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Functions: Number Solutions Equation Cases Cases

JEE Maths question with a full step-by-step solution.

Question
Number of solutions of the equation f(x−1)+f(x+1)=sin⁡αf(x-1)+f(x+1) = \sin\alpha, 0<α<π20 < \alpha < \dfrac{\pi}{2}, where
f(x)={1−∣x∣;∣x∣≤10;∣x∣>1f(x) = \begin{cases} 1-|x|; & |x| \leq 1 \\ 0; & |x| > 1 \end{cases}
is.
Solution
Answer: 4
Step 1: Compute g(x)=f(x−1)+f(x+1)g(x) = f(x-1)+f(x+1) f(x−1)f(x-1) is a tent centered at x=1x=1, zero outside [0,2][0,2]. f(x+1)f(x+1) is a tent centered at x=−1x=-1, zero outside [−2,0][-2,0].
g(x)={x+2−2≤x≤−1−x−1≤x≤0x0≤x≤12−x1≤x≤20otherwiseg(x) = \begin{cases} x+2 & -2 \leq x \leq -1 \\ -x & -1 \leq x \leq 0 \\ x & 0 \leq x \leq 1 \\ 2-x & 1 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}
 gg has two peaks of height 1 at x=−1x = -1 and x=1x = 1. Step 2: Count intersections with y=sin⁡αy = \sin\alpha Since sin⁡α∈(0,1)\sin\alpha \in (0,1), the horizontal line y=sin⁡αy = \sin\alpha intersects each of the four rising/falling segments exactly once:
g(x)=sin⁡α has exactly 4 solutionsg(x) = \sin\alpha \text{ has exactly } 4 \text{ solutions}
 Answer: 4
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