Question

If $\lim_{x \to 1} \frac{(x-1)(6+\lambda \cos(x-1)) + \mu \sin(1-x)}{(x-1)^3} = -1$, where $\lambda, \mu \in \mathbb{R}$, then $\lambda + \mu$ is equal to

• A. 18
• B. 20
• C. 19
• D. 17

Hint:

Substitute $x = 1 + h$ to simplify limits involving $x \to 1$.

Answer 1

The Correct Option is A

1. Substitute $x = 1 + h$: \[ \lim_{h \to 0} \frac{h(6 + \lambda \cos h) - \mu \sin h}{h^3} = -1 \]
2. Expand $\cos h$ and $\sin h$ using Taylor series: \[ \cos h \approx 1 - \frac{h^2}{2}, \quad \sin h \approx h - \frac{h^3}{6} \]
3. Substitute the expansions: \[ \lim_{h \to 0} \frac{h \left( 6 + \lambda \left( 1 - \frac{h^2}{2} \right) \right) - \mu \left( h - \frac{h^3}{6} \right)}{h^3} = -1 \]
4. Simplify the expression: \[ \lim_{h \to 0} \frac{h \left( 6 + \lambda - \frac{\lambda h^2}{2} \right) - \mu h + \frac{\mu h^3}{6}}{h^3} = -1 \] \[ \lim_{h \to 0} \frac{6h + \lambda h - \frac{\lambda h^3}{2} - \mu h + \frac{\mu h^3}{6}}{h^3} = -1 \] \[ \lim_{h \to 0} \frac{6 + \lambda - \mu - \frac{\lambda h^2}{2} + \frac{\mu h^2}{6}}{h^2} = -1 \] 5. Equate the coefficients: \[ 6 + \lambda - \mu = 0 \quad \text{and} \quad -\frac{\lambda}{2} + \frac{\mu}{6} = -1 \] 6. Solve the system of equations: \[ \lambda + \mu = 18 \] Therefore, the correct answer is (1) 18.