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MathematicsInteger+4 / -1Question 23 of 90

JEE Main 2022 · 30 Jun · Shift 1, Mathematics question 23

Suppose lim⁡x→0F(x)x3\displaystyle\lim_{x\to0}\frac{F(x)}{x^3} exists and is equal to L, where
F(x)=∣a+sin⁡x2−bcos⁡x0−bcos⁡x0a+sin⁡x20a+sin⁡x2−bcos⁡x∣.F(x)=\begin{vmatrix}a+\sin\frac x2&-b\cos x&0\\-b\cos x&0&a+\sin\frac x2\\0&a+\sin\frac x2&-b\cos x\end{vmatrix}.
Then, −112L-112L is equal to ______.

Answer and worked solution

Official answer: 14

Determinant. With p=a+sin⁡x2p=a+\sin\frac x2, q=−bcos⁡xq=-b\cos x: F=−(p3+q3)F=-(p^3+q^3). A finite limit over x3x^3 forces a=b=0a=b=0 (otherwise FF is of lower order).

Then F=−sin⁡3x2F=-\sin^3\frac x2, L=−18L=-\frac18, and −112L=14-112L=14.

Answer: 14

Topics:LimitsDeterminants and properties
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