If $$y = {\cos ^{ - 1}}\left( {\frac{{5\cos \,x - 12\sin \,x}}{{13}}} \right),\,x\, \in \left( {0,\,\frac{\pi }{2}} \right),$$ then $$\frac{{dy}}{{dx}}$$ is equal to :
A.
1
B.
$$-1$$
C.
0
D.
none of these
Answer :
1
Solution :
Let $$\cos \,\alpha = \frac{5}{{13}}.$$ Then $$\sin \,\alpha = \frac{{12}}{{13}}.$$
So, $$y = {\cos ^{ - 1}}\left\{ {\cos \,\alpha .\cos \,x - \sin \,\alpha .\sin \,x} \right\}$$
$$\therefore \,y = {\cos ^{ - 1}}\left\{ {\cos \left( {x + \alpha } \right)} \right\} = x + \alpha $$ ($$\because x + \alpha $$ is in the first or the second quadrant).
Releted MCQ Question on Calculus >> Differentiability and Differentiation
Releted Question 1
There exist a function $$f\left( x \right),$$ satisfying $$f\left( 0 \right) = 1,\,f'\left( 0 \right) = - 1,\,f\left( x \right) > 0$$ for all $$x,$$ and-
A.
$$f''\left( x \right) > 0$$ for all $$x$$
B.
$$ - 1 < f''\left( x \right) < 0$$ for all $$x$$
C.
$$ - 2 \leqslant f''\left( x \right) \leqslant - 1$$ for all $$x$$
If $$f\left( a \right) = 2,\,f'\left( a \right) = 1,\,g\left( a \right) = - 1,\,g'\left( a \right) = 2,$$ then the value of $$\mathop {\lim }\limits_{x \to a} \frac{{g\left( x \right)f\left( a \right) - g\left( a \right)f\left( x \right)}}{{x - a}}$$ is-
Let $$f:R \to R$$ be a differentiable function and $$f\left( 1 \right) = 4.$$ Then the value of $$\mathop {\lim }\limits_{x \to 1} \int\limits_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt$$ is-