Let a be a real number such that the function f ( x ) = a x ^ 2 + 6x - 15, x ∈ R is increasing in (- ∞ , (3)/(4)) and decreasing in ((3)/(4), ∞) . Then the function g ( x ) = a x ^ 2 - 6x + 15, x ∈ R has a

Correct answer: A. local maximum at x = - (3)/(4) We have, f(x) = a x ^ 2 + 6x - 15 Graph of f(x) is shown below as it is increasing in (- ∞ , (3)/(4)) and decreasing in ((3)/(4), ∞) . Then, abscissa of vertex is (3)/(4) . (- B)/(2A) = (3)/(4) ⇒ (- ( 6 ))/(2a) = (3)/(4) ⇒ a = - 4 ∴ g ( x ) = - 4 x ^ 2 - 6x + 15 Local maxima exists at the vertex of g(x), whose abscissa is x = (- B)/(2A) ⇒ x = - ( (- 6) - 8 ) = - (3)/(4)

JEE Main 202120 JulyMorning Shift

Exam: JEE Main
Year: 2021
Date: 20 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2021 · 20 July · Morning Shift · Mathematics · Application of Derivatives

Let a be a real number such that the function f ( x ) = a x ^ 2 +...

Let

a

be a real number such that the function

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

is increasing in \left(- \infty , \dfrac{3}{4}\right)

and decreasing in \left(\dfrac{3}{4}, \infty\right)

. Then the function

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

has a

Let a be a real number such that the function f ( x ) = a x ^ 2 + 6x - 15, x ∈ R is increasing in (- ∞ , (3)/(4)) and

Options

Alocal maximum at $x = - \dfrac{3}{4}$
Blocal minimum at $x = - \dfrac{3}{4}$
Clocal maximum at $x = \dfrac{3}{4}$
Dlocal minimum at $x = \dfrac{3}{4}$

Correct answer

A. local maximum at $x = - \dfrac{3}{4}$

Step-by-step solution

We have,

$f\left(x\right) = a{x}^{2} + 6x - 15$

Graph of $f\left(x\right)$ is shown below as it is increasing in $\left(- \infty , \dfrac{3}{4}\right)$ and decreasing in \left(\dfrac{3}{4}, \infty\right) $.<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/a1ad4593-f1e8-49d9-a19d-9941e3bb8310-image.png" style="width: 328px; height: 250px;" />Then, abscissa of vertex is$ \dfrac{3}{4} $.$ \dfrac{- B}{2A} = \dfrac{3}{4} $$ ⇒ \dfrac{- ( 6 )}{2a} = \dfrac{3}{4} $$ ⇒ a = - 4 $$ ∴ g ( x ) = - 4{x}^{2} - 6x + 15 $Local maxima exists at the vertex of$ g\left(x\right), whose abscissa is

$x =$ \dfrac{- B}{2A}

$⇒ x = - \left(\dfrac{\left(- 6\right)}$ {- 8}\right) = - $\dfrac{3}{4}$

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