QJEEM22Maa361: JEE Main 2022 Mathematics - Class 12 JEEM22Maa361 Solution: Suppose exists and is equal to L, where F ( x ) =

Given,

$$F(x)=\left|\begin{array}{ccc}a+\sin \frac{x}{2}& -b\cos x& 0\\ -b\cos x& 0& a+\sin \frac{x}{2}\\ 0& a+\sin \frac{x}{2}& -b\cos x\end{array}\right|$$

$= \left( {a + \sin {x \over 2}} \right)\left( { - {{\left( {a + \sin {x \over 2}} \right)}^2}} \right) + b\cos x \times {b^2}{\cos ^2}x$

$= - {\left( {a + \sin {x \over 2}} \right)^3} - {b^3}{\cos ^3}x$

Now,

$\mathop {\lim }\limits_{x \to 0} {{F(x)} \over {{x^3}}}$

$= \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {a + \sin {x \over 2}} \right)}^3} - {b^3}{{\cos }^3}x} \over {{x^3}}}$

Given limit exists, it only possible when a = 0 and b = 0.

$= \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {\sin {x \over 2}} \right)}^3}} \over {{x^3}}}$

$= \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2} \times \left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)} \right)^3}$

$= \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2}} \right)^3} \times {\left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)^3}$

$= - {1 \over 8} \times 1 = L$

$\therefore$ $- 112L = - 112 \times - {1 \over 8} = 14$