2 Marks Questions

Question 12 Marks

Evaluate: $\int \frac{x^2 \tan ^{-1} x}{\left(1+x^2\right)} d x$

Answer

Let $I =\int \frac{x^2 \tan ^{-1} x}{\left(1+x^2\right)} d x$
Now let $\tan ^{-1} x = t$ and $x =\tan t$
Differentiating both sides, we get
$\frac{1}{1+x^2} d x=d t$
Now we have
$I=\int \frac{x^2 \tan ^{-1} x}{\left(1+x^2\right)} d x=\int \tan ^2 t \cdot t d t=\int t\left(\sec ^2 t-1\right) dt$
$=\int tsec^2 tdt-\int tdt$
Here $t$ is the first function and $\sec ^2 t$ as the second function.
$I=\int tsec^2 tdt-\int tdt=t \int \sec ^2 tdt-\int\left(\frac{d t}{d t} \cdot \int \sec ^2 t d t\right) d t-\frac{t^2}{2}$
$=t \cdot \tan t-\int \tan tdt-\frac{t^2}{2}$
$=t \cdot \tan t-\ln |\sec t|-\frac{t^2}{2}+c$
We know that $\sec t =\sqrt{\tan ^2 t+1}$
$I=\tan ^{-1} x \cdot x-\ln \left|\sqrt{\tan ^2 t+1}\right|-\frac{\tan ^2 x}{2}+c$
$=x \tan ^{-1} x-\ln \left|\sqrt{x^2+1}\right|-\frac{\tan ^2 x}{2}+c$

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Question 22 Marks

Find the least value of a such that the function $f$ given by $f(x)=x^2+a x+1$ is strictly increasing on $(1,2)$.

Answer

$ f(x)=x^2+a x+1$
$\Rightarrow f^{\prime}(x)=2 x+a $
Since $f(x)$ is strictly increasing on $(1,2)$,
therefore $f^{\prime}(x)=2 x+a;0$ for all $x$ in $(1,2)$
$\therefore \text { On }(1,2) 1 < x < 2$
$\Rightarrow 2 < 2 x < 4$
$\Rightarrow 2+ a < 2 x + a < 4+ a $
$\therefore$ Minimum value of $f^{\prime}(x)$ is $2+a$ and maximum value is $4+a$.
Since $f^{\prime}(x);0$ for all $x$ in $(1,2)$
$ \therefore 2+a;0 \text { and } 4+a;0$
$\Rightarrow a;-2 \text { and } a;-4 $
Therefore least value of a is $- 2.$
Which is the required solution.

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Question 32 Marks

Find the maximum and minimum values of $f ( x )=\sin x$ in the interval $[\pi, 2 \pi]$.

Answer

The given function is, $f(x)=\sin x$
Therefore, $f^{\prime}(x)=\cos x$
At stationary points, we must have
$f^{\prime}(x)=0 \Rightarrow \cos x=0 \Rightarrow x=\frac{3 \pi}{2}$
Let us now compute the values of $f(x)$ at $x=\pi, \frac{3 \pi}{2}, 2 \pi$.
Now, $f (\pi)=\sin \pi=0, f\left(\frac{3 \pi}{2}\right)=\sin \frac{3 \pi}{2}=-1$ and $f (2 \pi)=\sin 2 \pi=0$.
The greatest and the least of these values are 0 and -1 respectively.
Hence, the maximum value of $f(x)$ is 0 which it attains at $x=\pi$ and $2 \pi$, and the minimum value is -1 which it attains at $x=\frac{3 \pi}{2}$.

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Question 42 Marks

The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x)=3 x^2+36 x+5$. Find the marginal revenue, when $x=5$, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at any instant.

Answer

Since Marginal Revenue is the rate of change of total revenue with respect to the number of units sold, we have
Marginal Revenue $( MR )=\frac{d R}{d x}=6 x +36$
When $x=5$, MR $=6(5)+36=66$
Hence, the required marginal revenue is 66.

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Question 52 Marks

Find the domain of $f(x)=\sin ^{-1}\left(-x^2\right)$.

Answer

The domain of $\sin ^{-1} x$ is $(-1,1)$.
Therefore, $f(x)=\sin ^{-1}\left(-x^2\right)$ is defined for all $x$ satisfying $-1 \leq-x^2 \leq 1$
$\Rightarrow 1 \geq x^2 \geq-1$
$\Rightarrow 0 \leq x^2 \leq 1$
$\Rightarrow x^2 \leq 1$
$\Rightarrow x^2-1 \leq 0$
$\Rightarrow(x-1)(x+1) \leq 0$
$\Rightarrow-1 \leq x \leq 1$
Hence, the domain of $f(x)=\sin ^{-1}\left(-x^2\right)$ is $(-1,1)$.

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Question 62 Marks

Find the principal value of $\operatorname{cosec}^{-1}(-1)$

Answer

We know that the range of principal value of $\operatorname{cosec}^{-1}$ is $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)-(0)$
Let $\operatorname{cosec}^{-1}(-1)=\theta$. Then we have, $\operatorname{cosec} \theta=-1$
$\operatorname{cosec} \theta=-1=-\operatorname{cosec} \frac{\pi}{2}=\operatorname{cosec}\left(\frac{-\pi}{2}\right)$
$\therefore \theta=\frac{-\pi}{2} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)-(0)$
Hence, the principal value of $\operatorname{cosec}^{-1}(-1)$ is equal to $\frac{-\pi}{2}$

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Question 72 Marks

The volume of a cube is increasing at the rate of $9 \ cm^3 / \sec$. How fast is the surface area increasing when the length of an edge is $10 \ cm$ ?

Answer

Let $x$ be the side and $V$ be the volume of the cube at any time $t$ Then,
$V=x^3$
Differentiating both sides with respect to $t,$
$\Rightarrow \frac{d V}{d t}=3 x^2 \frac{d x}{d t}$
$\Rightarrow 9=3(10)^2 \frac{d x}{d t}\ (\because x=10 \ cm$ and $\frac{d V}{d t}=9 \ cm^3 / \sec)$
$\Rightarrow \frac{d x}{d t}=0.03 \ cm / \sec $
Let $S$ be the surface area of the cube at any time then $t,$
$S=6 x^2$
Differentiating both sides with respect to $t,$
$\Rightarrow \frac{d S}{d t}=12 x \frac{d x}{d t}$
$\Rightarrow \frac{d S}{d t}=12 \times 10 \times 0.03\ (\because x=10 \ cm$ and $\frac{d x}{d t}=0.03 \ cm / \sec)$
$\Rightarrow \frac{d S}{d t}=3.6 \ cm^2 / \sec $

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