Question

Column I		Column II
A	Let $f:R\to R$ defined as $f(x)=e^{\mathrm{sgn}x}+e^{x^2}$, where $\mathrm{sgn}x$ denotes signum function of $x$, then $f(x)$ is	P	Odd
B	Let $f:(-1,1)\to R$ defined as$f(x)=x\left[x^4\right]+\frac{1}{\sqrt{1-x^2}}$where $[x]$ denotes greatest integer less than or equal to $x$, then $f(x)$ is	Q	Even
C	Let $f:R\to R$ defined as$f(x)=\frac{x(x+1)\left(x^4+1\right)+2x^4+x^2+2}{x^2+x+1}$then $f(x)$ is	R	Neither odd nor even
D	Let $f:R\to R$ defined as$f(x)=x+3x^3+5x^5+\dots \dots .+101x^{101}$then $f(x)$ is	S	One-One
		T	Many-One

• A. (A) R, T ; (B) Q, T ; (C) R, T ; (D) P, P
• B. (A) R, T ; (B) Q, T ; (C) R, T ; (D) P, Q
• C. (A) R, T ; (B) Q, T ; (C) R, T ; (D) P, R
• D. (A) R, T ; (B) Q, T ; (C) R, T ; (D) P, S

Answer 1

Answer: (D)

(A)$f(x)=e^{\mathrm{sgn}x}+e^{x^2}$
when $x=0f(0)=2$
when $x>0f(x)=e+e^{x^2}$
when $x<0f(x)=\frac{1}{e}+e^{x^2}$
Hence, $f(x)$ is many-one and neither odd nor even.
(B) $D_f:1-x^2>0$
$x^2-1<0\Rightarrow x\in (-1,1)$
$\left[x^4\right]\text{ when }x\in (-1,1)\text{ is equal to }0.$
$\therefore f(x)=\frac{1}{\sqrt{1-x^2}}$
Hence, $f(x)$ is even and many-one.
(C) $f(x)=x^4+1+\left(\frac{x^4+x^2+1}{x^2+x+1}\right)=x^4+1+x^2-x+1=x^4+x^2-x+2=x\left(x^3+x-1\right)+2$
Hence, $f(x)$ is many-one and neither odd nor even.
(D) $f(x)=x+3x^3+5x^5+\dots \dots .+101x^{101}$
Clearly $f(x)$ is an odd function.
Now $f^{\prime }(x)=1^2+3^2{x}^2+5^2{x}^4+$ $+(101{)}^2{x}^{100}$.
$\Rightarrow f^{\prime }(x)>0\forall x\in R$
Hence, $f(x)$ is one-one function.