Question

If $f\left(x\right)$ is twice differentiable function in $[c_1-1,c_2+1]$ and $f^{\prime }\left(c_1\right)=f^{\prime }\left(c_2\right)=0,f^{\prime \prime }\left(c_1\right)\cdot f^{\prime \prime }\left(c_2\right)<0,f\left(c_1\right)=9,f\left(c_2\right)=0$. Let $k$ and $m$ be the minimum number of the roots of $f\left(x\right)=0$ and $f^{\prime }\left(x\right)=0$ respectively,in $[c_1-1,c_2+1]$

$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(I) If}{f}^{\prime \prime }\left(c_1\right)-f^{\prime \prime }\left(c_2\right)>0,\text{then k =}& \text{(P) 1}\\ \text{(II) If}{f}^{\prime \prime }\left(c_1\right)-f^{\prime \prime }\left(c_2\right)<0,\text{then k =}& \text{(Q) 2}\\ \text{(III) If}{f}^{\prime \prime }\left(c_1\right)-f^{\prime \prime }\left(c_2\right)>0,\text{then m =}& \text{(R) 3}\\ \text{(IV) If}{f}^{\prime \prime }\left(c_1\right)-f^{\prime \prime }\left(c_2\right)<0,\text{then m =}& \text{(S) 4}\end{array}$

Which of the following is only CORRECT combination?

• A. $\left(I\right)\to \left(P\right)$
• B. $\left(II\right)\to \left(P\right)$
• C. $\left(III\right)\to \left(R\right)$
• D. $\left(IV\right)\to \left(S\right)$

Answer 1

The correct option is B $\left(II\right)\to \left(P\right)$

For $f^{\prime \prime }\left(c_1\right)-f^{\prime \prime }\left(c_2\right)>0$
$k=2,m=4$

For $f^{\prime \prime }\left(c_1\right)-f^{\prime \prime }\left(c_2\right)<0$
$k=1m=2$