Question
कोई फलन, $f(x)\, = \,\left\{ {\begin{array}{*{20}{c}}{{{\left( {{x^2} + {e^{\frac{1}{{2 - x}}}}} \right)}^{ - 1}}}&,&{x \ne 2}\\k&,&{x = 2}\end{array}} \right.$, यदि दाँयी ओर से $x = 2$ पर सतत् है, तो $k$ का मान होगा
यदि $f(x) , x = 2$ पर दाँयी ओर से सतत् है, तो
$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = f(2) = k$
==> $\mathop {\lim }\limits_{x \to {2^ + }} {\left[ {{x^2} + {e^{\frac{1}{{2 - x}}}}} \right]^{ - 1}} = k$
==> $k = \mathop {\lim }\limits_{h \to 0} f(2 + h)$
==> $k = \mathop {\lim }\limits_{h \to 0} \,{\left[ {{{(2 + h)}^2} + {e^{\frac{1}{{2 - (2 + h)}}}}} \right]^{\, - 1}}$
==> $k = \mathop {\lim }\limits_{h \to 0} \,{\left[ {\,4 + {h^2} + 4h + {e^{ - 1/h}}\,} \right]^{\, - 1}}$
==> $k = {[4 + 0 + 0 + {e^{ - \infty }}]^{\, - 1}}$==> $k = \frac{1}{4}$.
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$\left(1-x^2\right) d y=\left(x y+\left(x^3+2\right) \sqrt{1-x^2}\right) d x,-1 < x < 1$
का हल $v$ है तथा $y (0)=0$ है यदि $\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-x^2} y ( x ) dx = k \text { है, तो } k ^{-1}$ है।