Limit Continuity and Differentiability 2 Question 2

2. limx1π2sin1x1x is equal to

(a) π2

(b) 2π

(c) π

(d) 12π

(2019 Main, 12 Jan II)

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Answer:

Correct Answer: 2. (c)

Solution:

  1. limxπ424f(t)dtx2π216

=limxπ/4f(sec2x)2secxsecxtanx2x

=2f(2)π/4=8πf(2)

[using L’ Hospital’s rule]



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