Question

For the functions $f\left(x\right)=x^4(12\ln x-7)$ match the following:

	Column-I		Column-II
(A)	If $(a,b)$ is the point of inflection then $a-b$ is equal to	(p)	$3$
(B)	If $e^t$ is a point of minima then $12t$ is equal to	(q)	$1$
(C)	If graph is concave downward in $(d,e)$ then $d+3e$ is equal to	(r)	$4$
(D)	If graph is concave upward in $(p,\infty )$ then the least value of $p$ is equal to	(s)	$8$

• A. A $\to$ p, B $\to$ r, C $\to$ q, D $\to$ s
  
• B. A $\to$ q, B $\to$ p, C $\to$ r, D $\to$ s
  
• C. A $\to$ s, B $\to$ r, C $\to$ p, D $\to$ q
• D. A $\to$ r, B $\to$ p, C $\to$ q, D $\to$ s
  

Answer 1

The correct option is C A $\to$ s, B $\to$ r, C $\to$ p, D $\to$ q

$f\left(x\right)=x^4(12\ln x-7)$

$f^{\prime }\left(x\right)=4x^3(12\ln x-7)+x^4\left(\frac{12}{x}\right)$

$f^{\prime }\left(x\right)=4x^3\left(12\ln x\right)-28x^3+12x^3$

$\Rightarrow f^{\prime }\left(x\right)=48x^3\ln x-16x^3$

$\Rightarrow f^{\prime }\left(x\right)=x^3(48\ln x-16)=16x^3(3\ln x-1)$

Now $f^{\prime }\left(x\right)=0$

$\Rightarrow 16x^3(3\ln x-1)=0$

$\Rightarrow x=0$ or $\ln x=\frac{1}{3}\Rightarrow x=e^{\frac{1}{3}}$

$f^{\prime \prime }\left(x\right)=48x^2(3\ln x-1)+16x^3\left(\frac{3}{x}\right)$

$=48x^2(3\ln x-1)+48x^2$

$\Rightarrow f^{\prime \prime }\left(x\right)=144x^2\ln x$

$x=1,f\left(x\right)=-7$

$(1,-7)=$point of inflection

$\Rightarrow a-b=8,e^t=e^{\frac{1}{3}}$

$\Rightarrow 12t=12\left(\frac{1}{3}\right)=4$

$(d,e)=(0,1)\Rightarrow 3e+d=3$

$(p,\infty )=(1,\infty )\Rightarrow p=1$