If , where , then is equal to [2025]

Question

If $\lim_{x \to 1^{+}} \frac{(x - 1) (6 + λ \cos (x - 1)) + μ \sin (1 - x)}{{(x - 1)}^{3}} = - 1$ , where $λ, μ \in R$ , then $λ + μ$ is equal to [2025]

Smart Achievers · Accepted Answer

(4)

We have,

$\lim_{x \to 1^{+}} \frac{(x - 1) (6 + λ \cos (x - 1)) + μ \sin (1 - x)}{{(x - 1)}^{3}} = - 1$

Put x – 1 = t

$\Rightarrow \lim_{t \to 0^{+}} \frac{t (6 + λ \cos t) - μ \sin t}{t^{3}} = - 1$

$\Rightarrow \lim_{t \to 0^{+}} \frac{t [6 + λ (1 - \frac{t^{2}}{2!} + . . . \infty)] - μ [t - \frac{t^{3}}{3!} . . . \infty]}{t^{3}} = - 1$

Now, $6 + λ - μ = 0$ ...(i)

$\frac{- λ}{2} + \frac{μ}{6} = - 1 \Rightarrow 3 λ - μ = 6$ ...(ii)

From (i) and (ii), we get $λ = 6$ , $μ = 12$

$∴ λ + μ = 18$ .

Solution

Q.
If $\lim_{x \to 1^{+}} \frac{(x - 1) (6 + λ \cos (x - 1)) + μ \sin (1 - x)}{{(x - 1)}^{3}} = - 1$ , where $λ, μ \in R$ , then $λ + μ$ is equal to [2025]

1 20

2 19

3 17

4 18

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