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JEE Mathematics 2023 Question with Solution

For the differentiable function f:R−{0}→Rf: \mathbb{R} - \{0\} \rightarrow \mathbb{R}, let 3f(x) + 2f\left(\frac{1{x\right) = \frac{1{x - 10, then ∣f(3)+f(14)∣\left| f(3) + f\left(\frac{1}{4}\right) \right| is equal to:

  • A

    1313

  • B

    295\frac{29}{5}

  • C

    335\frac{33}{5}

  • D

    77

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

3f(x)+2f(1x)=1x−103f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10

Find: ∣f(3)+f′(14)∣\left|f(3) + f'\left(\frac{1}{4}\right)\right|

Replace xx by 1x\frac{1}{x} to get a second equation:

3f(1x)+2f(x)=x−103f\left(\frac{1}{x}\right) + 2f(x) = x - 10

Now solve the two linear equations in f(x)f(x) and f(1x)f\left(\frac{1}{x}\right).

Multiply the first equation by 33 and the second by 22:

9f(x)+6f(1x)=3x−309f(x) + 6f\left(\frac{1}{x}\right) = \frac{3}{x} - 30 4f(x)+6f(1x)=2x−204f(x) + 6f\left(\frac{1}{x}\right) = 2x - 20

Subtracting,

5f(x)=3x−2x−105f(x) = \frac{3}{x} - 2x - 10

So,

f(x)=15(3x−2x−10)f(x) = \frac{1}{5}\left(\frac{3}{x} - 2x - 10\right)

Evaluate f(3)f(3):

f(3)=15(33−2⋅3−10)=15(1−6−10)=−3f(3) = \frac{1}{5}\left(\frac{3}{3} - 2\cdot 3 - 10\right) = \frac{1}{5}(1 - 6 - 10) = -3

Differentiate:

f′(x)=15(−3x2−2)=−35x2−25f'(x) = \frac{1}{5}\left(-\frac{3}{x^2} - 2\right) = -\frac{3}{5x^2} - \frac{2}{5}

Now substitute x=14x = \frac{1}{4}:

f′(14)=−35(14)2−25=−10f'\left(\frac{1}{4}\right) = -\frac{3}{5\left(\frac{1}{4}\right)^2} - \frac{2}{5} = -10

Therefore,

f(3)+f′(14)=−3+(−10)=−13f(3) + f'\left(\frac{1}{4}\right) = -3 + (-10) = -13

Hence,

∣f(3)+f′(14)∣=13\left|f(3) + f'\left(\frac{1}{4}\right)\right| = 13

So the correct option is A.

Note: The source question text appears corrupted and shows f(14)f\left(\frac{1}{4}\right), but the solution working consistently uses f′(14)f'\left(\frac{1}{4}\right). The answer follows the solution.

Solving the functional system carefully

The key idea is to create two equations by using both xx and 1x\frac{1}{x}. Since the domain is R−{0}\mathbb{R} - \{0\}, this substitution is valid.

From

3f(x)+2f(1x)=1x−103f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10

and

3f(1x)+2f(x)=x−103f\left(\frac{1}{x}\right) + 2f(x) = x - 10

we eliminate f(1x)f\left(\frac{1}{x}\right) by matching its coefficient as 66.

That gives:

9f(x)+6f(1x)=3x−304f(x)+6f(1x)=2x−20\begin{aligned} 9f(x) + 6f\left(\frac{1}{x}\right) &= \frac{3}{x} - 30 \\ 4f(x) + 6f\left(\frac{1}{x}\right) &= 2x - 20 \end{aligned}

Subtracting,

5f(x)=3x−2x−105f(x) = \frac{3}{x} - 2x - 10

so

f(x)=15(3x−2x−10)f(x) = \frac{1}{5}\left(\frac{3}{x} - 2x - 10\right)

Now compute the required values exactly as shown:

f(3)=−3f(3) = -3 f′(14)=−10f'\left(\frac{1}{4}\right) = -10

Therefore the absolute value becomes

∣−3+(−10)∣=13|-3 + (-10)| = 13

Thus the final answer is 1313.

Common mistakes

  • Using the corrupted question text blindly and evaluating f(14)f\left(\frac{1}{4}\right) instead of f′(14)f'\left(\frac{1}{4}\right). This is wrong because the solution clearly differentiates f(x)f(x) and substitutes into f′(14)f'\left(\frac{1}{4}\right). Follow the working shown in the solution.

  • Replacing xx by 1x\frac{1}{x} incorrectly. The right substitution in the original equation gives 3f(1x)+2f(x)=x−103f\left(\frac{1}{x}\right) + 2f(x) = x - 10, not another expression in 1x\frac{1}{x} on the right-hand side. Be careful while simplifying 1(1/x)=x\frac{1}{(1/x)} = x.

  • Differentiating 3x\frac{3}{x} incorrectly. Since 3x=3x−1\frac{3}{x} = 3x^{-1}, its derivative is −3x−2=−3x2-3x^{-2} = -\frac{3}{x^2}. Do not treat it as a constant or differentiate it as 3x\frac{3}{x} again.

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