A) \[14\]
B) \[21\]
C) \[42\]
D) \[84\]
Correct Answer: C
Solution :
[c] \[L=\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{1-\sin x\sin 3x\sin 5x\sin 7x}{{{\left( \frac{\pi }{2}-x \right)}^{2}}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-\cosh \cos 3h\cos 5h\cos 7h}{{{h}^{2}}}\] \[\left( Putting\,\,\frac{\pi }{2}-x=h \right)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-\cosh \cos 3h\cos 5h\cos 7h}{{{h}^{2}}}\,\,\,\left( \frac{0}{0}form \right)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{\begin{align} & \sinh \cos 3h\cos 5h\cos 7h+\sin 3h\cosh \cos 5h\cos 7h \\ & +\sin 5h\cosh \cos 3h\cos 7h+\sin 7h\operatorname{coshcos}3hcos5h \\ \end{align}}{2h}\] (Using L' Hospital Rule) \[=\frac{1}{2}\,\,\left( 1+{{3}^{2}}+{{5}^{2}}+{{7}^{2}} \right)=42\]
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