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 $\mathrm{g}(\mathrm{x})=\mathrm{ax}^{2}-6 \mathrm{x}+15, x \in \mathrm{R}$ has a :

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

(A) local minimum at $x=\frac{3}{4}$

(B) local maximum at $x=\frac{3}{4}$

(C) local minimum at $x=-\frac{3}{4}$

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · 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

(- \infty, \frac{3}{4})

and decreasing in

(\frac{3}{4}, \infty)

. Then the function

g (x) = ax^{2} - 6 x + 15, x \in R

has a :

Answer: (D)

A quadratic rises to the left of its vertex and falls to the right of it only when its leading coefficient is negative, so

a < 0

and the vertex of

f

sits at

x = \frac{3}{4}

f^{'} (x) = 2 a x + 6 = 0 \Rightarrow x = - \frac{3}{a} = \frac{3}{4} \Rightarrow a = - 4

With

a = - 4

the second function is

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

, so

g^{'} (x) = - 8 x - 6 = 0 \Rightarrow x = - \frac{3}{4}

and

g^{''} (x) = - 8 < 0

for every

x

. The single stationary point of

g

is therefore a local maximum at

x = - \frac{3}{4}

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option