Let $k$ and $m$ be positive real numbers such that the function $f(x) = \begin{cases} 3x^2 + k\sqrt{x + 1}, & 0 0$. Then $\frac{8f'(8)}{f'\left(\frac{1}{8}\right)}$ is equal to ___________.

JEE Main 2023 · MathsNumerical answerNumerical valueMediumMulti-step

JEE Main 8 April 2023, Shift 2, Maths Q25

Question 25 of 90 in this shift, Maths question 25 of 30, Section B.

Let

k

and

m

be positive real numbers such that the function

f(x) = \begin{cases} 3x^2 + k\sqrt{x + 1}, & 0 < x < 1 \\ mx^2 + k^2, & x \geq 1 \end{cases}

is differentiable for all

x > 0

. Then

\frac{8f'(8)}{f'\left(\frac{1}{8}\right)}

is equal to ___________.

Official answer

309

NTA final key.