Subjective Type

The number of real solutions of the equation $$\sin^{-1}\left(\displaystyle\sum^{\infty}_{i=1}x^{i+1}-x\displaystyle\sum^{\infty}_{i=1}\left(\displaystyle\dfrac{x}{2}\right)^i\right)=\dfrac{\pi}{2}-\cos^{-1}\left(\displaystyle\sum^{\infty}_{i=1}\left(-\displaystyle\frac{x}{2}\right)^i-\displaystyle\sum^{\infty}_{i=1}(-x)^i\right)$$ lying in the interval $$\left(-\displaystyle\frac{1}{2}, \frac{1}{2}\right)$$ is

Solution

∞ ∑ i=1 xi+1= x2 1−x ∞ ∑ i=1 ( x 2 )i= x 2−x ∞ ∑ i=1 (− x 2 )i= −x 2+x ∞ ∑ i=1 (−x)i= −x 1+x To have real solutions ∞ ∑ i=1 xi+1−x ∞ ∑ i=1 ( x 2 )i= ∞ ∑ i=1 ( −x 2 )i− ∞ ∑ i=1 (−x)i x2 1−x − x2 2−x = −x 2+x + x 1+x x(x3+2x2+5x−2)=0 ∴x=0 and let f(x)=x3+2x2+5x−2 f( 1 2 )⋅f(− 1 2 )<0 Hence two solutions exist.

SIMILAR QUESTIONS

Inverse Trigonometric Functions

The solution set of the equation $$\tan^{-1}x -\cot^{-1}x =\cos^{-1}(2-x)$$ is

Inverse Trigonometric Functions

If $$\cos ^{ -1 }{ x } -\sin ^{ -1 }{ x } =0$$, then $$x$$ is equal to-

Inverse Trigonometric Functions

If $$\tan ^{ -1 }{ x } +\tan ^{ -1 }{ y } =\cfrac { 2\pi }{ 3 } $$, then $$\cot ^{ -1 }{ x } +\cot ^{ -1 }{ y } $$ is equal to

Inverse Trigonometric Functions

If $$4\sin^{-1}x+\cos^{-1}x=\pi$$, then $$x$$ is equal to:

Inverse Trigonometric Functions

Prove: $$\displaystyle { \tan }^{ -1 }\frac { 2 }{ 11 } +{ \tan }^{ -1 }\frac { 7 }{ 24 } ={ \tan }^{ -1 }\frac { 1 }{ 2 } $$

Inverse Trigonometric Functions

If $$\displaystyle \sin { \left( { \sin }^{ -1 }\frac { 1 }{ 5 } +{ \cos }^{ -1 }x \right) } =1$$, then find the value of $$x$$.

Inverse Trigonometric Functions

Prove: $$\displaystyle \frac { 9\pi }{ 8 } -\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 1 }{ 3 }=\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 } $$

Inverse Trigonometric Functions

If sin $$^{-1}(x-\dfrac{x^2}{2}+\dfrac{x^3}{4}+..............\infty)+$$ cos $$^{-1}(x^2-\dfrac{x^4}{2}+\dfrac{x^6}{4}-...........\infty)$$$$=\dfrac {\pi}{2}$$ and $$0 < x < \sqrt{2}$$ then x =

Inverse Trigonometric Functions

If $$\tan^{-1}\dfrac{a+x}{a}+\tan^{-1}\dfrac{a-x}{a}=\dfrac{\pi}{6}$$, then $$x^2=$$?

Inverse Trigonometric Functions

Solve the equation for $$x : {\sin}^{-1}\dfrac{5}{x} + {\sin}^{-1}\dfrac{12}{x}=\dfrac{\pi}{2}, x \ne 0$$.

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