If y = cos ^ - 1( cos x ) then dydx at x = 5 4 is equal to :

y = cos ^ - 1cos ( + overline x - ) = cos ^ - 1cos ( - overline x - ). When x is around 5 4, - overline x - is in the second quadrant. So, y = - ( x - ); dydx = - 1

if y cos pow 1 cos x then dy dx at x 5pi 4 is equal to

If $$y = {\cos ^{ - 1}}\left( {\cos \,x} \right)$$ then $$\frac{{dy}}{{dx}}$$ at $$x = \frac{{5\pi }}{4}$$ is equal to :

A. 1

B. $$-1$$

C. $$\frac{1}{{\sqrt 2 }}$$

D. none of these

Answer : $$-1$$

Solution :
$$y = {\cos ^{ - 1}}\cos \left( {\pi + \overline {x - \pi } } \right) = {\cos ^{ - 1}}\cos \left( {\pi - \overline {x - \pi } } \right).$$ When $$x$$ is around $$\frac{{5\pi }}{4},\,\pi - \overline {x - \pi } $$ is in the second quadrant. So, $$y = \pi - \left( {x - \pi } \right)\,;\,\,\,\,\,\,\therefore \,\,\frac{{dy}}{{dx}} = - 1$$

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 $$x$$

D. $$f''\left( x \right) < - 2$$ for all $$x$$

View Answer

Releted Question 2

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-

A. $$-5$$

B. $$\frac{1}{5}$$

C. $$5$$

D. none of these

View Answer

Releted Question 3

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-

A. $$8f'\left( 1 \right)$$

B. $$4f'\left( 1 \right)$$

C. $$2f'\left( 1 \right)$$

D. $$f'\left( 1 \right)$$

View Answer

Releted Question 4

Let [.] denote the greatest integer function and $$f\left( x \right) = \left[ {{{\tan }^2}x} \right],$$ then:

A. $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$ does not exist

B. $$f\left( x \right)$$ is continuous at $$x = 0$$

C. $$f\left( x \right)$$ is not differentiable at $$x =0$$

D. $$f'\left( 0 \right) = 1$$

View Answer

If $$y = {\cos ^{ - 1}}\left( {\cos \,x} \right)$$ then $$\frac{{dy}}{{dx}}$$ at $$x = \frac{{5\pi }}{4}$$ is equal to :

Releted MCQ Question on
Calculus >> Differentiability and Differentiation

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-

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-

Let [.] denote the greatest integer function and $$f\left( x \right) = \left[ {{{\tan }^2}x} \right],$$ then:

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If $$y = {\cos ^{ - 1}}\left( {\cos \,x} \right)$$ then $$\frac{{dy}}{{dx}}$$ at $$x = \frac{{5\pi }}{4}$$ is equal to :

Releted MCQ Question on Calculus >> Differentiability and Differentiation

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-

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-

Let [.] denote the greatest integer function and $$f\left( x \right) = \left[ {{{\tan }^2}x} \right],$$ then:

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Releted MCQ Question on
Calculus >> Differentiability and Differentiation

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