Question

$\begin{array}{cccc}& \text{List -I}& & \text{List -I}\\ \left(P\right)& Letf:R\to R\text{defined us}& \left(1\right)& odd\\ & f\left(x\right)=e^{sgn\left(x\right)}+e^{x^2},\text{where sgn (x) represents}& & \\ & \text{signum function of x, then f(x) is}& & \\ \left(Q\right)& Letf:(-1,1)\to R\text{defined as}& \left(2\right)& Even\\ & f\left(x\right)=x\left[x^4\right]+\frac{1}{\sqrt{1-x^2}},where\left[x\right]denotes& & \\ & \text{greatest integer function, then f(x) is}& & \\ \left(R\right)& Letf:R\to R\text{defined as}& \left(3\right)& \text{Neither even}\\ & f\left(x\right)=\frac{x(x+1)(x^4+1)+2x^4+x^2+2}{x^2+x+1},\text{then f(x) is}& & \text{non odd}\\ \left(S\right)& Letf:R\to R\text{defined as}& \left(4\right)& \text{one-one}\\ & f\left(x\right)=x+3x^3+5x^5+....+101x^{101}\text{then f(x) is}& & \\ & & \left(5\right)& \text{Many -one}\end{array}$

• A. $P\to 2,3,Q\to 3,4,R\to 5,1,S\to 4,2$
• B. $P\to 3,Q\to 4,5,R\to 3,4,S\to 5$
• C. $P\to 1,2,Q\to 2,5,R\to 4,5,S\to 2,4$
• D. $P\to 3,5,Q\to 2,R\to 3,5,S\to 1,4$

Answer 1

The correct option is D $P\to 3,5,Q\to 2,R\to 3,5,S\to 1,4$

(P) $f\left(x\right)=\left\{\begin{array}{cc}e+e^{x^2}& x>0\\ \frac{1}{e}+e^{x^2}& x<0\\ 2& x=0\end{array}\right\}$
So f(x) is many one but neither odd nor even
(Q) $x\in (-1,1)$
$\left[x^4\right]=0$
So $f\left(x\right)=\frac{1}{\sqrt{1-x^2}}$
So f(x) is even and many one.
(R) $f\left(x\right)=\frac{x(x+1)(x^4+1)+2x^4+x^2+2}{x^2+x+1}$
$f\left(x\right)=\frac{(x^2+x+1-1)(x^4+1)+2x^4+x^2+2}{x^2+x+1}$
$f\left(x\right)=\frac{(x^2+x+x)(x^4+1)-x^4-1+2x^4+x^2+2}{x^2+x+1}$
$f\left(x\right)=x^4+1+\left(\frac{x^4+x^2+1}{x^2+x+1}\right)$
$f\left(x\right)=x(x^3+x-1)+2$
f(x) is many one & neither odd nor even.
(S) $f\left(x\right)=x+3x^3+5x^5+---+101x^{101}$
So f(x) is odd function.
$f^{\prime }\left(x\right)>0\forall x\in R$
So f(x) is one - one function