⇒Application Of Derivative

Correct Option is (C)

Applying Lagrange's mean value theorem on interval $[0,2]$, we get

there exist atleast one '$c$' $\in (0,2)$ such that

$\begin{array}{ll}& \frac{\mathrm{f}(2)-\mathrm{f}(0)}{2-0}={\mathrm{f}}^{\prime }(\mathrm{c})\\ \therefore & \mathrm{f}(2)-\mathrm{f}(0)=2{\mathrm{f}}^{\prime }(\mathrm{c})\\ \therefore & \mathrm{f}(2)=\mathrm{f}(0)+2{\mathrm{f}}^{\prime }(\mathrm{c})\\ \therefore & \mathrm{f}(2)=-3+2{\mathrm{f}}^{\prime }(\mathrm{c})\\ & \text{ Given that }{\mathrm{f}}^{\prime }(x)\le 5\text{ for all }x\\ \therefore & \mathrm{f}(2)\le -3+10\\ \therefore & \mathrm{f}(2)\le 7\end{array}$

$\therefore$ Largest possible value of $\mathrm{f}(2)$ is 7.

Correct Option is (A)

Volume of parallelopiped is $[\overrightarrow{\mathrm{a}}\overrightarrow{\mathrm{b}}\overrightarrow{\mathrm{c}}]$

$\begin{array}{rl}\therefore V& =\left|\begin{array}{lll}1& \alpha & 1\\ 0& 1& \alpha \\ \alpha & 0& 1\end{array}\right|\\ & =1-\alpha (-{\alpha }^2)-\alpha \\ & =1+{\alpha }^3-\alpha \end{array}$

Differentiating w.r.t. $\alpha$, we get

$\begin{array}{ll}& \frac{dV}{d\alpha }=3{\alpha }^2-1\\ \therefore & \frac{d^{2}V}{d{\alpha }^2}=6\alpha \\ & \text{ Let }\frac{dV}{d\alpha }=0\\ \therefore & 3{\alpha }^2-1=0\\ \therefore & \alpha =\pm \frac{1}{\sqrt{3}}\\ & \text{ at }\alpha =\frac{-1}{\sqrt{3}},\\ & \frac{d^{2}V}{d{\alpha }^2}=\frac{-6}{\sqrt{3}}<0\end{array}$

$\therefore \mathrm{V}$ is maximum at $\alpha =\frac{-1}{\sqrt{3}}$

Correct Option is (D)

$\begin{array}{rl}& x+2y=8\\ \therefore & 2y=8-x\\ \therefore & y=\frac{8-x}{2}\end{array}$

Let $\mathrm{f}(x)=xy$

$\therefore \mathrm{f}(x)=x\cdot \frac{(8-x)}{2}$

Differentiating w.r.t $x$, we get

$\begin{array}{rl}& f^{\prime }(x)=\frac{(8-x)-x}{2}\\ & f^{\prime }(x)=4-x\end{array}$

To find critical points,

$\begin{array}{rl}& {\mathrm{f}}^{\prime }(x)=0\\ \therefore & 4-x=0\\ \therefore & x=4\end{array}$

critical point at $x=4$

$\therefore \mathrm{f}(4)=\frac{4(8-4)}{2}=8$

$\therefore$ Maximum value of the given function is 8.

Correct Option is (A)

We have $f(x)=\frac{1-x+x^2}{1+x+x^2}$

$\begin{array}{rl}\therefore f^{\prime }(x)& =\frac{(1+x+x^2)(2x-1)-(1-x+x^2)(2x+1)}{{(1+x+x^2)}^2}\\ & =\frac{(2x+2x^2+2x^3-x-1-x^2)-(2x-2x^2+2x^3+1-x+x^2)}{{(1+x+x^2)}^2}\\ & =\frac{(x+x^2+2x^3-1)-(x-x^2+2x^3+1)}{{(1+x+x^2)}^2}\\ & =\frac{2x^2-2}{{(1+x+x^2)}^2}\text{ and when }{f}^{\prime }(x)=0,\text{ we get }\end{array}$

$2(x^2-1)=0\Rightarrow x=\pm 1$

When $x=1,f(x)=\frac{1}{3}$ and when $x=-1,f(x)=3$

Hence minimum value of $f(x)$ is $\frac{1}{3}$.

Correct Option is (B)

Let $x$ be the one part and $y$ be the other part.

We have $x+y=20\Rightarrow y=20-x$

As per condition given, we write

$\begin{array}{rl}& f(x)=(20-x)x^3\\ & =20x^3-x^4\\ & \therefore f^{\prime }(x)=60x^2-4x^3\end{array}$

When $f^{\prime }(x)=0$, we get

$\begin{array}{rl}& 4x^2(15-x)=0\Rightarrow x=0,15\\ & f^{\prime \prime }(x)=120x-12x^2\\ & {[f^{\prime \prime }(x)]}_{x=15}=(120)(15)-(12)(15{)}^2=-900<0\\ & \therefore f(x)\text{ is maximum when }x=15\end{array}$

$\therefore \mathrm{f}(\mathrm{x})$ is maximum when $\mathrm{x}=15$.

$\therefore y=5\Rightarrow xy=(15)(5)=75$