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← JEE Advanced 2014 · Paper 1
MathematicsMultiple correct+3Question 46 of 60

JEE Advanced 2014 · Paper 1, Mathematics question 46

Let f:(−π2,π2)→Rf:\left(-\frac\pi2,\frac\pi2\right)\to\mathbb R be given by
f(x)=(log⁡(sec⁡x+tan⁡x))3.f(x)=\left(\log(\sec x+\tan x)\right)^3.
Then

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Answer and worked solution

Official answer: A, B, C

The inner function. Let u(x)=log⁡(sec⁡x+tan⁡x)u(x)=\log(\sec x+\tan x) on (−π2,π2)(-\frac\pi2,\frac\pi2).

  • u′(x)=sec⁡x>0u'(x)=\sec x>0, so uu is strictly increasing.
  • sec⁡(−x)+tan⁡(−x)=sec⁡x−tan⁡x=1sec⁡x+tan⁡x\sec(-x)+\tan(-x)=\sec x-\tan x=\dfrac1{\sec x+\tan x}, so u(−x)=−u(x)u(-x)=-u(x): uu is odd.
  • As x→±π2x\to\pm\frac\pi2, u→±∞u\to\pm\infty, so uu maps onto R\mathbb R.

Cubing keeps all three properties.

  • (A) Odd. True.
  • (B) Strictly increasing, so one-one. True.
  • (C) Range is R\mathbb R, so onto. True.
  • (D) Not even (it is odd and not identically zero). False.

(A), (B), (C)