Question

Which of the following options is the only CORRECT combination?

• A. $\left(II\right)\left(iii\right)\left(S\right)$
• B. $\left(I\right)\left(ii\right)\left(R\right)$
• C. $\left(III\right)\left(iv\right)\left(P\right)$
• D. $\left(IV\right)\left(i\right)\left(S\right)$

Answer 1

Answer: (A)

$\left(II\right)\left(iii\right)\left(S\right)$

In column 1

$f\left(1\right)=1+0-\left(1\right)\left(0\right)=1$

$f\left(e^2\right)=e^2+2-\left(2\right)e^2=2-e^2<0$

Since $f\left(x\right)$ is a continuous function $f\left(x\right)$ will be zero for some $x$ where $1<x<e^2$

Therefore (I) is correct

$f^{\prime }\left(x\right)=\frac{1}{x}-\ln x$

$f^{\prime }\left(1\right)=1$

$f^{\prime }\left(e\right)=\frac{1}{e}-1<0$

Since $f^{\prime }\left(x\right)$ is a continuous function $f^{\prime }\left(x\right)$ will be zero for some $x$ where $1<x<e$

Therefore (II) is correct

$f^{\prime }\left(0\right)\to \infty -(-\infty )\to \infty +\infty >0$

$f^{\prime }\left(1\right)=1$

Since $f^{\prime }\left(x\right)$ is a continuous function $f^{\prime }\left(x\right)$ will not be zero for some $x$ where $0<x<1$

Therefore (III) is false

$f^{\prime \prime }\left(x\right)=\frac{-1}{x^2}-\frac{1}{x}$

Therefore $f^{\prime \prime }\left(x\right)<0$ for $x>1$

Therefore (IV) is false

Therefore option C & D can be eliminated

In column 3

$f^{\prime }\left(x\right)>0$ for $0<x<1$

Therefore $f\left(x\right)$ will be increasing in $(0,1)$

Therefore (P) is correct

$f^{\prime }\left(x\right)<0$ for $e<x<e^2$

Therefore $f\left(x\right)$ will be decreasing in $(e,e^2)$

Therefore (Q) is correct

$f^{\prime \prime }\left(x\right)=\frac{-1}{x^2}-\frac{1}{x}$

$f^{\prime \prime }\left(x\right)<0forx>0$

Therefore $f^{\prime }\left(x\right)$ is decreasing for $x>0$

Therefore S is correct and R is false

Therefore option B can be eliminated

In column 2

$\underset{x\to \infty }{lim}{f}^{\prime }\left(x\right)=\underset{x\to \infty }{lim}(\frac{1}{x}-lnx)=0-\infty =-\infty$

Therefore (iii) is true

Therefore answer is option A