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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Continuity and Differentiability

BITSAT 2014 · Mathematics · Continuity and Differentiability

Let f(x)=\ array l 1- ^ 3 x 3 ^ 2 x , x (π)/(2) array . If f(x) is...

Let \mathrm{f}(\mathrm{x})=\left\{1−sin⁡3x3cos⁡2x,x<π2p,x=π2q(1−sin⁡x)(π−2x)2,x>π2\begin{array}{l}\frac{1-\sin ^{3} \mathrm{x}}{3 \cos ^{2} \mathrm{x}}, \mathrm{x}< \frac{\pi}{2} \\ \mathrm{p}, \quad \mathrm{x}=\frac{\pi}{2} \\ \frac{\mathrm{q}(1-\sin \mathrm{x})}{(\pi-2 \mathrm{x})^{2}}, \mathrm{x} > \frac{\pi}{2}\end{array}\right. If f(x)f(x) is continuous at x=π2,(p,q)=x=\frac{\pi}{2},(p, q)=

Options

  1. A(1,4)
  2. B(12,2)\left(\frac{1}{2}, 2\right)
  3. C(12,4)\left(\frac{1}{2}, 4\right)
  4. DNone of these

Correct answer

C. (12,4)\left(\frac{1}{2}, 4\right)

Step-by-step solution

f[(π/2)−]=lim⁡h→01−sin⁡3[(π/2)−h]3cos⁡2[(π/2)−h]=lim⁡h→01−cos⁡3h3sin⁡2h=12f[(π/2)+]=lim⁡h→0q[1−sin⁡{(π/2)+h}][π−2{(π/2)+h}]2=lim⁡h→0q(1−cosh⁡)4h2=q8∴p=12=q8⇒p=12,q=4\mathrm{f}\left[(\pi / 2)^{-}\right]=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\sin ^{3}[(\pi / 2)-\mathrm{h}]}{3 \cos ^{2}[(\pi / 2)-\mathrm{h}]}\quad=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos ^{3} \mathrm{ h}}{3 \sin ^{2} \mathrm{ h}}=\frac{1}{2}\mathrm{f}\left[(\pi / 2)^{+}\right]=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{q}[1-\sin \{(\pi / 2)+\mathrm{h}\}]}{[\pi-2\{(\pi / 2)+\mathrm{h}\}]^{2}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{q}(1-\cosh )}{4 \mathrm{ h}^{2}}=\frac{\mathrm{q}}{8}\therefore \mathrm{p}=\frac{1}{2}=\frac{\mathrm{q}}{8} \Rightarrow \mathrm{p}=\frac{1}{2}, \mathrm{q}=4
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