Question

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

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& f\left(x\right)={\sin }^{-1}\left(\frac{\sin x+\cos x}{2}\right)& \text{(P)}& \text{Domain is}R\\ \text{(B)}& g\left(x\right)={\sin }^{-1}\left(\frac{2}{\pi }{\tan }^{-1}x\right)& \text{(Q)}& \text{Range contains only one integer}\\ \text{(C)}& h\left(x\right)={\tan }^{-1}\left(\frac{2}{\pi }(2{\tan }^{-1}x-{\sin }^{-1}x+{\cot }^{-1}x-{\cos }^{-1}x)\right)& \text{(R)}& \text{Odd function}\\ \text{(D)}& j\left(x\right)={\tan }^{-1}(x^3+x)& \text{(S)}& \text{No vertical tangent}\end{array}$

Which of the following is a CORRECT combination?

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

Answer 1

The correct option is C $\text{C}\to \text{(P)},\text{D}\to \text{(S)}$

$h\left(x\right)={\tan }^{-1}\left(\frac{2}{\pi }(2{\tan }^{-1}x-{\sin }^{-1}x+{\cot }^{-1}x-{\cos }^{-1}x)\right)$
$\Rightarrow h\left(x\right)={\tan }^{-1}\left(\frac{2}{\pi }({\tan }^{-1}x+({\tan }^{-1}x+{\cot }^{-1}x)-({\sin }^{-1}x+{\cos }^{-1}x\left)\right)\right)$
$\Rightarrow h\left(x\right)={\tan }^{-1}\left(\frac{2}{\pi }{\tan }^{-1}x\right)$
Domain of $h\left(x\right)$ is $R.$
${\tan }^{-1}x\in (-\frac{\pi }{2},\frac{\pi }{2})\forall x\in R$
$\Rightarrow \frac{2}{\pi }{\tan }^{-1}x\in (-1,1)$
Range of $h\left(x\right)$ is $({\tan }^{-1}\left(-1\right),{\tan }^{-1}\left(1\right))=(-\frac{\pi }{4},\frac{\pi }{4})$
which contains only $1$ integer i.e., $0$

$h(-x)=-h\left(x\right),$ therefore $h\left(x\right)$ is an odd function.

We have, $h\left(x\right)={\tan }^{-1}\left(\frac{2}{\pi }{\tan }^{-1}x\right)$
Differentiating w.r.t. $x,$ we get
$h^{\prime }\left(x\right)=\frac{1}{1+\frac{4{\left({\tan }^{-1}(x+\frac{\pi }{4})\right)}^2}{{\pi }^2}}\left(\frac{2}{\pi (1+x^2)}\right)\ne \pm \infty$
$\therefore h\left(x\right)$ has no vertical tangent.


We have $j\left(x\right)={\tan }^{-1}(x^3+x)$
Domain is $R$ (cubic polynomial)
$\therefore$ Range is $(-\frac{\pi }{2},\frac{\pi }{2})$
Since $j(-x)=-j\left(x\right),$ therefore $j\left(x\right)$ is an odd function.
and $j\left(x\right)$ has no vertical tangent.