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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Differentiation

BITSAT 2023 · Paper 2 · Mathematics · Differentiation

If f(x)= 1+ ^2(x^2) , then the value of f^ ( π 2 )

If f(x)=1+cos⁡2(x2)\mathrm{f}(\mathrm{x})=\sqrt{1+\cos ^2\left(\mathrm{x}^2\right)}, then the value of f′(π2)f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)If f(x)= 1+ ^2(x^2) , then the value of f^ ( π 2 )If f(x)= 1+ ^2(x^2) , then the value of f^ ( π 2 )

Options

  1. Aπ6\frac{\sqrt{\pi}}{6}
  2. B−π6-\sqrt{\frac{\pi}{6}}
  3. C16\frac{1}{\sqrt{6}}
  4. Dπ6\frac{\pi}{\sqrt{6}}

Correct answer

B. −π6-\sqrt{\frac{\pi}{6}}

Step-by-step solution


If f(x)= 1+ ^2(x^2) , then the value of f^ ( π 2 )

On differentiating (i) w.r.t.x, we get
f′(x)=−2sin⁡x2cos⁡x21+cos⁡2x2(x)f^{\prime}(x)=\frac{-2 \sin x^2 \cos x^2}{\sqrt{1+\cos ^2 x^2}}(x)If f(x)= 1+ ^2(x^2) , then the value of f^ ( π 2 )

Put, x=π2x=\frac{\sqrt{\pi}}{2} in (ii), we get
<br/>f′(π2)=−π2⋅sin⁡2(π4)1+12<br/>=−π2⋅sin⁡π232=−π6<br/>\begin{aligned}<br />& \mathrm{f}^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)=-\frac{\sqrt{\pi}}{2} \cdot \frac{\sin 2\left(\frac{\pi}{4}\right)}{\sqrt{1+\frac{1}{2}}} \\<br />& =-\frac{\sqrt{\pi}}{2} \cdot \frac{\sin \frac{\pi}{2}}{\sqrt{\frac{3}{2}}}=-\sqrt{\frac{\pi}{6}}<br />\end{aligned}
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