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.

Answer 2

We need to evaluate the limit

\[ \lim_{x \to 1} \frac{(x-1)\big(6+\lambda \cos(x-1)\big) + \mu \sin(1-x)}{(x-1)^3} = -1, \]

and find \(\lambda+\mu\), where \(\lambda,\mu\in\mathbb{R}\).

Concept Used:

Set \(h = x-1 \to 0\). Use the Taylor expansions near 0:

\[ \cos h = 1 - \frac{h^2}{2} + O(h^4), \qquad \sin(-h) = -\sin h = -h + \frac{h^3}{6} + O(h^5). \]

Step-by-Step Solution:

Step 1: Rewrite the numerator in terms of \(h\):

\[ N(h) = h\big(6+\lambda \cos h\big) + \mu \sin(1-x) = h\big(6+\lambda \cos h\big) + \mu(-\sin h). \]

Step 2: Substitute the series:

\[ \cos h = 1 - \frac{h^2}{2} + O(h^4), \quad \sin h = h - \frac{h^3}{6} + O(h^5). \] \[ N(h) = h\Big(6+\lambda\big(1-\tfrac{h^2}{2}+O(h^4)\big)\Big) + \mu\Big(-h+\tfrac{h^3}{6}+O(h^5)\Big). \]

Step 3: Collect terms up to \(h^3\):

\[ N(h) = h(6+\lambda) - \frac{\lambda}{2}h^3 - \mu h + \frac{\mu}{6}h^3 + O(h^5) = h\big((6+\lambda)-\mu\big) + h^3\Big(-\frac{\lambda}{2}+\frac{\mu}{6}\Big) + O(h^5). \]

Step 4: For the limit \(\dfrac{N(h)}{h^3}\) to be finite, the \(h\)-term must vanish:

\[ (6+\lambda)-\mu = 0 \;\;\Rightarrow\;\; \mu = 6+\lambda. \]

Step 5: The limit equals the coefficient of \(h^3\):

\[ \lim_{h\to 0}\frac{N(h)}{h^3} = -\frac{\lambda}{2}+\frac{\mu}{6} = -1. \] Substitute \(\mu=6+\lambda\): \[ -\frac{\lambda}{2} + \frac{6+\lambda}{6} = -1 \;\Rightarrow\; -\frac{\lambda}{2} + 1 + \frac{\lambda}{6} = -1 \;\Rightarrow\; 1 - \frac{\lambda}{3} = -1 \;\Rightarrow\; \lambda = 6. \] \[ \mu = 6 + \lambda = 12. \]

Final Computation & Result

\[ \lambda + \mu = 6 + 12 = \boxed{18}. \]