Q79 / 150 · MHT CET 2022 (06 Aug Shift 2) · Mathematics

If the function $f(x)=x^3-3(a-2) x^2+3 a x+7$, for some $a \in R$ is increasing in $(0,1]$ and decreasing in $[0,5)$, then a root of the equation $\frac{f(x)-14}{(x-1)^2}=0(x \neq 1)$ is

  • A
    $-7$
  • B
    $-14$
  • Correct answer
    C
    $7$
  • D
    $14$

Solution

$\begin{aligned} & f(x)=x^3-3(a-2) x^2+3 a x+7 \\ & f^{\prime}(x)=3 x^2-6(a-2) x+3 a\end{aligned}$ $f^{\prime}(x)$ changes its behaving at $x=1$ $\begin{aligned} & f^{\prime}(1)=0 \\ & \Rightarrow 3 \times 1^2-6(a-2) \times 1+3 a=0 \\ & \Rightarrow a=b \Rightarrow f(x)=x^3-9 x^2+15 x+7 \\ & \Rightarrow \frac{f(x)-14}{(x-1)^2}=0 \Rightarrow \frac{\left(x^3-9 x^2+15 x+7\right)-14}{(x-1)^2}=0 \\ & \Rightarrow x^3-9 x^2+15 x-7=0 \\ & \Rightarrow(x-1)(x-1)(x-7) \\ & \Rightarrow x=1,1,7\end{aligned}$

Hence, the correct answer is option: 'C'.

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