JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Limits

Putting $h=x-1\to 0^{+}$ gives $\sin (1-x)=-\sin h$, so the quotient becomes $\frac{6h+\lambda h\cos h-\mu \sin h}{h^3}$. With $\cos h=1-\frac{h^2}{2}+\cdots$ and $\sin h=h-\frac{h^3}{6}+\cdots$, the numerator is $(6+\lambda -\mu )h+\left(-\frac{\lambda }{2}+\frac{\mu }{6}\right)h^3+\cdots$. A finite limit forces the coefficient of $h$ to vanish, so $\mu =6+\lambda$, and the coefficient of $h^3$ must equal $-1$, giving $-\frac{\lambda }{2}+\frac{\mu }{6}=-1$. Substituting $\mu =6+\lambda$ yields $-3\lambda +6+\lambda =-6$, so $\lambda =6$ and $\mu =12$. Hence $\lambda +\mu =18$.