MCQ
Let $f(x) = \,\left\{ {\begin{array}{*{20}{c}} {\frac{{\sin \pi x}}{{5x}},}&{x \ne 0} \\ {k,}&{x = 0} \end{array}} \right.$ if $f(x)$ is continuous at $x = 0,$ then $k=$
- ✓$\frac{\pi }{5}$
- B$\frac{5}{\pi }$
- C$1$
- D$0$
therefore $\mathop {{\rm{lim}}}\limits_{x \to 0} f(x) = f(0)$
==> $\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\sin \pi \,x}}{{5x}} = k$
==> $\mathop {{\rm{lim}}}\limits_{x \to 0} \left( {\frac{{\sin \pi \,x}}{{\pi x}}} \right)\,.\,\frac{\pi }{5} = k$
==> $(1)\,.\,\frac{\pi }{5} = k$ ==> $k = \frac{\pi }{5}$.
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$\sqrt {5\,x\,\, - \,\,6\,\, - \,\,{x^2}} \,\, + \,\,\frac{\pi }{2}\,\,\int\limits_0^x {} $$dz > x \int\limits_0^\pi {} sin^2 x \,dx$ is :
$1.$ The real number $s$ lies in the interval
$(A)$ $\left(-\frac{1}{4}, 0\right)$ $(B)$ $\left(-11,-\frac{3}{4}\right)$
$(C)$ $\left(-\frac{3}{4},-\frac{1}{2}\right)$ $(D)$ $\left(0, \frac{1}{4}\right)$
$2.$ The area bounded by the curve $y=f(x)$ and the lines $x=0, y=0$ and $x=t$, lies in the interval
$(A)$ $\left(\frac{3}{4}, 3\right)$ $(B)$ $\left(\frac{21}{64}, \frac{11}{16}\right)$
$(C)$ $(9,10)$ $(D)$ $\left(0, \frac{21}{64}\right)$
$3.$ The function $f^{\prime}(x)$ is
$(A)$ increasing in $\left(-t,-\frac{1}{4}\right)$ and decreasing in $\left(-\frac{1}{4}, t\right)$
$(B)$ decreasing in $\left(-t,-\frac{1}{4}\right)$ and increasing in $\left(-\frac{1}{4}, t\right)$
$(C)$ increasing in (-t, t) $(D)$ decreasing in ( $-\mathrm{t}, \mathrm{t})$
Give the answer question $1,2$ and $3.$