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MHTCET 10 May 2024 Shift 2 Paper

Question : 143 of 150

Marks: +1, -0

\tan ^{-1}(x+2)+\tan ^{-1}(x-2)-\tan ^{-1}\left(\frac{1}{2}\right)=0

x

Solution:

Concept:

Use the inverse tangent addition formula and verify the principal-value domain.

Explanation:

Given:

\tan^{-1}(x+2)+\tan^{-1}(x-2)-\tan^{-1}\left(\frac{1}{2}\right)=0

\Rightarrow \tan^{-1}(x+2)+\tan^{-1}(x-2)=\tan^{-1}\left(\frac{1}{2}\right)

Using

\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right)

with

a=x+2

b=x-2

\tan^{-1}\left(\frac{2x}{5-x^2}\right)=\tan^{-1}\left(\frac{1}{2}\right)

Taking tangent on both sides:

\frac{2x}{5-x^2}=\frac{1}{2}

\Rightarrow 4x=5-x^2

\Rightarrow x^2+4x-5=0

\Rightarrow (x+5)(x-1)=0

\Rightarrow x=-5 \text{ or } x=1

Check the principal value branch.

For

x=-5

, the sum

\tan^{-1}(-3)+\tan^{-1}(-7)

does not lie in

\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

, so

x=-5

is extraneous.

For

x=1

\tan^{-1}3+\tan^{-1}(-1)-\tan^{-1}\left(\frac{1}{2}\right)=0

Hence, the valid value of

x

1

Answer:

1

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