ExamHabit
Practise free

JEE Main 2019 · 12 Jan · Shift I

JEE Main 2019 Mathematics question on Limits Continuity and Differentiability

lim⁡x→π4cot⁡3x−tan⁡xcos⁡(x+π4)\lim_{x \to \frac{\pi}{4}} \dfrac{\cot^3 x - \tan x}{\cos\left(x + \dfrac{\pi}{4}\right)} is
  1. A.
    424\sqrt{2}
  2. B.
    44
  3. C.
    88
  4. D.
    828\sqrt{2}
Show answer and solution

Correct option: C

Solution

Factor the numerator: cot³x - tan x = (cos³x/sin³x) - (sin x/cos x) = (cos⁴x - sin⁴x)/(sin³x · cos x).

cos⁴x - sin⁴x = (cos²x - sin²x)(cos²x + sin²x) = cos 2x.

Also, cos(x + π/4) = (cos x - sin x)/√2 = -√2 · sin(x - π/4), and cos 2x = cos²x - sin²x = (cos x - sin x)(cos x + sin x).

So the expression becomes: cos 2x / [sin³x · cos x · cos(x + π/4)] = (cos x - sin x)(cos x + sin x) / [sin³x · cos x · (cos x - sin x)/√2] = √2(cos x + sin x) / (sin³x · cos x).

At x = π/4: cos x + sin x = √2, sin³x = (1/√2)³ = 1/(2√2), cos x = 1/√2.

Result = √2 · √2 / [(1/(2√2)) · (1/√2)] = 2 / (1/4) = 8.

[JEE Main 2019 · 12 Jan · Shift I]

Practise Limits Continuity and Differentiability the way it is asked

A timed set of real JEE Main Limits Continuity and Differentiability questions, graded as the exam grades. No sign-up.

Start a Limits Continuity and Differentiability test

More: all JEE Main Limits Continuity and Differentiability questions · JEE Main 2019 · Mathematics chapter list