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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Mathematics
Chapter
Binomial Theorem

BITSAT 2009 · Mathematics · Binomial Theorem

The general solution of (8 ^2 (x)/(2)= 1+ x) is

The general solution of (8 \tan ^2 x2=\frac{x}{2}=1+\sec x) is

Options

  1. A(2 n $\pi \pm \cos ^{-1} \left(\frac{-1}{3}\right)
  2. B(2 nπ±π6\mathrm{n} \pi \pm \frac{\pi}{6})
  3. C(2 $\mathrm{n} \pi \pm \cos ^{-1} \left(\frac{1}{3}\right))
  4. DNone of these

Correct answer

C. (2 $\mathrm{n} \pi \pm \cos ^{-1} \left(\frac{1}{3}\right))

Step-by-step solution

We have, (8 \tan ^2 x2=\frac{x}{2}=1+\sec x)⇒8(1−cos⁡x1+cos⁡x)=1+1cos⁡x=1+cos⁡xcos⁡x⇒8cos⁡x−8cos⁡2x=(1+cos⁡x)2⇒9cos⁡2x−6cos⁡x+1=0⇒(3cos⁡x−1)2=0⇒3cos⁡x−1=0⇒cos⁡x=13=cos⁡α (say) ⇒x=2nπ±α∴x=2nπ±cos⁡−1(13) where n∈I\begin{aligned}& \Rightarrow 8 \left(\frac{1-\cos x}{1+\cos x}\right)=1+\frac{1}{\cos x}=\frac{1+\cos x}{\cos x}\\ & \Rightarrow 8 \cos x-8 \cos ^2 x=(1+\cos x)^2 \\ & \Rightarrow 9 \cos ^2 \mathrm{x}-6 \cos \mathrm{x}+1=0 \\ & \Rightarrow(3 \cos \mathrm{x}-1)^2=0 \Rightarrow 3 \cos \mathrm{x}-1=0 \\ & \Rightarrow \cos \mathrm{x}=\frac{1}{3}=\cos \alpha \text{ (say) } \Rightarrow \mathrm{x}=2 \mathrm{n} \pi \pm \alpha\\ & \therefore \mathrm{x}=2 \mathrm{n} \pi \pm \cos ^{-1}\left(\frac{1}{3}\right) \text{ where } \mathrm{n} \in \mathrm{I} \end{aligned}
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