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

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Continuity and Differentiability

Let k and m be positive real numbers such that the function

f (x) = {3 x^{2} + k x + 1, m x^{2} + k^{2}, 0 < x < 1 x \geq 1

is differentiable for all

x > 0

. Then

\frac{8 f ^{'} ( 8 )}{f ^{'} ( \frac{1}{8} )}

is equal to ______.

Answer: 309

Continuity at

x = 1

requires

3 + k 2 = m + k^{2} . (1)

The derivative is

f^{'} (x) = 6 x + \frac{k}{2 x + 1}

0 < x < 1

and

f^{'} (x) = 2 m x

x > 1

; matching at

x = 1

6 + \frac{k}{2 2} = 2 m . (2)

From

(2)

m = 3 + \frac{k}{4 2}

. Substituting into

(1)

3 + k 2 = 3 + \frac{k}{4 2} + k^{2} \Rightarrow k^{2} = k (2 - \frac{1}{4 2}) = k \cdot \frac{7}{4 2},

k = \frac{7}{4 2}

and

m = 3 + \frac{7}{32} = \frac{103}{32}

. Then

f^{'} (8) = 16 m = \frac{103}{2}

, and since

\frac{1}{8} < 1

f^{'} (\frac{1}{8}) = \frac{6}{8} + \frac{k}{2 9/8} = \frac{3}{4} + \frac{k 2}{3} = \frac{3}{4} + \frac{7}{12} = \frac{4}{3} .

Therefore

\frac{8 f ^{'} ( 8 )}{f ^{'} ( 1/8 )} = \frac{8 \cdot \frac{103}{2}}{\frac{4}{3}} = \frac{412 \cdot 3}{4} = 309.

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Integer answer