⇒Application Of Derivative

Correct Option is (B)

Let height and breadth of the sheet be '$y$' m and '$x$' m respectively.

$\therefore xy=180000{\mathrm{cm}}^2$

$\therefore y=\frac{180000}{x}$

$\therefore$ The area available for printing is

$\begin{array}{rl}\mathrm{A}& =(y-150)(x-100)\\ & =(\frac{180000}{x}-150)(x-100)\\ & =180000-\frac{18000000}{x}-150x-15000\\ & =165000-150x-\frac{18000000}{x}\\ \therefore \frac{\mathrm{dA}}{\mathrm{d}x}& =0-150+\frac{18000000}{x^2}\\ \therefore \frac{\mathrm{dA}}{\mathrm{d}x}& =0\Rightarrow x^2=\frac{18000000}{150}=120000\end{array}$

$\begin{array}{rl}& \Rightarrow x=200\sqrt{3}\mathrm{cm}\\ & \Rightarrow y=\frac{180000}{200\sqrt{3}}=300\sqrt{3}\mathrm{cm}\end{array}$

Now, $\frac{{\mathrm{d}}^2\mathrm{A}}{\mathrm{d}{x}^2}=\frac{-36000000}{x^3}$

$\therefore \text{ At }x=200\sqrt{3}\mathrm{cm},\frac{{\mathrm{d}}^2\mathrm{A}}{\mathrm{d}{x}^2}<0$

$\therefore$ Area is maximum at $x=200\sqrt{3}\mathrm{cm}$ and $\begin{array}{rl}y& =300\sqrt{3}\mathrm{cm}\\ \therefore y& =3\sqrt{3}\mathrm{m}\text{ and }x=2\sqrt{3}\mathrm{m}\end{array}$

Correct Option is (D)

$\begin{array}{rl}& \text{ let }\mathrm{f}(\theta )=\mathrm{a}\sec \theta -\mathrm{b}\tan \theta \\ & \therefore {\mathrm{f}}^{\prime }(\theta )=\mathrm{a}\sec \theta \tan \theta -\mathrm{b}{\sec }^2\theta \\ & =\sec \theta (a\tan \theta -b\sec \theta )\\ & \therefore {\mathrm{f}}^{\prime }(\theta )=0\Rightarrow \sec \theta (\mathrm{a}\tan \theta -\mathrm{b}\sec \theta )=0\\ & \Rightarrow \mathrm{a}\tan \theta -\mathrm{b}\sec \theta =0\\ & \dots [\text{ As }0<\pi <\frac{\pi }{2},\sec \theta \ne 0]\\ & \Rightarrow \mathrm{a}\sin \theta -\mathrm{b}=0\\ & \dots [\text{ As }0<\pi <\frac{\pi }{2},\cos \theta \ne 0]\end{array}$

$\begin{array}{rl}& \Rightarrow \sin \theta =\frac{b}{a}\\ & \Rightarrow \sec \theta =\frac{a}{\sqrt{a^2-b^2}}\text{ and }\tan \theta =\frac{b}{\sqrt{a^2-b^2}}\text{... (i)}\end{array}$

Now,

$\begin{array}{rl}& f^{\prime \prime }(\theta )=\sec \theta \tan \theta (a\tan \theta -b\sec \theta )\\ & +\sec \theta (a{\sec }^2\theta -b\sec \theta \tan \theta )\\ & =a{\tan }^2\theta \sec \theta -b{\sec }^2\theta \tan \theta \\ & +a{\sec }^3\theta -b{\sec }^2\theta \tan \theta \\ & =a\sec \theta ({\tan }^2\theta +{\sec }^2\theta )\\ & =a\sec \theta (1+2{\tan }^2\theta )\\ & >0\dots [\because \text{ a is positive and }0<\theta <\frac{\pi }{2}]\end{array}$

$\therefore \mathrm{f}(\theta )$ is minimum when $\sin \theta =\frac{\mathrm{b}}{\mathrm{a}}$.

$\therefore$ Minimum value of $f(\theta )$

$\begin{array}{rl}& =a\left(\frac{a}{\sqrt{a^2-b^2}}\right)-b\left(\frac{b}{\sqrt{a^2-b^2}}\right)\dots [\text{ From (i) }]\\ & =\frac{a^2-b^2}{\sqrt{a^2-b^2}}\\ & =\sqrt{a^2-b^2}\end{array}$

Correct Option is (C)

$\begin{array}{rl}& \text{ Volume of parallelopiped }=\left|\begin{array}{lll}1& \alpha & 1\\ 0& 1& \alpha \\ \alpha & 0& 1\end{array}\right|\\ & \therefore \mathrm{V}=1+{\alpha }^3-\alpha \end{array}$

For maxima or minima,

$\begin{array}{rl}& \frac{\mathrm{dV}}{\mathrm{d}\alpha }=0\\ & \Rightarrow 3{\alpha }^2-1=0\\ & \Rightarrow \alpha =\pm \frac{1}{\sqrt{3}}\end{array}$

Now, $\frac{d^{2}V}{d{\alpha }^2}=6\alpha$

For $\alpha =\frac{1}{\sqrt{3}}$,

$\frac{d^2\mathrm{V}}{d{x}^2}>0$

$\therefore V$ is minimum at $\alpha =\frac{1}{\sqrt{3}}$.

Correct Option is (B)

Let the length, breadth and depth of open tank be $x,x$ and $y$ respectively.

Volume $(\mathrm{V})=x^{2}y$

$\therefore 500=x^{2}y$ ..... (i)

Total surface area of open tank is given by

$\mathrm{S}=x^2+4xy$ ..... (ii)

From (i), $y=\frac{500}{x^2}$

From (ii), $\mathrm{S}=x^2+4x\times \frac{500}{x^2}$

$=x^2+\frac{2000}{x}$

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

$\frac{\mathrm{dS}}{\mathrm{d}x}=2x-\frac{2000}{x^2}$

For minimum area, $\frac{\mathrm{dS}}{\mathrm{d}x}=0$

$\begin{array}{rl}\therefore & 2x-\frac{2000}{x^2}=0\\ & \Rightarrow 2000=2x^3\\ & \Rightarrow x^3=1000\end{array}$

$\begin{array}{rl}& \Rightarrow x=10\mathrm{m}\\ & \frac{{\mathrm{d}}^2\mathrm{S}}{\mathrm{d}{x}^2}=2+\frac{4000}{x^3}\\ & \Rightarrow {\left(\frac{{\mathrm{d}}^2\mathrm{S}}{\mathrm{d}{x}^2}\right)}_{x=10}>0\end{array}$

$\mathrm{S}$ is minimum when $x=10\mathrm{m}$ and $y=5\mathrm{m}$ ...[From (i)]

Correct Option is (A)

$\begin{array}{rl}& \mathrm{S}=\{x\in \mathbb{R}/x^2+30\le 11x\}\\ & x^2+30\le 11x\\ & x^2-11x+30\le 0\\ & (x-5)(x-6)\le 0\\ & x\in [5,6]\\ & \text{ Now, }\mathrm{f}(x)=3x^3-18x^2+27x-40\\ & {\mathrm{f}}^{\prime }(x)=9x^2-36x+27\\ & {\mathrm{f}}^{\prime }(x)=9(x^2-4x+3)\\ & =9[(x^2-4x+4)-1]\\ & =9(x-2{)}^2-9\\ & \therefore {\mathrm{f}}^{\prime }(x)>0\mathrm{\forall }x\in [5,6]\end{array}$

$\therefore \mathrm{f}(x)$ is strictly increasing in the interval $[5,6]$

$\therefore$ Maximum value of $\mathrm{f}(x)$ when $x\in [5,6]$ is $f(6)=122$