2 Marks Questions

Question 12 Marks

Prove that the function $f$ given by $f(x)=x^2-x+1$ is neither strictly increasing nor strictly decreasing on $(-1,1)$.

Answer

Given: $f(x)=x^2-x+1 f ( x )= x ^2- x +1$
$\Rightarrow f^{\prime}(x)=2 x-1$
$f ( x )$ is strictly increasing if $f ( x )<0$
$\Rightarrow 2 x-1>0$
$\Rightarrow x>\frac{1}{2} $
i.e., increasing on the interval $\left(\frac{1}{2}, 1\right)$ $f(x)$ is strictly decreasing if $f(x)<0$
$ \Rightarrow 2 x-1<0$
$\Rightarrow x<\frac{1}{2}$
i.e., decreasing on the interval $\left(-1, \frac{1}{2}\right)$
hence, $f(x)$ is neither strictly increasing nor decreasing on the interval $(-1, 1).$

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Question 22 Marks

Integrate the function $e^x(\sin x+\cos x)$

Answer

$I=\int e^x(\sin x+\cos x) d x$
Now,
Let $\sin x = f ( x ) \Rightarrow f ^{\prime}( x )=\cos x$
We know that,
$\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+c$
Thus,
$\int e^x(\sin x+\cos x) d x=e^x \sin x+C$

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Question 32 Marks

Find the intervals in which the function $f$ given by $f(x)=4 x^3-6 x^2-72 x+30$ is
$i.$ increasing
$ii.$ decreasing.

Answer

$f(x)=4 x^3-6 x^2-72 x+30$
$f^{\prime}(x)=12 x^2-12 x-72$
$=12\left(x^2-x-6\right)$
$=12\left(x^2-3 x+2 x-6\right)$
$=12[x(x-3)+2(x-3)]$
$=12(x-3)(x+2)$

$x = - 2, 3$

int	Sign of $f\ '(x)$	Result
$(-\infty,-2)$	$+$ tive	Increase
$(-2,3)$	$+$ tive	Decrease
$(3, \infty)$	$+$ tive	increase

Hence function is
$i.$ increasing in $(-\infty,-2)$
$ii. (3, \infty)$ decreasing in $(-2,3)$

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Question 42 Marks

Determine whether $f(x)=-\frac{\pi}{2}+ \sin x$ is increasing or decreasing on $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$

Answer

$f(x)=-\frac{x}{2}+\sin x$
$\Rightarrow f^{\prime}(x)=-\frac{1}{2}+\cos x$
Now
$x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$
$\Rightarrow-\frac{\pi}{3} < x < \frac{\pi}{3}$
$\Rightarrow \cos \left(-\frac{\pi}{3}\right) < \cos x < \cos \frac{\pi}{3}$
$\Rightarrow \cos \left(\frac{\pi}{3}\right) < \cos x < \cos \frac{\pi}{3}$
$\Rightarrow \frac{1}{2} < \cos x < \frac{1}{2}$
$\Rightarrow-\frac{1}{2}+\cos x > 0$
$\Rightarrow f^{\prime}(x) > 0$
Hence, $f(x)$ is an increasing function on $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.

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Question 52 Marks

The volume of a spherical balloon is increasing at the rate of $25 \ cm^3 / \sec$. Find the rate of change of its surface area at the instant when radius is $5 \ cm.$

Answer

Let $r$ be the radius and $V$ be the volume of the sphere at any $t$ then,
$ V=\frac{4}{3} \pi r^3 $
Differentiating both sides with respect to $t$,
$ \Rightarrow \frac{d V}{d t}=4 \pi r^2 \frac{d r}{d r}$
$\Rightarrow \frac{d r}{d r}=\frac{1}{4 \pi r^2} \frac{d V}{d t}$
$\Rightarrow \frac{d r}{d t}=\frac{25}{4 \pi(5)^2}$
$\Rightarrow \frac{d r}{d t}=\frac{1}{4 \pi} \ cm / \sec $
Now, let $S$ be the surface area of the sphere area at any $t$ then
$S=4 \pi r^2$
Differentiating both sides with respect to $t$,
$ \Rightarrow \frac{d S}{d t}=8 \pi r \frac{d r}{d t}$
$\Rightarrow \frac{d S}{d t}=8 \pi(5) \times \frac{1}{4 \pi}$
$\Rightarrow \frac{d S}{d t}=10 \ cm^2 / \sec $

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Question 62 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 72 Marks

$\sin ^{-1}\left(\frac{-1}{2}\right)$

Answer

Let $\sin ^{-1}\left(\frac{-1}{2}\right)=y$
$\Rightarrow \sin y=-\frac{1}{2}$
$\Rightarrow \sin y=-\sin \frac{\pi}{6}$
$\Rightarrow \sin y=\sin \left(-\frac{\pi}{6}\right)$
Since, the principal value branch of $\sin ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Therefore, principal value of $\sin ^{-1}\left(\frac{-1}{2}\right)$ is $-\frac{\pi}{6}$.

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