Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}f\left(x\right)=\underset{0}{\overset{g\left(x\right)}{\int }}\frac{dt}{\sqrt{1+t^3}}\text{where}g\left(x\right)=\underset{0}{\overset{\cos x}{\int }}(1+\sin t^2)dt,\text{then the value of}{f}^{\prime }\left(\pi /2\right)\text{is equal to}& \text{(P)}& 3\\ \text{(B)}& \text{If}f\left(x\right)\text{is a non-zero differentiable function such that}\underset{0}{\overset{x}{\int }}f\left(t\right)dt=\left(f\right(x){)}^2\text{for all}x,\text{then}f(2)\text{equals}& \text{(Q)}& 2\\ \text{(C)}& \text{If}\underset{a}{\overset{b}{\int }}(2+x-x^2)dx,(a<b)\text{is maximum, then the value of}(a+b)\text{is equal to}& \text{(R)}& 1\\ \text{(D)}& \text{If}\underset{x\to 0}{lim}(\frac{\sin 2x}{x^3}+a+\frac{b}{x^2})=0,\text{then the value of}(3a+b)\text{is equal to}& \text{(S)}& -1\end{array}$

Which of the following is a CORRECT combination?

• A. $\text{C}\to \text{(P)},\text{D}\to \text{(R)}$
• B. $\text{C}\to \text{(R)},\text{D}\to \text{(Q)}$
• C. $\text{C}\to \text{(Q)},\text{D}\to \text{(R)}$
• D. $\text{C}\to \text{(R)},\text{D}\to \text{(R)}$

Answer 1

The correct option is B $\text{C}\to \text{(R)},\text{D}\to \text{(Q)}$

Let $f\left(x\right)=\int (2+x-x^2)dx$
$f^{\prime }\left(x\right)=2+x-x^2=(2-x)(x+1)$

$f\left(x\right)$ is increasing in interval $[-1,2]$
So, $\underset{a}{\overset{b}{\int }}(2+x-x^2)dx$ is maximum in interval $[-1,2]$
Hence, $a=-1,b=2$
$\Rightarrow a+b=-1+2=1$


We have, $\underset{x\to 0}{lim}\frac{\sin 2x+a{x}^3+bx}{x^3}=0$
$\Rightarrow \underset{x\to 0}{lim}\frac{a{x}^3+bx+2x-\frac{8x^3}{6}+\cdots }{x^3}=0$
$\Rightarrow \underset{x\to 0}{lim}\frac{x(b+2)+x^3(a-\frac{4}{3})+\cdots }{x^3}=0$
For existence of limit,
$b+2=0\Rightarrow b=-2$
Now, $\underset{x\to 0}{lim}\frac{x^3(a-\frac{4}{3})+\cdots }{x^3}=0$
$\Rightarrow a=\frac{4}{3}$
Hence, $3a+b=4-2=2$