MCQ
Differential coefficient of ${\sin ^{ - 1}}x$ $w.r.t$ ${\cos ^{ - 1}}\sqrt {1 - {x^2}} $ is
- ✓$1$
- B${1 \over {1 + {x^2}}}$
- C$2$
- DNone of these
Differentiating w.r.t. $x$ of ${y_1}$ and ${y_2}$, we get
$\frac{{d{y_1}}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }}$
$\frac{{d{y_2}}}{{dx}} = - \frac{1}{{\sqrt {1 - (1 - {x^2})} }}\frac{{1( - 2x)}}{{2\sqrt {1 - x} }} $
$= \frac{1}{{\sqrt {1 - {x^2}} }} \Rightarrow \frac{{d{y_2}}}{{d{y_1}}} = 1.$
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$ \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R} $
$ \mathrm{L}_2: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{k}, \mu \in \mathbb{R}$
is $\frac{\mathrm{m}}{\sqrt{\mathrm{n}}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then the value of $\mathrm{m}+\mathrm{n}$ equals.
$f(x)=x \cos \frac{1}{x}, \quad x \geq 1,$
$(A)$ for at least one $x$ in the interval $[1, \infty), f(x+2)-f(x)<2$
$(B)$ $\lim _{x \rightarrow \infty} f^{\prime}(x)=1$
$(C)$ for all $x$ in the interval $[1, \infty), f(x+2)-f(x)>2$
$(D)$ $f^{\prime}(x)$ is strictly decreasing in the interval $[1, \infty)$