Let S = {x ∈ [-π, π] : sin x (sin x + cos x) = a, a ∈ Z}. Then n(S) is equal to :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q12 – JEE-MAIN 2026

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Home JEE-MAIN 2026 (2nd April 2026 Shift 1)MathsQuestion #12

Let

S = {x \in [- π, π] : \sin x (\sin x + \cos x) = a, a \in Z}

. Then

n (S)

is equal to :

Clue 1 of 3: Function Simplification

Rewrite the given expression $f (x) = \sin x (\sin x + \cos x)$ using double angle formulas to simplify it into a form $A + B \sin (C x + D)$ .

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● Video Walkthrough

Step 1: Simplify the expression for

f (x)

and determine its range✦ Active

The given expression is $f (x) = \sin x (\sin x + \cos x) = \sin^{2} x + \sin x \cos x$ . Using the identities $\sin^{2} x = \frac{1 - \cos (2 x)}{2}$ and $\sin x \cos x = \frac{\sin (2 x)}{2}$ , we can rewrite $f (x)$ as:

f (x) = \frac{1 - \cos (2 x)}{2} + \frac{\sin (2 x)}{2} = \frac{1}{2} + \frac{1}{2} (\sin (2 x) - \cos (2 x))

We can express $\sin (2 x) - \cos (2 x)$ as $\sqrt{2} \sin (2 x - \frac{π}{4})$ . So, $f (x)$ becomes:

f (x) = \frac{1}{2} + \frac{1}{\sqrt{2}} \sin (2 x - \frac{π}{4})

Given $x \in [- π, π]$ , we have $2 x \in [- 2 π, 2 π]$ . Therefore, $2 x - \frac{π}{4} \in [- 2 π - \frac{π}{4}, 2 π - \frac{π}{4}] = [- \frac{9 π}{4}, \frac{7 π}{4}]$ . The range of $\sin (2 x - \frac{π}{4})$ is $[- 1, 1]$ . Thus, the range of $f (x)$ is $[\frac{1}{2} - \frac{1}{\sqrt{2}}, \frac{1}{2} + \frac{1}{\sqrt{2}}]$ . Numerically, this range is approximately $[- 0.207, 1.207]$ . Since $a \in Z$ , the possible integer values for $a$ are $0$ and $1$ .

Step 2: Solve for

a = 0

and

a = 1

○ Expand

Case 1: $a = 0$ . Set $f (x) = 0$ :

\frac{1}{2} + \frac{1}{\sqrt{2}} \sin (2 x - \frac{π}{4}) = 0 ⟹ \sin (2 x - \frac{π}{4}) = - \frac{1}{\sqrt{2}}

Let $Y = 2 x - \frac{π}{4}$ . We need to solve $\sin Y = - \frac{1}{\sqrt{2}}$ for $Y \in [- \frac{9 π}{4}, \frac{7 π}{4}]$ . The values of $Y$ are $- \frac{9 π}{4}, - \frac{3 π}{4}, - \frac{π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ . Each of these $Y$ values corresponds to a unique $x$ value in $[- π, π]$ . Thus, there are 5 solutions for $a = 0$ .

Case 2: $a = 1$ . Set $f (x) = 1$ :

\frac{1}{2} + \frac{1}{\sqrt{2}} \sin (2 x - \frac{π}{4}) = 1 ⟹ \sin (2 x - \frac{π}{4}) = \frac{1}{\sqrt{2}}

Let $Y = 2 x - \frac{π}{4}$ . We need to solve $\sin Y = \frac{1}{\sqrt{2}}$ for $Y \in [- \frac{9 π}{4}, \frac{7 π}{4}]$ . The values of $Y$ are $- \frac{7 π}{4}, - \frac{5 π}{4}, \frac{π}{4}, \frac{3 π}{4}$ . Each of these $Y$ values corresponds to a unique $x$ value in $[- π, π]$ . Thus, there are 4 solutions for $a = 1$ .

Step 3: Calculate the total number of solutions○ Expand

The total number of solutions $n (S)$ is the sum of solutions for $a = 0$ and $a = 1$ . Therefore, $n (S) = 5 + 4 = 9$ .

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Maths Question 12 – JEE-MAIN 2026

Clue 1 of 3: Function Simplification

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