Let 'a' be a real number such that the function{f}({x})={ax}^{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)=a x^{2}-6 x+15, x \\in{R} has a: [20 Jul 2021 Shift 1]

Correct Answer is : local maximum at x=−34

local maximum at {x}=-\frac{3}{4}

local minimum at x=-\frac{3}{4}

local maximum at {x}=\frac{3}{4}

local minimum at {x}=\frac{3}{4}

JEE Main 20 July 2021 Shift 1 Solved Paper

Section: Mathematics

Question : 76 of 90

Marks: +1, -0

Let 'a' be a real number such that the functionf(x)=ax2+6x−15,x∈R is increasing in(−∞,

) and decreasing in (

,∞). Then the function g(x)=ax2−6x+15,x∈R has a:

[20 Jul 2021 Shift 1]

Go to Question:

Prev Question

Next Question