Question

Match the following items of column I with column II
$\begin{array}{cccc}& \text{Column I}& & \text{column II}\\ \left(a\right)& \text{If f (x)}={\int }_0^{g\left(x\right)}\frac{dt}{\sqrt{1+t^3}}\text{where}g\left(x\right)=& \left(p\right)& 2\\ & {\int }_0^{\cos x}(1+\sin t^2)dt,\text{then the value of}f\left(\frac{\pi }{2}\right)\text{is}& & \\ \left(b\right)& \text{If}{\int }_a^b(2+x-x^2)dx\text{is maximum, then (a + b)}\text{is}& \left(q\right)& -2\\ & \text{equal to}& & \\ \left(c\right)& \text{If}{lim}_{x\to 0}(\frac{\sin 2x}{x^3}+a+\frac{b}{x^2})=0,\text{then (3a + b)}\text{is equal}& \left(r\right)& 1\\ & \text{to}& & \\ \left(d\right)& \text{Number of points of maxima for}f\left(x\right)=\frac{x}{1+x\tan x}\text{in}& \left(s\right)& -1\\ & xϵ(0,\frac{\pi }{2})& & \end{array}$

• A. (a) – (r); (b) – (s); (c) – (p); (d) – (q)
• B. (a) – (s); (b) – (r); (c) – (p); (d) – (r)
• C. (a) – (s); (b) – (q); (c) – (p); (d) – (r)
• D. (a) – (r); (b) – (p); (c) – (q); (d) – (s)

Answer 1

The correct option is B (a) – (s); (b) – (r); (c) – (p); (d) – (r)

(A) $f^{\prime }\left(x\right)=\frac{g^{\prime }\left(x\right)}{\sqrt{1+\left(g\right(x){)}^3}}$ and $g^{\prime }\left(x\right)=[1+\sin ({\cos }^{2}x\left)\right](-\sin x)$

Hence, $f^{\prime }\left(x\right)=\frac{[1+\sin ({\cos }^{2}x\left)\right(-\sin x\left)\right]}{\sqrt{1+\left(g\right(x){)}^3}}$

$f^{\prime }\left(\frac{\pi }{2}\right)=\frac{(1+0)(-1)}{\sqrt{1+{\left(g\left(\frac{\pi }{2}\right)\right)}^3}}=\frac{-1}{1+0}=-1\text{as}g\left(\frac{\pi }{2}\right)=0$

$\therefore f^{\prime }\left(\frac{\pi }{2}\right)=-1$

(B) $f\left(x\right)=2+x-x^2$
Area under the graph is maximum when $a=-1$ and $b=2$

$\Rightarrow a+b=1$

(C) Given, $\underset{x\to 0}{lim}(\frac{\sin 2x}{x^3}+a+\frac{b}{x^2})=0$

$\Rightarrow \underset{x\to 0}{lim}\frac{\sin 2x+a{x}^3+bx}{x^3}=0$

For limit to exist, $2+b=0\Rightarrow b=-2$

$\therefore \underset{x\to 0}{lim}\frac{\sin 2x+a{x}^3-2x}{x^3}=0$

Applying L'Hospital Rule,
$\underset{x\to 0}{lim}\frac{2\cos 2x+3a{x}^2-2}{3x^2}=0$

$\Rightarrow \underset{x\to 0}{lim}\frac{-4\sin 2x+6ax}{6x}=0$

$\Rightarrow \underset{x\to 0}{lim}\frac{-8\cos 2x+6a}{6}=0$

$\Rightarrow -8+6a=0$

$\Rightarrow a=\frac{4}{3}$

$\Rightarrow 3a+b=3\times \frac{4}{3}-2=2$

(D) $f\left(x\right)=\frac{x}{1+x\tan x}$

$f^{\prime }\left(x\right)=\frac{1+x\tan x-x(\tan x+x{\sec }^{2}x)}{(1+x\tan x{)}^2}$

$=\frac{1+x\tan x-x\tan x-x^2{\sec }^x}{(1+x\tan x{)}^2}=\frac{1-x^2{\sec }^{2}x}{(1+x\tan x{)}^2}$

To find maxima, $f^{\prime }\left(x\right)=0$

$\Rightarrow x^2{\sec }^{2}x=1$

$\Rightarrow x=\pm \cos x$

$f\left(x\right)$ has maxima at $x=\cos x$

Hence, number of points of maxima is 1.