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JEE Mains · Maths · STD 12 - 6. Application of derivatives

Let ' \(a\) ' be a real number such that the function \(f(x)=a x^{2}+6 x-15, x \in R\) is increasing in \(\left(-\infty, \frac{3}{4}\right)\) and decreasing in \(\left(\frac{3}{4}, \infty\right) .\) Then the function \(g(x)=a x^{2}-6 x+15, x \in R\) has a:

  1. A local minimum at \(x=-\frac{3}{4}\)
  2. B local maximum at \(x=\frac{3}{4}\)
  3. C local minimum at \(\mathrm{x}=\frac{3}{4}\)
  4. D local maximum at \(\mathrm{x}=-\frac{3}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(D) local maximum at \(\mathrm{x}=-\frac{3}{4}\)

Step-by-step Solution

Detailed explanation

\(f(x)=a x^{2}+6 x-15\) \(f^{\prime}=2 a x+6=2 a\left(x+\frac{3}{a}\right)\) \(\Rightarrow-\frac{3}{a}=\frac{3}{4} \Rightarrow a=-4\) Now \(g(x)=-4 x^{2}-6 x+15\) \(g^{\prime}(x) =-8 x-6\) \(\quad\quad=-2\{4 x+3\}\)
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