If 2^1/xx² dx=k·2^1/x+C, then k is equal to (A) (-1)/( 2) (B) - 2 (C) -1 (D) 1/2

MathsCBSE Class 12 · Board exam 2025, set 65/1/1 · 1 mark · MCQ

CBSE Class 12 Mathematics 2025: question 10

Set in the CBSE Class 12 board exam of 2025, paper 65/1/1, section A, question 10, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q10 · 1 mark · MCQIntegrals

If $\int \frac{2 ^{\frac{1}{x}}}{x ^{2}} d x = k \cdot 2^{\frac{1}{x}} + C$ , then k is equal to

(A) $\frac{- 1}{l o g 2}$
(B) $- lo g 2$
(C) $- 1$
(D) $\frac{1}{2}$

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 65/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Notice that $\frac{1}{x ^{2}} d x$ is closely related to $d (\frac{1}{x})$ .

Hint 2

Use $\int a^{t} d t = \frac{a ^{t}}{l o g a}$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Put $t = \frac{1}{x}$ , so $d t = - \frac{1}{x ^{2}} d x$ .
The integral becomes $- \int 2^{t} d t = - \frac{2 ^{t}}{l o g 2} + C = - \frac{1}{l o g 2} 2^{1/ x} + C$ , so $k = - \frac{1}{l o g 2}$ , option (A).
1 mark

Answer: $k = - \frac{1}{l o g 2}$ , option (A)

Where the marks are

Substitutes $t = 1/ x$ and integrates $2^{t}$ to get $k$ (1 mark)

Revise the chapter

Chapter: Integrals (notes, formulas, where marks are lost and every past question)
Topic: Integration by Substitution

Practise this chapterBoard-style questions from Integrals, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 65/1/1: every question, by section.

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CBSE Class 12 Mathematics 2025: question 10

Hints

Step-by-step solution

Revise the chapter

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