MCQ
If $y = {\sin ^{ - 1}}\sqrt {1 - {x^2}} $, then $dy/dx = $
- A${1 \over {\sqrt {1 - {x^2}} }}$
- B${1 \over {\sqrt {1 + {x^2}} }}$
- ✓$ - {1 \over {\sqrt {1 - {x^2}} }}$
- D$ - {1 \over {\sqrt {{x^2} - 1} }}$
Let $\sqrt {1 - {x^2}} = \sin \theta \Rightarrow 1 - {x^2} = {\sin ^2}\theta $
==> ${x^2} = 1 - {\sin ^2}\theta = {\cos ^2}\theta $
$\therefore x = \cos \theta $ or $\theta = {\cos ^{ - 1}}x$
$ \Rightarrow y = {\cos ^{ - 1}}x$
Differentiating w.r.t. $x $ of $y,$ we get
$\frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {x^2}} }}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ Which of the following is true?
$(A)$ $(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=0$
$(B)$ $(2-a)^2 f^{\prime}(1)-(2+a)^2 f^{\prime \prime}(-1)=0$
$(C)$ $f^{\prime}(1) f^{\prime}(-1)=(2-a)^2$
$(D)$ $f^{\prime}(1) f^{\prime}(-1)=-(2+a)^2$
$2.$ Which of the following is true?
$(A)$ $f(x)$ is decreasing on $(-1,1)$ and has a local minimum at $x=1$
$(B)$ $f(x)$ is increasing on $(-1,1)$ and has a local maximum at $x=1$
$(C)$ $f(x)$ is increasing on $(-1,1)$ but has neither a local maximum nor a local minimum at $x=1$
$(D)$ $f(x)$ is decreasing on $(-1,1)$ but has neither a local maximum nor a local minimum at $x=1$
$3.$ Let $g(x)=\int_0^{e^x} \frac{f^{\prime}(t)}{1+t^2} d t$ which of the following is true?
$(A)$ $g^{\prime}(x)$ is positive on $(-\infty, 0)$ and negative on $(0, \infty)$
$(B)$ $g^{\prime}(x)$ is negative on $(-\infty, 0)$ and positive on $(0, \infty)$
$(C)$ $\mathrm{g}^{\prime}(\mathrm{x})$ changes sign on both $(-\infty, 0)$ and $(0, \infty)$
$(D)$ $g^{\prime}(x)$ does not change sign on $(-\infty, \infty)$
Give the answer question $1,2$ and $3.$