) represent an ellipse with major axis along \(y\)-axis

Therefore, \(f(3 a+15)>f\left(a^2+7 a+3\right)\)

Given that \(f\) is a strictly decreasing positive function

So, \(3 a+15<a^2+7 a+3\)

\(a^2+4 a-12>0\)

\((a+2)^2-16>0\)

\(\Rightarrow(a+2)^2>16\)

\(|a+2|>4 \Rightarrow a+2>4\) or \(a+2<-4\).

So, \(a>2\) or \(a<-6\)

i.e. \(a \in \mathbb{R}-[-6,2]\)

Hence, \(\alpha=-6\) and \(\beta=2\).

Therefore,

\(\alpha^2+\beta^2=(-6)^2+(2)^2=36+4=40\)

\(E=16 \sin \frac{x}{2} \cos ^3 \frac{x}{2} \)
\(\mathrm{E}=4 \sin \mathrm{x}[1+\cos \mathrm{x}]\)
\(\frac{\mathrm{dE}}{\mathrm{dx}}=4[\cos \mathrm{x}+\cos 2 \mathrm{x}] \)
\(=8 \cos \frac{3 x}{2} \cos \frac{x}{2}=0\)
\( \Rightarrow \cos \frac{3 x}{2}=0 \) or \( \cos \frac{x}{2}=0 \)
\( \Rightarrow \mathrm{x}=\left\{\frac{\pi}{3}, \pi\right\} \)are critical points of the function
\( \mathrm{E}(0)=0 \)
\( \mathrm{E}(\pi)=0\)

$E\left(\frac{\mathrm{\pi }}{3}\right)=3\sqrt{3}$

Hence, maximum value of \(\mathrm{E}=3 \sqrt{3}\)

\(f(x)=e^{x^3-3x+1}\)

\(f'(x)=e^{x^3-3x+1}\cdot(3x^2-3)\)

\(=e^{x^3-3x+1}\cdot 3(x-1)(x+1)\)

For \(f'(x)\geq 0\)

\(\therefore f(x)\) is increasing function.

\(\therefore a=e^{-\infty}=0=f(-\infty)\)

\(b=e^{-1+3+1}=e^3=f(-1)\)

\(P(2b+4,a+2)\)

\(\therefore P\left(2e^3+4,2\right)\)

\(d=\frac{\left(2e^3+4\right)+2e^{-3}-4}{\sqrt{1+e^{-6}}}=2\sqrt{1+e^6}\)

Given $\mathrm{f}\left(\mathrm{x}\right)=2{\log }_{\mathrm{e}}\left(\mathrm{x}-2\right)-{\mathrm{x}}^2+\mathrm{ax}+1$

On differentiating both sides, we get

${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=\frac{2}{\mathrm{x}-2}-2\mathrm{x}+\mathrm{a} \ge 0$

For increasing \(f(x)\)

Again,differentiating both sides, we get

$f^{"}\left(\mathrm{x}\right)=\frac{-2}{(\mathrm{x}-2{)}^2}-2<0$

\(f'(x)\) is decreasing function

$$\begin{gathered} {\mathrm{f}}^{\text{'}}\left(3\right)\ge 0 \\ 2-6+\mathrm{a}\ge 0 \\ \mathrm{a}\ge 4 \\ {\mathrm{a}}_{\min }=4 \\ \mathrm{g}\left(\mathrm{x}\right)={\left(\mathrm{x}-1\right)}^3{\left(\mathrm{x}+2-\mathrm{a}\right)}^2 \\ \mathrm{g}\left(\mathrm{x}\right)={\left(\mathrm{x}-1\right)}^3{\left(\mathrm{x}-2\right)}^2 \end{gathered}$$

On differentiating, we get

$$\begin{gathered} {\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)={\left(x-1\right)}^{3}2\left(x-2\right)+{\left(\mathrm{x}-2\right)}^{2}3{\left(x-1\right)}^2 \\ ={\left(x-1\right)}^2\left(\mathrm{x}-2\right)\left(2\mathrm{x}-2+3\mathrm{x}-6\right) \\ ={\left(x-1\right)}^2\left(\mathrm{x}-2\right)\left(5\mathrm{x}-8\right)<0 \\ \mathrm{x}\in \left(\frac{8}{5},2\right) \end{gathered}$$

Hence $100\left(\mathrm{a}+\mathrm{b}-\mathrm{c}\right)=100\left(4+\frac{8}{5}-2\right)=360$

\(f^{\prime}(x)=e^x\left(2 x^2-9 x+11\right)+e^x(4 x-9) \)
\(=e^x\left(2 x^2-5 x+2\right)\)
\(=e^x(2 x-1)(x-2)\)

Critical points:

\(x=\frac{1}{2},x=2\)

Since \(e^x>0\), sign depends on

\((2 x-1)(x-2)\)

For \(x<\frac{1}{2}\) : positive
For \(\frac{1}{2}<x<2\) : negative
For \(x>2\) : positive

Hence:
at \(x=\frac{1}{2}\) : \(f^{\prime}\) changes \(+\rightarrow-\), so local maximum.
at \(x=2\) : \(f^{\prime}\) changes \(-\rightarrow+\), so local minimum.

local maximum at \(x=\frac{1}{2}\) and local minimum at \(x=2\)

Since \(f:\{1,2,3,4\} \rightarrow\left\{1, e, e^2, e^3\right\}\) is bijective and strictly decreasing,

we match the largest element of the domain to the smallest element of the codomain.

So,

\(f(1)=e^3, f(2)=e^2, f(3)=e, f(4)=1\)

Hence

\(f^{-1}(1)=4, f^{-1}(e)=3, f^{-1}\left(e^2\right)=2, f^{-1}\left(e^3\right)=1\)

Now \(g:\left\{1, e, e^2, e^3\right\} \rightarrow\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}\) is bijective and strictly increasing

for \(g\) to be strictly increasing, we must assign

\(g(1)=\frac{1}{4}, g(e)=\frac{1}{3}, g\left(e^2\right)=\frac{1}{2}, g\left(e^3\right)=1\)

\(g^{-1}\left(\frac{1}{2}\right)=e^2\)

Now, \(f^{-1}\left(g^{-1}\left(\frac{1}{2}\right)\right)=f^{-1}\left(e^2\right)=2\)

So, \(\phi(x)=\left[f^{-1}\left\{g^{-1}\left(\frac{1}{2}\right)\right\}\right]^x=2^x\)

Thus region is

\(R=\left\{(x, y): x^2 \leq y \leq 2^x, 0 \leq x \leq 1\right\}\)

So area is:

\(=\int_0^1\left(2^x-x^2\right) d x\)

\(=\left[\frac{2^x}{\ln 2}\right]_0^1-\left[\frac{x^3}{3}\right]_0^1\)

\(=\frac{1}{\ln 2}-\frac{1}{3}\)

\(=\frac{3-\ln 2}{3 \ln 2}\)

For maximum and minima, \(f'(x)=0\)

\(\Rightarrow 6x^2-18ax+12a^2=0\) and \(f''(x)=12x-18a\)

\(f'(x)=0\)

\(x=a,2a\) and \(f''(a)<0\) and \(f''(2a)>0\)

Now \(p=a\) and \(q=2a\), and \(p^2=q\)

\(\Rightarrow a^2=2a\)

\(\Rightarrow a^2-2a=0\)

\(\Rightarrow a(a-2)=0\)

\(\Rightarrow a=0,\ a=2\)

\(f(3)=37\)

\(f^{\prime}(x)=3 \sqrt{2}(2 x+1)(2 x-1)>0 \quad \forall x \in\left(\frac{1}{2}, 1\right]\)

\(\Rightarrow \mathrm{f}(\mathrm{x})\) is strictly increasing in \(\left[\frac{1}{2}, 1\right]\)

\(f\left(\frac{1}{2}\right)<0, f(1)>0\)

\(\Rightarrow \mathrm{f}(\mathrm{x})=0\) has exactly one solution in \(\left[\frac{1}{2}, 1\right]\)
Also \(f(x)=0 \Rightarrow 4 x^3-3 x=\frac{1}{\sqrt{2}}\)
Now put \(x=\cos \theta \Rightarrow \cos 3 \theta=\frac{1}{\sqrt{2}}\)

\(\Rightarrow x=\cos \frac{\pi}{12} .\)

\(f^{\prime}(x)=3 \sqrt{2}(2 x+1)(2 x-1)>0 \quad \forall x \in\left(\frac{1}{2}, 1\right]\)

\(\Rightarrow \mathrm{f}(\mathrm{x})\) is strictly increasing in \(\left[\frac{1}{2}, 1\right]\)

\(f\left(\frac{1}{2}\right)<0, f(1)>0\)

\(\Rightarrow \mathrm{f}(\mathrm{x})=0\) has exactly one solution in \(\left[\frac{1}{2}, 1\right]\)
Also \(f(x)=0 \Rightarrow 4 x^3-3 x=\frac{1}{\sqrt{2}}\)
Now put \(x=\cos \theta \Rightarrow \cos 3 \theta=\frac{1}{\sqrt{2}}\)

\(\Rightarrow x=\cos \frac{\pi}{12} .\)

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\frac{|\mathrm{x}-1|}{{\mathrm{x}}^2} \\ = \left\{\begin{array}{l}\frac{x-1}{x^2} ; x\ge 1\\ \frac{1-x}{x^2} ; x<1\\ \end{array}\right. \\ =\left\{\begin{array}{l}\frac{1}{x}-\frac{1}{x^2}, x\ge 1\\ \frac{1}{x^2}-\frac{1}{x}, x<1\\ \end{array}\right. \end{gathered}$$

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{l}\frac{-1}{x^2}+\frac{2}{x^3}, x\ge 1\\ \frac{-2}{x^3}+\frac{1}{x^2}, x<1\end{array}\right.$

for decreasing

$$\begin{gathered} {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)<0, \\ \text{for }x\ge 1, \\ \frac{2}{x^3}-\frac{1}{x^2}<0 \\ \Rightarrow 2-x<0 \\ \Rightarrow x>2 \\ \text{ For }x<1, \\ \frac{1}{x^2}<\frac{2}{x^3} \\ \Rightarrow \frac{x-2}{x}<0 \\ \mathrm{x}\in (0,2) \\ \therefore \mathrm{x}\in (0,1)\cup (2,\infty ) \end{gathered}$$

\(f(x)=\frac{x}{x^2-6 x-16}\)

Now,\( f^{\prime}(x)=\frac{-\left(x^2+16\right)}{\left(x^2-6 x-16\right)^2} \)
\( f^{\prime}(x)<0\)

Thus \(f(x)\) is decreasing in\((-\infty,-2) \cup(-2,8) \cup(8, \infty) .\)

\(f(x)=\frac{x}{x^2-6 x-16}\)

Now,\( f^{\prime}(x)=\frac{-\left(x^2+16\right)}{\left(x^2-6 x-16\right)^2} \)
\( f^{\prime}(x)<0\)

Thus \(f(x)\) is decreasing in\((-\infty,-2) \cup(-2,8) \cup(8, \infty) .\)

$f^{\prime}(x)=18x^{2}-90\;ax+108a^{2}=0$

x = 2a & x = 3a

${\mathrm{x}}_1=2\mathrm{a} , {\mathrm{x}}_2=3\mathrm{a}$

${\mathrm{x}}_1{\mathrm{x}}_2=54$

$6{\mathrm{a}}^2=54$

a = 3

$\mathrm{a}+{\mathrm{x}}_1+{\mathrm{x}}_2$=

3 + 2 × 3 + 3 × 3 = 18

Given \( f(x)=3 x+\cos 3 x \)

$f^{\text{'}}\left(x\right)=3-3\sin 3x = 3\left(1-\sin 3x\right) \ge 0$

since, \(-1 \leq \sin 3x \leq 1\) $\Rightarrow 0\le 1-\sin 3x\le 2$

$\therefore f^{\text{'}}\left(x\right) \ge 0$or all $x \in R \Rightarrow f\left(x\right)$ is increasing on R

\(f'(x)=\frac{g'(x)-g'(2-x)}{2}\), \(f'\left(\frac{3}{2}\right)=\frac{g'\left(\frac{3}{2}\right)-g'\left(\frac{1}{2}\right)}{2}=0\)

Also \(f'\left(\frac{1}{2}\right)=\frac{g'\left(\frac{1}{2}\right)-g'\left(\frac{3}{2}\right)}{2}=0\), \(f'\left(\frac{1}{2}\right)=0\)

\(\Rightarrow f'\left(\frac{3}{2}\right)=f'\left(\frac{1}{2}\right)=0\)

\(\Rightarrow\) roots in \(\left(\frac{1}{2},1\right)\) and \(\left(1,\frac{3}{2}\right)\)

\(\Rightarrow f''(x)\) is zero at least twice in \(\left(\frac{1}{2},\frac{3}{2}\right)\)

\(f'(x)=\frac{g'(x)-g'(2-x)}{2}\), \(f'\left(\frac{3}{2}\right)=\frac{g'\left(\frac{3}{2}\right)-g'\left(\frac{1}{2}\right)}{2}=0\)

Also \(f'\left(\frac{1}{2}\right)=\frac{g'\left(\frac{1}{2}\right)-g'\left(\frac{3}{2}\right)}{2}=0\), \(f'\left(\frac{1}{2}\right)=0\)

\(\Rightarrow f'\left(\frac{3}{2}\right)=f'\left(\frac{1}{2}\right)=0\)

\(\Rightarrow\) roots in \(\left(\frac{1}{2},1\right)\) and \(\left(1,\frac{3}{2}\right)\)

\(\Rightarrow f''(x)\) is zero at least twice in \(\left(\frac{1}{2},\frac{3}{2}\right)\)

Given \( f(x)=3 x+\cos 3 x \)

$f^{\text{'}}\left(x\right)=3-3\sin 3x = 3\left(1-\sin 3x\right) \ge 0$

since, \(-1 \leq \sin 3x \leq 1\) $\Rightarrow 0\le 1-\sin 3x\le 2$

$\therefore f^{\text{'}}\left(x\right) \ge 0$or all $x \in R \Rightarrow f\left(x\right)$ is increasing on R

\(f(x)=\sin x+3x-\frac{2}{\pi}(x^2+x),\ x\in\left[0,\frac{\pi}{2}\right]\)

\(f'(x)=\cos x+3-\frac{2}{\pi}(2x+1)>0\)

\(f(x)\) increasing

\(f'(x)=-\sin x+0-\frac{\pi}{2}(2)\)

\(=-\sin x-\frac{4}{\pi}<0\)

\(f'(x)\) decreasing

\(0<x<\frac{\pi}{2}\)

\(\Rightarrow -\frac{2}{\pi}(0+1)<-\frac{2}{\pi}(2x+1)<-\frac{2}{\pi}(\pi+1)\)

\(-\frac{2}{\pi}>-\frac{2}{\pi}(2x+1)>-\frac{2}{\pi}(\pi+1)\)

\(+3\)

\(3-\frac{2}{\pi}>3-\frac{2}{\pi}(2x+1)>3-\frac{2}{\pi}(\pi+1)\)

Positive

\(
g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \text { and } f^{\prime \prime}(x)>0 \forall x \in(0,3)
\)

\(\Rightarrow f^{\prime}(x)\) is increasing function

\(
\begin{aligned}
& g^{\prime}(x)=3 \times \frac{1}{3} \cdot f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \\
& =\mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})
\end{aligned}
\)

If g is decreasing in \((0, \alpha)\)

\(
\begin{aligned}
& \mathrm{g}^{\prime}(\mathrm{x})<0 \\
& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})<0 \\
& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)<\mathrm{f}^{\prime}(3-\mathrm{x}) \\
& \Rightarrow \frac{\mathrm{x}}{3}<3-\mathrm{x} \\
& \Rightarrow \mathrm{x}<\frac{9}{4}
\end{aligned}
\)

\(x^4-a x^2+9=0 \ldots(1)\)

let \(x^2=t\)

\(t^2-a t+9=0\)

for roots of equation (1) to be real & distinct roots of equation (2) must be positive & distinct.
(i) \(D>0 \Rightarrow a^2-36>0\) $\Rightarrow a\in (-\infty ,-6)\cup (6,\infty )$
(ii) \(\frac{-b}{2 a}>0 \Rightarrow \frac{a}{2}>0 \Rightarrow a>0\)
(iii) \(f(0)>0 \Rightarrow 9>0 \Rightarrow a \in R\)
\(B y(i) \cap(i i) \cap(i i i)\)

\(\therefore a \in(6, \infty)\)

$\therefore$ least integral value of a is \(7\)

$f^{\prime}(x)=18x^{2}-90\;ax+108a^{2}=0$

x = 2a & x = 3a

${\mathrm{x}}_1=2\mathrm{a} , {\mathrm{x}}_2=3\mathrm{a}$

${\mathrm{x}}_1{\mathrm{x}}_2=54$

$6{\mathrm{a}}^2=54$

a = 3

$\mathrm{a}+{\mathrm{x}}_1+{\mathrm{x}}_2$=

3 + 2 × 3 + 3 × 3 = 18

\(f'(x)=(x+3)^2\cdot 3(x-2)^2+(x-2)^3\cdot 2(x+3)\)

\(=5(x+3)(x-2)^2(x+1)\)

\(f'(x)=0,\ x=-3,-1,2\)

\(f(-4)=-216\)

\(f(-3)=0,\ f(4)=49\times 8=392\)

\(M=392,\ m=-216\)

\(M-m=392+216=608\)

\(5f(x)+4f\left(\frac{1}{x}\right)=x^2-2\) \(\tag{1}\)

Substitute \(x\rightarrow \frac{1}{x}\)

\(5f\left(\frac{1}{x}\right)+4f(x)=\frac{1}{x^2}-2\) \(\tag{2}\)

\((1)\times 5-(2)\times 4\)

\(\Rightarrow 9f(x)=5x^2-10-\frac{4}{x^2}+8\)

\(\Rightarrow 9f(x)=\frac{5x^4-2x^2-4}{x^2}\)

\(\Rightarrow 9x^2f(x)=5x^4-2x^2-4\)

\(y=5x^4-2x^2-4\)

\(\frac{dy}{dx}=20x^3-4x>0\) given

\(\Rightarrow 4x(5x^2-1)>0\)

\(\Rightarrow y\) is increasing in \(\left(-\frac{1}{\sqrt{5}},0\right)\cup\left(\frac{1}{\sqrt{5}},\infty\right)\).

\(f(x)=\cos x-x+1\)

\(f'(x)=-\sin x-1\)

\(f\) is decreasing \(\forall x\in\mathbb{R}\).

\(f(x)=0\)

\(f(0)=2,\ f(\pi)=-\pi\)

\(f\) is strictly decreasing in \([0,\pi]\) and \(f(0)\cdot f(\pi)<0\).

\(\Rightarrow\) only one solution of \(f(x)=0\).

\(S_1\) is correct and \(S_2\) is incorrect.

$\frac{dy}{dx}=\frac{2(1+\sin t)\times \cos t}{1+\cos 2t}$

$$\begin{gathered} \Rightarrow \frac{2(1+\sin t)\cos t}{2{\cos }^{2}t}=\tan \frac{\pi }{3} \\ \Rightarrow \frac{2(1+\sin t)\cos t}{2{\cos }^{2}t}=\sqrt{3} \end{gathered}$$

$\Rightarrow t=\frac{\pi }{6},y_0=27$

Since \(f(x)\) has extrema at \(x=1\) and \(x=-1,\)

\(f^{\prime}(x)=(x-1)(x+1)\left(a x^2+b x+c\right)\)
\( f^{\prime}(x)=\left(x^2-1\right)\left(a x^2+b x+c\right) \)
\( f^{\prime}(x)=a x^4+b x^3+c x^2-a x^2-b x-c \)
\( f^{\prime}(x)=a x^4+b x^3+(c-a) x^2-b x-c\)
\( f(x)=\frac{a x^5}{5}+\frac{b x^4}{4}+\frac{(c-a) x^3}{3}-\frac{b x^2}{2}-c x+d \)
\( \lim _{x \rightarrow 0} \frac{f(x)}{x^3}=-5 \)

Then \( b=c=d=0\)
and \(\frac{c-a}{3}=-5\)
\(a=15 \)
\( f(x)=3 x^5-5 x^3 \)
\( f(x)-f(-x)=f(x)+f(x) \)
\( =2 f(x) \)
\( =2\left[3(2)^5-5(2)^3\right] \)
\( =112\)

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\frac{|\mathrm{x}-1|}{{\mathrm{x}}^2} \\ = \left\{\begin{array}{l}\frac{x-1}{x^2} ; x\ge 1\\ \frac{1-x}{x^2} ; x<1\\ \end{array}\right. \\ =\left\{\begin{array}{l}\frac{1}{x}-\frac{1}{x^2}, x\ge 1\\ \frac{1}{x^2}-\frac{1}{x}, x<1\\ \end{array}\right. \end{gathered}$$

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{l}\frac{-1}{x^2}+\frac{2}{x^3}, x\ge 1\\ \frac{-2}{x^3}+\frac{1}{x^2}, x<1\end{array}\right.$

for decreasing

$$\begin{gathered} {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)<0, \\ \text{for }x\ge 1, \\ \frac{2}{x^3}-\frac{1}{x^2}<0 \\ \Rightarrow 2-x<0 \\ \Rightarrow x>2 \\ \text{ For }x<1, \\ \frac{1}{x^2}<\frac{2}{x^3} \\ \Rightarrow \frac{x-2}{x}<0 \\ \mathrm{x}\in (0,2) \\ \therefore \mathrm{x}\in (0,1)\cup (2,\infty ) \end{gathered}$$

\(V=\frac{4}{3}\pi r^3\)

\(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\)

\(81=4\pi r^2\times \frac{1}{4\pi}\)

\(r^2=81\)

\(r=9\)

Surface area of chocolate \(=4\pi(r-1)^2=256\pi\)

$f^{\prime}(x)=18x^{2}-90\;ax+108a^{2}=0$

$\mathrm{x}=2\mathrm{a} , \mathrm{x}=3\mathrm{a}$

${\mathrm{x}}_1=2\mathrm{a} , {\mathrm{x}}_2=3\mathrm{a}$

${\mathrm{x}}_1{\mathrm{x}}_2=54$

$6{\mathrm{a}}^2=54$

$a=3$

$\mathrm{a}+{\mathrm{x}}_1+{\mathrm{x}}_2=3+2\times 3+3\times 3=18$

$\frac{dy}{dx}=\frac{2(1+\sin t)\times \cos t}{1+\cos 2t}$

$$\begin{gathered} \Rightarrow \frac{2(1+\sin t)\cos t}{2{\cos }^{2}t}=\tan \frac{\pi }{3} \\ \Rightarrow \frac{2(1+\sin t)\cos t}{2{\cos }^{2}t}=\sqrt{3} \end{gathered}$$

$\Rightarrow t=\frac{\pi }{6},y_0=27$

Equation of line is

\(y+9=m(x-4)\)

\(\therefore A=\left(\frac{9+4m}{m},0\right)\)

\(B=(0,-9-4m)\)

\(\therefore OA+OB=\frac{9+4m}{m}+9+4m\)

\(\because m>0\)

\(=13+\frac{9}{m}+4m\)

\(\because \frac{4m+\frac{9}{m}}{2}\geq \sqrt{36}\Rightarrow 4m+\frac{9}{m}\geq 12\)

\(\therefore OA+OB\geq 25\)