⇒Application Of Derivative

Correct Option is (D)

According to the figure, $x^2+y^2=25$ .... (i)

Note that $\cos \theta =\frac{\mathrm{OB}}{\mathrm{AB}}=\frac{x}{5}$

$\begin{array}{rl}& \therefore x=5\cos \theta \\ & \therefore \text{ (i) }\Rightarrow 25{\cos }^2\theta +y^2=25\end{array}$

Differentiating w.r.t. '$t$', we get

$\begin{array}{rl}& -50\cos \theta \sin \theta \frac{\mathrm{d}\theta }{\mathrm{dt}}+2y\frac{\mathrm{d}y}{\mathrm{dt}}=0\\ & 25\sin \theta \cos \theta \frac{\mathrm{d}\theta }{\mathrm{dt}}=y\frac{\mathrm{d}y}{\mathrm{dt}}\\ & \therefore 25\sin \theta \cos \theta \frac{\mathrm{d}\theta }{\mathrm{dt}}=y(-0.1)\\ & \dots [\because \frac{\mathrm{d}y}{\mathrm{d}x}=-10\mathrm{cm}/\mathrm{s}=-0.1\mathrm{m}/\mathrm{s}]\end{array}$

$\begin{array}{l}\therefore 25\sin \theta \cos \theta \frac{\mathrm{d}\theta }{\mathrm{dt}}=-(0.1)y\text{.... (ii)}\\ \text{ at }x=4,\cos \theta =\frac{4}{5},\sin \theta =\frac{3}{5}\text{ and }y=3\\ \therefore \text{ (ii) }\Rightarrow 25\times \frac{3}{5}\times \frac{4}{5}\times \frac{\mathrm{d}\theta }{\mathrm{dt}}=-0.3\\ \Rightarrow \frac{\mathrm{d}\theta }{\mathrm{dt}}=-0.025\end{array}$

i.e., the angle is decreasing at the rate of $0.025\mathrm{rad}/\mathrm{s}$

Correct Option is (B)

Let length and breadth of the tank be '$x$' $\mathrm{m}$ and '$y$' m respectively.

Height of the tank is $4\mathrm{m}$.

$\begin{array}{ll}& \text{ Height of the tank }\\ & \text{ Volume }=36{\mathrm{m}}^3\\ \therefore & 4xy=36\\ \therefore & xy=9\text{... (i)}\\ \therefore & y=\frac{9}{x}\text{... (ii)}\end{array}$

$\therefore$ Total area of the tank including sides and base $=xy+2(4x)+2(4y)$

$\begin{array}{rl}& \therefore \mathrm{f}(x)=9+8x+8\left(\frac{9}{x}\right)\text{...[From (i) and (ii)]}\\ & =9+8x+\frac{72}{x}\\ \\ & \therefore f^{\prime }(x)=8-\frac{72}{x^2}\\ \\ & \therefore {\mathrm{f}}^{\prime }(x)=0\Rightarrow x=3\\ \\ & \Rightarrow y=3\\ \\ & \therefore \text{ Required cost }=100\times (3\times 3)+50\times (2\times 4\times 3+2\times 4\times 3)\\ \\ & =900+2400\\ \\ & =3300\\ \end{array}$

Correct Option is (A)

$\frac{\mathrm{dP}}{\mathrm{d}x}=100-12\sqrt{x}$

Integrating both sides, we get

$\begin{array}{rl}& \int \mathrm{dp}=\int (100-12\sqrt{x})\mathrm{d}x\\ \therefore & \mathrm{P}=100x-8x\sqrt{x}+\mathrm{c}\end{array}$

Given that $\mathrm{P}=1000$, when $x=0$

$\begin{array}{rl}& \Rightarrow 1000=100(0)-8(0)+c\\ & \Rightarrow c=1000\\ \therefore & P=100x-8x\sqrt{x}+1000\end{array}$

When $x=9$, we get

$\mathrm{P}=900-216+1000=1684$

$\therefore$ The new level of production of items is 1684.

Correct Option is (B)

Surface area, $S=4\pi r^2$

$\begin{array}{rl}\therefore & \frac{\mathrm{dS}}{\mathrm{dt}}=8\pi \mathrm{r}\frac{\mathrm{dr}}{\mathrm{dt}}\\ \Rightarrow & 2=8\pi \mathrm{r}\frac{\mathrm{dr}}{\mathrm{dt}}\\ & \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{4\pi \mathrm{r}}\text{... (i)}\end{array}$

Volume, $V=\frac{4}{3}\pi r^3$

$\begin{array}{rl}\therefore \frac{\mathrm{dV}}{\mathrm{dt}}& =\frac{4}{3}\times 3\pi {\mathrm{r}}^2\times \frac{\mathrm{dr}}{\mathrm{dt}}\\ & =4\pi {\mathrm{r}}^2\times \frac{1}{4\pi \mathrm{r}}...[\text{From (i)}]\\ & =\mathrm{r}\\ & =6{\mathrm{cm}}^3/\sec \end{array}$

Correct Option is (A)

Let $x$ be the number of bacteria present at time $t$.

$\begin{array}{ll}\therefore & \frac{\mathrm{d}x}{\mathrm{dt}}\propto x\\ \therefore & \frac{\mathrm{d}x}{\mathrm{dt}}=\mathrm{k}x\\ \therefore & \frac{\mathrm{d}x}{x}=\mathrm{kdt}\end{array}$

Integrating on both sides, we get

$\log x=\mathrm{kt}+\mathrm{c}$ .... (i)

When $\mathrm{t}=3,x=10,000$

Equation (i) becomes

$\log (10,000)=3\mathrm{k}+\mathrm{c}$ .... (ii)

When $\mathrm{t}=5,x=40,000$

Equation (i) becomes

$\log (40,000)=5k+c$ .... (iii)

Subtracting (ii) from (iii), we get

$\mathrm{k}=\log 2$

From equation (ii),

$\begin{array}{rl}& \log (10,000)=3\log 2+c\end{array}$

$\therefore c=\log (1250)$

Now, Initially $\mathrm{t}=0$

From (i),

$\begin{array}{rl}& \log x=\mathrm{k}\times 0+\log (1250)\\ \therefore & \log x=\log 1250\\ \therefore & x=1250\end{array}$