Let tan-1x∈-π2, π2, for x∈ℝ. Then the number of real solutions of the equation 1+cos2x=2tan-1tanx in the set -3π2, -π2∪-π2, π2∪π2, 3π2 is equal to

JEE Advanced · Mathematics · 8. Trigonometric Equations

Let $\tan^{- 1} (x) \in (- \frac{π}{2}, \frac{π}{2})$ , for $x \in ℝ$ . Then the number of real solutions of the equation $\sqrt{1 + \cos (2 x)} = \sqrt{2} \tan^{- 1} (\tan x)$ in the set $(- \frac{3 π}{2}, - \frac{π}{2}) \cup (- \frac{π}{2}, \frac{π}{2}) \cup (\frac{π}{2}, \frac{3 π}{2})$ is equal to

Verified Solution

Answer & Solution

Correct Answer

(C) 3

Step-by-step Solution

Detailed explanation

Given,

\sqrt{1 + \cos 2 x} = \sqrt{2} \tan^{- 1} (\tan x)

\Rightarrow \sqrt{2 \cos^{2} x} = \sqrt{2} \tan^{- 1} (\tan x)

\Rightarrow |\cos x| = \tan^{- 1} (\tan x)

Now plotting the diagram of

|\cos x| & \tan^{- 1} (\tan x)

we get,

Now from above diagram we can say that,
Number of solutions

=

Number of intersection points

= 3

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Column I		Column II
$(A)$	In a triangle $∆ X Y Z$ let $a, b$ and $c$ be the lengths of the angles $X, Y$ and $Z$ , respectively. If $2 (a^{2} - b^{2}) = c^{2}$ and $λ = \frac{\sin (X - Y)}{sinZ}$ then possible values of $n$ for which $\cos (n π λ) = 0$ is are	$(P)$	$1$
$(B)$	In a triangle $Δ X Y Z,$ let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$ , respectively. If $1 + \cos 2 x - 2 \cos 2 y$ $= 2 \sin x \sin y$ , then possible value(s) of $\frac{a}{b}$ is (are)	$(Q)$	$2$
$(C)$	In $R^{2},$ let $\sqrt{3} \hat{i} + \hat{j}, \hat{i} + \sqrt{3} \hat{j}$ and $β \hat{i} + (1 - β) \hat{j}$ be the position vectors of $X, Y$ and $Z$ with respect to the origin $O$ , respectively. If the distance of $Z$ from the bisector of the acute angle of $\vec{O X}$ with $\vec{O Y}$ is $\frac{3}{\sqrt{2}}$ , then possible value (s) of $\|β\|$ is (are)	$(R)$	$3$
$(D)$	Suppose that $F (α)$ denotes the area of the region bounded by $x = 0, x = 2, y^{2} = 4 x$ and $y = \|a x - 1\| +$ $\|α x - 2\| + α x,$ where $α \in \{0, 1\} .$ Then the value (s) of $F (α) + \frac{8}{3} \sqrt{2},$ when $α = 0$ and $α = 1$ , is (are)	$(S)$	$5$
		$(T)$	$6$

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