JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

For the differentiable function $f : \mathbb{R} - \{0\} \to \mathbb{R}$ . Let $3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$ , then $\left|f(3) + f'\left(\frac{1}{4}\right)\right|$ is equal to:

Choose the correct answer:

Explanation

Solution

Step 1: $f(x)$ ki value nikalna

Hume equation di gayi hai:

(1) \quad 3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10

Ab $x$ ki jagah $\frac{1}{x}$ rakhne par:

(2) \quad 3f\left(\frac{1}{x}\right) + 2f(x) = x - 10

In dono equations ko solve karne ke liye, equation (1) ko $3$ se aur equation (2) ko $2$ se multiply karte hain:

9f(x) + 6f\left(\frac{1}{x}\right) = \frac{3}{x} - 30

4f(x) + 6f\left(\frac{1}{x}\right) = 2x - 20

Subtract karne par:

5f(x) = \frac{3}{x} - 2x - 10

f(x) = \frac{1}{5} \left( \frac{3}{x} - 2x - 10 \right)

Step 2: $f(3)$ ki value nikalna

f(3) = \frac{1}{5} \left( \frac{3}{3} - 2(3) - 10 \right)

f(3) = \frac{1}{5} (1 - 6 - 10) = \frac{1}{5} (-15) = -3

Step 3: $f'(x)$ nikalna aur $f'\left(\frac{1}{4}\right)$ calculate karna

f(x) = \frac{3}{5x} - \frac{2x}{5} - 2

Differentiate karne par:

f'(x) = -\frac{3}{5x^2} - \frac{2}{5}

Ab $x = \frac{1}{4}$ rakhne par:

f'\left(\frac{1}{4}\right) = -\frac{3}{5(1/4)^2} - \frac{2}{5}

f'\left(\frac{1}{4}\right) = -\frac{3 \times 16}{5} - \frac{2}{5} = -\frac{48}{5} - \frac{2}{5} = -\frac{50}{5} = -10

Step 4: Final value calculate karna

\left| f(3) + f'\left(\frac{1}{4}\right) \right| = | -3 + (-10) |

= | -13 | = 13

Correct Answer

Sahi option (1) 13 hai.