JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 12)
Let 'a' be a real number such that the function f(x) = ax2 + 6x $$-$$ 15, x $$\in$$ R is increasing in $$\left( { - \infty ,{3 \over 4}} \right)$$ and decreasing in $$\left( {{3 \over 4},\infty } \right)$$. Then the function g(x) = ax2 $$-$$ 6x + 15, x$$\in$$R has a :
local maximum at x = $$-$$ $${{3 \over 4}}$$
local minimum at x = $$-$$$${{3 \over 4}}$$
local maximum at x = $${{3 \over 4}}$$
local minimum at x = $${{3 \over 4}}$$
Explanation
$${{ - B} \over {2A}} = {3 \over 4}$$
$$ \Rightarrow {{ - (6)} \over {2a}} = {3 \over 4}$$
$$ \Rightarrow a = {{ - 6 \times 4} \over 6} \Rightarrow a = - 4$$
$$\therefore$$ $$g(x) = 4{x^2} - 6x + 15$$
Local max. at $$x = {{ - B} \over {2A}} = - {{( - 6)} \over {2( - 4)}}$$
$$ = {{ - 3} \over 4}$$
$$ \Rightarrow {{ - (6)} \over {2a}} = {3 \over 4}$$
$$ \Rightarrow a = {{ - 6 \times 4} \over 6} \Rightarrow a = - 4$$
$$\therefore$$ $$g(x) = 4{x^2} - 6x + 15$$
Local max. at $$x = {{ - B} \over {2A}} = - {{( - 6)} \over {2( - 4)}}$$
$$ = {{ - 3} \over 4}$$
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