Question

Which of the following options is the only INCORRECT combination?

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

Answer 1

The correct option is D $\left(III\right)\left(i\right)\left(R\right)$

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

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

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

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

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

Since we need to find the incorrect option only option D has (III)

so option D will be our answer