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← JEE Main 2026 · 5 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 17 of 75

JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 17

The product of all possible values of α\alpha, for which
lim⁡x→0(1−cos⁡(αx)cos⁡((α+1)x)cos⁡((α+2)x)sin⁡2((α+1)x))=2,\lim_{x\to0}\left(\frac{1-\cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}\right)=2,
is:

Choose one option
Answer and worked solution

Official answer: C

Small x. With a=αa=\alpha, b=α+1b=\alpha+1, c=α+2c=\alpha+2:
1−cos⁡axcos⁡bxcos⁡cx≈(a2+b2+c2)x22,sin⁡2bx≈b2x2.1-\cos ax\cos bx\cos cx\approx\frac{(a^2+b^2+c^2)x^2}2,\qquad\sin^2bx\approx b^2x^2.

Condition. a2+b2+c22b2=2\dfrac{a^2+b^2+c^2}{2b^2}=2. With u=α+1u=\alpha+1: (u−1)2+u2+(u+1)2=4u2(u-1)^2+u^2+(u+1)^2=4u^2, so u2=2u^2=2 and α=−1±2\alpha=-1\pm\sqrt2.

Product: (−1)2−2=−1(-1)^2-2=-1.

(C)

Topics:Limits
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