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  • Let 'a' be a real number such that the function <img alt="\mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+6 \mathrm{x}-15, \mathrm{x} \in \mathbf{R}" src="/latex-image/?%5Cmathrm%7Bf%7D%28%5Cmathrm%7Bx%7D%29%3D%5Cmathrm%7Bax%7D%5E%7B2%7D&plus;6%20%5Cma

Let 'a' be a real number such that the function \mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+6 \mathrm{x}-15, \mathrm{x} \in \mathbf{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 \mathbf{R} has a
 
Option: 1 \text {local maximum at}\; x=-\frac{3}{4}
Option: 2 \text {local minimum at }\mathrm{x}=-\frac{3}{4}
Option: 3 \text {local maximum at} \; x=\frac{3}{4}
Option: 4 \text {local minimum at} x=\frac{3}{4}

Answers (1)

best_answer

\begin{aligned} &f(x)=a x^{2}+6 x-15 . \\ &f^{\prime}(x)=2 a x+6 . \end{aligned}

f(x) increasing in \left(-\infty, \frac{3}{4}\right) & decreasing in \left(\frac{3}{4}, \infty\right)

f^{\prime}(x)>0 \text { in }\left(-\infty, \frac{3}{4}\right) \& f^{\prime}(x)<0 \text { in }\left(\frac{3}{4}, \infty\right)

As f'(x) is linear, so f^{\prime}(x)=0 \text { at } x=\frac{3}{4}

\begin{aligned} &\Rightarrow \quad 2 \cdot a \cdot\left(\frac{3}{4}\right)+6=0 \\ &\Rightarrow \quad a=-4 . \end{aligned}

Now, g(x)=-4 x^{2}-6 x+15

It is a downward parabola, so it will have maximum at x=\frac{-b}{2 a} \text { (vertex) }

So maximum at x=\frac{-(-6)}{2 \cdot(-4)}=-\frac{3}{4}

Hence, the correct answer is option (1)

Posted by

Kuldeep Maurya

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