MCQ
$cos\left[tan^{-1}\left\{cot\left(sin^{-1\frac{1}{2}}\right)\right\}\right]=..............$
- A$1$
- B$\frac{1}{4}$
- C$\frac{1}{8}$
- ✓$\frac{1}{2}$
$=cos[tan^{-1}{\left\{cot(sin^{-1}\frac{1}{2})\right\}}]$
$=cos[tan^{-1}{\left\{cot \frac{\pi}{6}\right\}}]$
$=cos[tan^{-1}\sqrt{3}]$
$=cos \frac{\pi}{3}$
$=\frac{1}{2}$
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$f(x) \rightarrow \frac{\lambda\left|x^{2}-5 x+6\right|}{\mu\left(5 x-x^{2}-6\right)}, x<2$
$\quad\quad\quad\quad e^{\frac{\tan (x-2)}{x-[x]}}, \quad x>2$
$\quad\quad\quad\quad \mu \quad\quad\quad\quad x=2$
કે જ્યાં $[x]$ એ મહતમ પૃણાંક વિધેય છે. જો $f$ એ $x=2$ આગળ સતત હોય તો $\lambda+\mu$ ની કિમંત મેળવો.
$(S1)$: $f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2$
$( S 2): \int_{-2}^{2} f ( x ) dx =12$ હોય તો .. .
$(I)$ $y=f(x)$ એ $x$-અક્ષને બરાબર એક બિંદુએ છેદ છે.
$(II)$ $y=f(x)$ એ $x$-અક્ષને $x=\cos \frac{\pi}{12}$ આગળ છેદ છે. તો.......