MHT CET · Maths · Functions
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 \(-14\)
- C \(7\)
- D \(14\)
Answer & Solution
Correct Answer
(C) \(7\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & f(x)=x^3-3(a-2) x^2+3 a x+7 \\ & f^{\prime}(x)=3 x^2-6(a-2) x+3 a\end{aligned}\)
\(f^{\prime}(x)\) changes its behaving at \(x=1\)
\(\begin{aligned} & f^{\prime}(1)=0 \\ & \Rightarrow 3 \times 1^2-6(a-2) \times 1+3 a=0 \\ & \Rightarrow a=b \Rightarrow f(x)=x^3-9 x^2+15 x+7 \\ & \Rightarrow \frac{f(x)-14}{(x-1)^2}=0 \Rightarrow \frac{\left(x^3-9 x^2+15 x+7\right)-14}{(x-1)^2}=0 \\ & \Rightarrow x^3-9 x^2+15 x-7=0 \\ & \Rightarrow(x-1)(x-1)(x-7) \\ & \Rightarrow x=1,1,7\end{aligned}\)
\(f^{\prime}(x)\) changes its behaving at \(x=1\)
\(\begin{aligned} & f^{\prime}(1)=0 \\ & \Rightarrow 3 \times 1^2-6(a-2) \times 1+3 a=0 \\ & \Rightarrow a=b \Rightarrow f(x)=x^3-9 x^2+15 x+7 \\ & \Rightarrow \frac{f(x)-14}{(x-1)^2}=0 \Rightarrow \frac{\left(x^3-9 x^2+15 x+7\right)-14}{(x-1)^2}=0 \\ & \Rightarrow x^3-9 x^2+15 x-7=0 \\ & \Rightarrow(x-1)(x-1)(x-7) \\ & \Rightarrow x=1,1,7\end{aligned}\)
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