MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \,\left( {\pi {{\cos }^2}\,x} \right)}}{{{x^2}}}$ equals
- A$-\pi $
- B$1$
- C$-1$
- ✓$\pi $
$\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}}$
$\, = \,\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi - \pi {{\sin }^2}x} \right)}}{{{x^2}}}$
[$\because $ $\sin \left( {\pi - \theta } \right) = \sin \theta $]
$ = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi {{\sin }^2}x} \right)}}{{\pi {{\sin }^2}x}} \times \frac{{\left( {\pi {{\sin }^2}x} \right)}}{{{x^2}}} = \pi $
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$\left| {1 - {{\log }_{\frac{1}{6}}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{\frac{1}{6}}}x + {{\log }_{\frac{1}{2}}}x} \right|$ is $\left[ {\frac{a}{b},a} \right],a,b, \in N,$ then the value of $(a + b)$ is
Statement $-1$:${s_3} = 55 \times {2^9}$
Statement $-2$: ${s_1} = 90 \times {2^8}\;$ and ${s_2} = 10 \times {2^8}$