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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Functions

BITSAT 2024 · Paper 1 · Mathematics · Functions

f(x)= x+ x, g(x)=x^2-1 , then g(f(x)) is invertible if

f(x)=sin⁡x+cos⁡x,g(x)=x2−1f(x)=\sin x+\cos x, g(x)=x^2-1, then g(f(x))g(f(x)) is invertible iff(x)= x+ x, g(x)=x^2-1 , then g(f(x)) is invertible if

Options

  1. A−π4≤x≤π4\frac{-\pi}{4} \leq x \leq \frac{\pi}{4}
  2. B−π2≤x≤0\frac{-\pi}{2} \leq x \leq 0
  3. C−π2≤x≤π\frac{-\pi}{2} \leq x \leq \pi
  4. D0≤x≤π20 \leq x \leq \frac{\pi}{2}

Correct answer

A. −π4≤x≤π4\frac{-\pi}{4} \leq x \leq \frac{\pi}{4}

Step-by-step solution

Given that, f(x)=sin⁡x+cos⁡xg(x)=x2−1g[f(x)]=(sin⁡x+cos⁡x)2−1=sin⁡2x+cos⁡2x+2sin⁡xcos⁡x−1=1+sin⁡2x−1=sin⁡2xf(x)=\sin x+\cos xg(x)=x^2-1g[f(x)]=(\sin x+\cos x)^2-1=\sin ^2 x+\cos ^2 x+2 \sin x \cos x-1=1+\sin 2 x-1=\sin 2 x\left(\because \sin ^22x+\cos ^2 x=1, \sin x=2 \sin x \cos x\right)<imgsrc="/api/proxy−image?url=httpsAmongthegivenoptions,<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fbitsat%252Fdas8psdkef8Jpeb0nqSuEub-8U_ijwhvC_pHDla6qhI.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2Fdas8psdkef8Jpeb0nqSuEub-8U_ijwhvC_pHDla6qhI.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2Fdas8psdkef8Jpeb0nqSuEub-8U_ijwhvC_pHDla6qhI.original.fullsize.png?alt=media"/> Among the given options,\sin 2 xismonotonous(herestrictlyincreasing)inis monotonous (here strictly increasing) in−π4\leqx≤π4.-\frac{\pi}{4} \leqx \leq \frac{\pi}{4}.So,So,g(f(x))isinvertibleinis invertible in−π4\leqx≤π4.-\frac{\pi}{4} \leqx \leq \frac{\pi}{4}.
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