MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}} = $
- ✓$\frac{3}{2}$
- B$ - \frac{1}{2}$
- C$1$
- DNone of these
Now expanding ${e^{{x^2}}}$ and $\cos x,$ we get
$\mathop {\lim }\limits_{x \to 0} \,\frac{{\frac{{3{x^2}}}{{2\,!}} + {x^4}\,\left( {\frac{1}{{2\,!}} - \frac{1}{{4\,!}}} \right) + .......}}{{{x^2}}} = \frac{3}{2}$
Aliter : Apply $L-$ Hospital’s rule,
$\mathop {\lim }\limits_{x \to 0} \,\frac{{2x{e^{{x^2}}} + \sin x}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \,{e^{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \,\frac{{\sin x}}{{2x}} $
$= 1 + \frac{1}{2} = \frac{3}{2}.$
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Statement $p$ : The value of $sin\,120^o$ can be divided by taking $\theta\, = 240^o$ in the equation $2\,\sin \frac{\theta }{2} = \sqrt {1 + \sin \theta } - \sqrt {1 - \sin \theta } $
Statement $q$ : The angles $A, B, C$ and $D$ of any quadrilateral $ABCD$ satisfy the equation $\cos \left( {\frac{1}{2}\left( {A + C} \right)} \right) + \cos \left( {\frac{1}{2}\left( {B + D} \right)} \right) = 0$
Then the truth values of $p$ and $q$ are respectively.