the set of values of k for which x 2 kx sin 1 sin 4 0 for

The set of values of $$k$$ for which $${x^2} - kx + {\sin ^{ - 1}}\left( {\sin 4} \right) > 0$$ for all real $$x$$ is

A. $$\phi $$

B. $$\left( { - 2,2} \right)$$

C. $$R$$

D. $$\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right)$$

Answer : $$\phi $$

Solution :
$$\because \frac{\pi }{2} < 4 < \frac{{3\pi }}{2},{\text{so }}{\sin ^{ - 1}}\sin 4 = {\sin ^{ - 1}}\sin \left( {\pi - 4} \right) = \pi - 4$$
The inequality becomes $${x^2} - kx + \pi - 4 > 0$$
The discriminant $$D = {k^2} - 4\left( {\pi - 4} \right) > 0$$ for all $$k,$$ that is $${x^2} - kx + \left( {\pi - 4} \right) > 0$$ can not hold for all $$x.$$

Releted MCQ Question on
Trigonometry >> Inverse Trigonometry Function

Releted Question 1

The value of $$\tan \left[ {{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right]$$ is

A. $$\frac{6}{{17}}$$

B. $$\frac{7}{{16}}$$

C. $$\frac{16}{{7}}$$

D. none

View Answer

Releted Question 2

If we consider only the principle values of the inverse trigonometric functions then the value of $$\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right)$$ is

A. $$\frac{{\sqrt {29} }}{3}$$

B. $$\frac{{29}}{3}$$

C. $$\frac{{\sqrt {3}}}{29}$$

D. $$\frac{{3}}{29}$$

View Answer

Releted Question 3

The number of real solutions of $${\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$$ is

A. zero

B. one

C. two

D. infinite

View Answer

Releted Question 4

If $${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$ for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals

A. $$ \frac{1}{2}$$

B. 1

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

D. $$- 1$$

View Answer

The set of values of $$k$$ for which $${x^2} - kx + {\sin ^{ - 1}}\left( {\sin 4} \right) > 0$$ for all real $$x$$ is

Releted MCQ Question on
Trigonometry >> Inverse Trigonometry Function

The value of $$\tan \left[ {{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right]$$ is

If we consider only the principle values of the inverse trigonometric functions then the value of $$\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right)$$ is

The number of real solutions of $${\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$$ is

If $${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$ for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals

Practice More Releted MCQ Question on
Inverse Trigonometry Function

Practice More MCQ Question on Maths Section

The set of values of $$k$$ for which $${x^2} - kx + {\sin ^{ - 1}}\left( {\sin 4} \right) > 0$$ for all real $$x$$ is

Releted MCQ Question on Trigonometry >> Inverse Trigonometry Function

The value of $$\tan \left[ {{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right]$$ is

If we consider only the principle values of the inverse trigonometric functions then the value of $$\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right)$$ is

The number of real solutions of $${\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$$ is

If $${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$ for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals

Practice More Releted MCQ Question on Inverse Trigonometry Function

Practice More MCQ Question on Maths Section

Releted MCQ Question on
Trigonometry >> Inverse Trigonometry Function

Practice More Releted MCQ Question on
Inverse Trigonometry Function