Question

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

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{The least positive integral value of}x\text{satisfying the inequality}& \text{(P)}& 1\\ & {\tan }^{-1}(x^3+5x-2)>{\tan }^{-1}(4-6x+6x^2),\text{is}& & \\ \text{(B)}& \text{If}f\left(x\right)=\frac{a\cos x-\cos bx}{x^2},x\ne 0\text{and}f\left(0\right)=4\text{is continuous at}x=0,\text{then}& \text{(Q)}& 2\\ & |a+b|\text{can be}& & \\ \text{(C)}& \text{Let}f\left(x\right)=\underset{n\to \infty }{lim}\frac{x^n(a+\sin (x^n\left)\right)+(b-\sin (x^n\left)\right)}{(1+x^n)\sec \left({\tan }^{-1}\right(x^n+x^{-n}\left)\right)}\text{be continuous at}x=1.& \text{(R)}& 3\\ & \text{Then}(a+b+1)\text{is}& & \\ \text{(D)}& \text{Let}f\left(x\right)=\underset{t\to 0}{lim}{\sin }^{-1}\left(\frac{e^{xt}-1}{t}\right).\text{Then}\underset{t\to 0}{lim}6\left(\frac{f\left(x\right)-x}{x^3}\right)\text{is}& \text{(S)}& 4\\ & & \text{(T)}& 6\end{array}$

Which of the following is the only CORRECT combination?

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

Answer 1

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

$\left(A\right)$
${\tan }^{-1}(x^3+5x-2)>{\tan }^{-1}(4-6x+6x^2)$
$\Rightarrow x^3+5x-2>4-6x+6x^2\Rightarrow x^3-6x^2+11x-6>0\Rightarrow (x-1)(x-2)(x-3)>0\Rightarrow x\in (1,2)\cup (3,\infty )$
Least positive integral value is $4.$
$\text{(A)}\to \text{(S)}$

$\left(B\right)$
$f\left(x\right)=\frac{a\cos x-\cos bx}{x^2},x\ne 0$
and $f\left(0\right)=4$
As $f\left(0\right)$ is finite, limit should be in $0/0$ form.
Numerator should be $0,$ at $x=0.$
$\Rightarrow a=1$

$\underset{x\to 0}{lim}\frac{-a\sin x+b\sin bx}{2x}=4$
$\Rightarrow \frac{-a}{2}+\frac{b^2}{2}=4$
$\Rightarrow \frac{b^2}{2}=4+\frac{1}{2}$
$\Rightarrow b^2=9$
$\Rightarrow b=\pm 3$
$\therefore |a+b|$ can be $4$ or $2.$
$\text{(B)}\to \text{(Q)},\text{(S)}$