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MHT CET 2018 Math Solved Paper

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Question : 35 of 50

Marks: +1, -0

If

\tan^{-1}

\tan^{-1}

\frac{\pi}{4}

then x =

-1
$\frac{1}{3}$
$\frac{1}{6}$
$\frac{1}{2}$

Solution:

(c)

\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}

⇒

\tan^{-1} \left( \frac{2x+3x}{1-6x^{2}} \right) = \frac{\pi}{4}

⇒

\frac{5x}{1-6x^{2}} = \tan \frac{\pi}{4}

⇒

\frac{5x}{1-6x^{2}} = 1

⇒ 5x = 1 - 6

x^2

⇒ 6

x^2

+ 5x - 1 = 0

⇒ 6

x^2

+ 6x - x - 1 = 0

⇒ 6x (x + 1) - 1(x +1) = 0

⇒ (x + 1)(6 x - 1) = 0

∴ x = -1, x =

\frac{1}{6}

When x =

\frac{1}{6}

, given equation is satisfied.

When x = -1, we get sum of two negative angles, hence discarded.

∴ x =

\frac{1}{6}

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