Question

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

Column-I	Column-II
(A) If $\left( a, b \right)$ 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 $\left( d,e \right)$ then $\left( d+3e \right)$ is equal to	(r) 4
(D) If the graph is concave upward in $\left( p,\infty \right)$ 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

Hint: We find the critical points at $x=c$ by equating the derivative of $f\left( x \right)={{x}^{4}}\left( 12\ln x-7 \right)$ to zero. We use the first derivative test to check whether there is a minimum at $x=c$. We find the point of inflection $\left( p.f\left( p \right) \right)$ by equation the second derivative of $f\left( x \right)={{x}^{4}}\left( 12\ln x-7 \right)$ to zero. We use the fact that $f\left( x \right)$ is concave downward at $x=a$ if ${{f}^{''}}\left( a \right)<0$ and concave upward if ${{f}^{''}}\left( a \right)>0$. We check what is value of second derivative at both side of $x=p$.

Complete step-by-step solution
We know that maxima or minima of a function occurs at critical points $x=c$ where the first derivative ${{f}^{'}}\left( c \right)=0$ or the first derivative does not exist at $x=c$. We are given the function$f\left( x \right)={{x}^{4}}\left( 12\ln x-7 \right)$. Let us differentiate the function with respect to $x$ using product rule of differentiation .We have;
\[\begin{align}
  & {{f}^{'}}\left( x \right)=\dfrac{d}{dx}{{x}^{4}}\left( 12\ln x-7 \right) \\
& \Rightarrow {{f}^{'}}\left( x \right)={{x}^{4}}\dfrac{d}{dx}\left( 12\ln x-7 \right)+\left( 12\ln x-7 \right)\dfrac{d}{dx}{{x}^{4}} \\
 & \Rightarrow {{f}^{'}}\left( x \right)={{x}^{4}}\dfrac{12}{x}+\left( 12\ln x-7 \right)4{{x}^{3}} \\
 & \Rightarrow {{f}^{'}}\left( x \right)=12{{x}^{3}}+48{{x}^{3}}\ln x-28{{x}^{3}} \\
 & \Rightarrow {{f}^{'}}\left( x \right)=16{{x}^{3}}\left( 3\ln x-1 \right) \\
\end{align}\]
Now let us equate the first derivative to find the critical point
\[\begin{align}
  & {{f}^{'}}\left( x \right)=0 \\
 & \Rightarrow 16{{x}^{3}}\left( 3\ln x-1 \right)=0 \\
 & \Rightarrow {{x}^{3}}=0\text{ or }\ln x=\dfrac{1}{3} \\
 & \Rightarrow x=0\text{ or }x={{e}^{\dfrac{1}{3}}} \\
\end{align}\]
So we have two critical points $x=0$ and$x={{e}^{\dfrac{1}{3}}}$. The function $\log x$ is not defined for $x<0$, so we need to check whether at $x={{e}^{\dfrac{1}{3}}}$ there is maxima or minima occurs for $f\left( x \right)$. We use first order derivative test take points $x=1$ on the left of $x={{e}^{\dfrac{1}{3}}}$ and $x=e$ on the right of $x={{e}^{\dfrac{1}{3}}}$ to check whether the functional value. We have;
\[\begin{align}
  & f\left( 1 \right)={{1}^{4}}\left( 12\ln 1-1 \right)=-1<0 \\
 & f\left( e \right)={{1}^{4}}\left( 12\ln e-1 \right)=11>0 \\
\end{align}\]
Since the function changes its value from negative to positive at $x={{e}^{\dfrac{1}{3}}}$ there exists a minima at $x={{e}^{\dfrac{1}{3}}}$.
We know that the point of inflection is a point on the graph of the function where the graph changes from concave up to concave down or vice versa. The point of inflection $\left( p,f\left( p \right) \right)$ is obtained from equating ${{f}^{''}}\left( x \right)$ to zero. Let us find ${{f}^{''}}\left( x \right)$ by differentiating ${{f}^{'}}\left( x \right)$ with respect to $x$. We have;
\[\begin{align}
  & {{f}^{''}}\left( x \right)=\dfrac{d}{dx}{{f}^{'}}\left( x \right) \\
 & \Rightarrow {{f}^{''}}\left( x \right)=\dfrac{d}{dx}16{{x}^{3}}\left( 3\ln x-1 \right) \\
 & \Rightarrow {{f}^{''}}\left( x \right)=16\left( {{x}^{3}}\dfrac{d}{dx}\left( 3\ln x-1 \right)+\left( 3\ln x-1 \right)\dfrac{d}{dx}{{x}^{3}} \right) \\
 & \Rightarrow {{f}^{''}}\left( x \right)=16\left( {{x}^{3}}\times \dfrac{3}{x}+\left( 3\ln x-1 \right)\times 3{{x}^{2}} \right) \\
 & \Rightarrow {{f}^{''}}\left( x \right)=16\left( 3{{x}^{2}}+9{{x}^{2}}\ln x-3{{x}^{2}} \right)=144{{x}^{2}}\ln x \\
 & \\
\end{align}\]
Let us equate ${{f}^{''}}\left( x \right)$ to zero and find the point of inflection.
\[\begin{align}
  & {{f}^{''}}\left( x \right)=0 \\
 & \Rightarrow 144{{x}^{2}}\ln x=0 \\
 & \Rightarrow {{x}^{2}}=0\text{ or }\ln x=0 \\
 & \Rightarrow x=0\text{ or }x=1 \\
\end{align}\]
Since we have already rejected the point of inflection is
\[\left( 1,f\left( 1 \right) \right)=\left( 1,{{1}^{4}}\left( 12\ln 1-7 \right) \right)=\left( 1,1\left( -7 \right) \right)=\left( 1,-7 \right)\]
We know that a graph at $x=a$ is concave downward if ${{f}^{''}}\left( a \right)<0$ and concave upward if${{f}^{''}}\left( a \right)>0$. Since we have $\ln x<0$ when $x\in \left( 0,1 \right)$ then ${{f}^{''}}\left( a \right)=144{{a}^{2}}\ln a<0$ for all $x=a$ in the interval $x\in \left( 0,1 \right)$ . We also have $\ln x > 0$ for all $x\in \left( 1,\infty \right)$ then ${{f}^{''}}\left( a \right)=144{{a}^{2}}\ln a > 0$ for all $x=a$ in the interval$x\in \left( 1,\infty \right)$. So the graph of $f\left( x \right)$ is concave downward in $\left( 0,1 \right)$ and concave upward in $\left( 1,\infty \right)$.\[\]
Now let us check the columns. \[\]
(A) Since the point of inflection is $\left( 1,-7 \right)$, we have $\left( a,b \right)=\left( 1,-7 \right)$ and $a-b=1-\left( -7 \right)=8$. So we have $A\to s$\[\]
(B) Since $x={{e}^{\dfrac{1}{3}}}$ is point of minima we have ${{e}^{t}}={{e}^{\dfrac{1}{3}}}\Rightarrow t=\dfrac{1}{3}\Rightarrow 12t=12\times \dfrac{1}{3}=4$. So we have $B\to r$.\[\]
(C)Since the graph of $f\left( x \right)$ is concave downward in $\left( 0,1 \right)$ we have$\left( d,e \right)=\left( 0,1 \right)\Rightarrow d+3e=0+3\times 1=3$. So we have $C\to p$\[\]
(D)Since the graph of $f\left( x \right)$ is concave upward in $\left( 1,\infty \right)$; we have $\left( 1,\infty \right)=\left( p,\infty \right)\Rightarrow p=1$. So we have $D\to q$. \[\]
Hence the correct choice is C. $A\to s,B\to r,C\to p,D\to q$

Note: We note that we can also find the minima using second derivative test that is ${{f}^{'}}\left( c \right)=0$ and ${{f}^{''}}\left( c \right) > 0$ then there is a minima at $x=c$. We choose the points $x=1,e$ for testing minima with first order because $e>1\Rightarrow {{e}^{\dfrac{1}{3}}}>0$. The function changes its sign at critical points and the second order derivative changes its sign at the point of inflection.