Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

Match List I with List II

LIST I LIST II
A. $\lim\limits_{x→0}\frac{(1-cos2x)sin5x}{x^2sin3x}$ I. 18
B. $\lim\limits_{x→∞}\frac{(3x-5)(2x-7)}{(4x-9)(5x-3)}$ II. $\frac{10}{3}$
C. $\lim\limits_{x→0}\frac{2sin^23x}{x^2}$ III. $\frac{3}{4}$
D. $\lim\limits_{x→\frac{\pi}{4}}\frac{1-cos^3x}{2-cotx-cot^3x}$ IV. $\frac{3}{10}$

Choose the correct answer from the options given below :

Options:

A-IV, B-I, C-III, D-II

A-III, B-II, C-IV, D-I

A-II, B-IV, C-I, D-III

A-I, B-III, C-II, D-IV

Correct Answer:

A-II, B-IV, C-I, D-III

Explanation:

A. $\displaystyle \lim_{x\to 0}\frac{(1-\cos 2x)\sin 5x}{x^2\sin 3x} =\lim_{x\to 0}\frac{\left(\frac{(2x)^2}{2}\right)\,(5x)}{x^2\,(3x)} =\frac{10}{3}$

B. $\displaystyle \lim_{x\to \infty}\frac{(3x-5)(2x-7)}{(4x-9)(5x-3)} =\lim_{x\to \infty}\frac{6x^2+\cdots}{20x^2+\cdots} =\frac{6}{20}=\frac{3}{10}$

C. $\displaystyle \lim_{x\to 0}\frac{2\sin^2 3x}{x^2} =\lim_{x\to 0}\,2\left(\frac{\sin 3x}{x}\right)^2 =2\cdot 3^2=18$

D. $\displaystyle \lim_{x\to \frac{\pi}{4}}\frac{1-\cos^3 x}{2-\cot x-\cot^3 x} =\frac{3}{4}$