JEE Mains · Maths · STD 12 - 6. Application of derivatives
If the function \(f\) given by \(f\,(x)\, = \,{x^3} - 3(a - 2){x^2} + 3ax\, + 7,\) for some \(a\in R\) is increasing in \((0, 1]\) and decreasing in \([1, 5),\) then a root of the equation, \(\frac{{f(x) - 14}}{{{{(x - 1)}^2}}} = 0\,(x\, \ne 1)\) is
- A \(-7\)
- B \(5\)
- C \(7\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(7\)
Step-by-step Solution
Detailed explanation
\(f^{\prime}(x)=3 x^{2}-6(a-2) x+3 a\) \(f^{\prime}(x) \geq 0 \forall x \in(0,1]\) \(f^{\prime}(x) \leq 0 \forall x \in[1,5)\) \(\Rightarrow f^{\prime}(x)=0\) at \(x=1 \Rightarrow a=5\) \(f(x)-14=(x-1)^{2}(x-7)\) \(\frac{f(x)-14}{(x-1)^{2}}=x-7\) Hence root of equation…
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