Question

$f\left(x\right)$ is cubic polynomial which has local maximum at $x=1,f\left(b\right)=18,f\left(a\right)=1$ and $f\text{'}\left(x\right)$ has local maximum at $x=1$. If $f\left(b\right)=18,f\left(a\right)=1$ and $f\text{'}\left(x\right)$ has local minimum at $x=0$, then

• A. The distance between $(-1,2)$and $(a,f(a\left)\right)$, where $x=0$ is the point of local minima is $2\sqrt{5}$
• B. $f\left(x\right)$ is increasing for $x\in [1,2\sqrt{5}]$
• C. $f\left(x\right)$ has local minima at $x=1$
• D. The value of $f\left(0\right)=5$
• E. B and C

Answer 1

The correct option is E

B and C

Explanation for the correct option:

Step 1. Evaluating the given conditions:

$\because f\left(x\right)$ has local maxima at $x=1$ and $f\text{'}\left(x\right)$ has local minima at $x=0$

$\therefore f\text{'}\text{'}\left(x\right)=\lambda x$

$\Rightarrow$ $f\text{'}\left(x\right)=\lambda \frac{x^2}{2}+c$

As $f\text{'}\left(1\right)=0$

$\Rightarrow$$\frac{\lambda }{2}+c=0$

$\Rightarrow$ $\lambda =-2c$ ...(1)

Step 2. integrating $f\text{'}\left(x\right)$ both sides, we get

$f\left(x\right)=\lambda \frac{x^3}{6}+cx+d$ ...(2)

$\Rightarrow$$\begin{array}{rcl}f\left(b\right)& =& \lambda \left(\frac{b^3}{6}\right)+bc+d\\ & =& 18\end{array}$ …(3)

$\Rightarrow$$\begin{array}{rcl}f\left(a\right)& =& \lambda \left(\frac{a^3}{6}\right)+ac+d\\ & =& 1\end{array}$ ...(4)

Step 3. By Using equation (1),( 2), (3) and (4), we get

$f\left(x\right)=\frac{1}{4}(19x^3-57x+34)$

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{1}{4}(57x^2-57)\\ & =& \frac{57}{4}(x-1)(x+1)\end{array}$

using number line rule

$\therefore f\left(x\right)$ is increasing for $[1,2\sqrt{5}]$ and $f\left(x\right)$ has local minimum at $x=1$

Hence, Option ‘E’ is Correct.