MCQ
If $x = \int\limits_0^y {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} $, then $\frac{{{d^2}y}}{{d{x^2}}}$ is equal to
- ✓$y$
- B$\sqrt {1 + {y^2}} $
- C$\frac{x}{{\sqrt {1 + {y^2}} }}$
- D$y^2$
$ \Rightarrow 1 = \frac{1}{{\sqrt {1 + {y^2}} }}.\frac{{dy}}{{dx}}$
[$\because $ If $1\left( x \right) = \int\limits_{\phi \left( x \right)}^{\psi \left( x \right)} {f\left( t \right)dt} ,$ then $\frac{{dI\left( x \right)}}{{dx}} = f\left\{ {\psi \left( x \right)} \right\}.$
$\left\{ {\frac{d}{{dx}}\psi \left( x \right)} \right\} - f\left\{ {\phi \left( x \right)} \right\}.\left\{ {\frac{d}{{dx}}\varphi \left( x \right)} \right\}$]
$\frac{{dy}}{{dx}} = \sqrt {1 - {y^2}} $
$ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{{2\sqrt {1 + {y^2}} }}.2y.\frac{{dy}}{{dx}} = \frac{y}{{\sqrt {1 + {y^2}} }}.\sqrt {1 + {y^2}} = y$
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