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JEE Mathematics 2025 Question with Solution

If lim⁡x→1(x−1)(6+λcos⁡(x−1))+μsin⁡(1−x)(x−1)3=−1\lim_{x \to 1} \frac{(x-1)(6+\lambda \cos(x-1)) + \mu \sin(1-x)}{(x-1)^3} = -1, where λ,μ∈R\lambda, \mu \in \mathbb{R}, then λ+μ\lambda + \mu is equal to

  • A

    1818

  • B

    2020

  • C

    1919

  • D

    1717

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

lim⁡x→1(x−1)(6+λcos⁡(x−1))+μsin⁡(1−x)(x−1)3=−1\lim_{x \to 1} \frac{(x-1)(6+\lambda \cos(x-1)) + \mu \sin(1-x)}{(x-1)^3} = -1

Find: λ+μ\lambda + \mu

Let

h=x−1h = x-1

so that h→0h \to 0 as x→1x \to 1. Then the limit becomes

lim⁡h→0h(6+λcos⁡h)−μsin⁡hh3=−1\lim_{h \to 0} \frac{h(6 + \lambda \cos h) - \mu \sin h}{h^3} = -1

Use the Taylor expansions near h=0h=0:

cos⁡h=1−h22+O(h4)\cos h = 1 - \frac{h^2}{2} + O(h^4) sin⁡h=h−h36+O(h5)\sin h = h - \frac{h^3}{6} + O(h^5)

Substituting these in the numerator,

h(6+λcos⁡h)−μsin⁡h=h(6+λ(1−h22+O(h4)))−μ(h−h36+O(h5))=h(6+λ)−λ2h3−μh+μ6h3+O(h5)=h((6+λ)−μ)+h3(−λ2+μ6)+O(h5)\begin{aligned} h(6+\lambda \cos h) - \mu \sin h &= h\left(6+\lambda\left(1-\frac{h^2}{2}+O(h^4)\right)\right) - \mu\left(h-\frac{h^3}{6}+O(h^5)\right) \\ &= 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\left(-\frac{\lambda}{2}+\frac{\mu}{6}\right) + O(h^5) \end{aligned}

Therefore,

h(6+λcos⁡h)−μsin⁡hh3=(6+λ)−μh2+(−λ2+μ6)+O(h2)\frac{h(6+\lambda \cos h) - \mu \sin h}{h^3} = \frac{(6+\lambda)-\mu}{h^2} + \left(-\frac{\lambda}{2}+\frac{\mu}{6}\right) + O(h^2)

For the limit to be finite, we must have

6+λ−μ=06+\lambda-\mu = 0

so,

μ=6+λ\mu = 6+\lambda

Now equate the constant term with the given limit −1-1:

−λ2+μ6=−1-\frac{\lambda}{2}+\frac{\mu}{6} = -1

Substitute μ=6+λ\mu = 6+\lambda:

−λ2+6+λ6=−1-\frac{\lambda}{2} + \frac{6+\lambda}{6} = -1 1−λ3=−11 - \frac{\lambda}{3} = -1 λ=6\lambda = 6

Hence,

μ=12\mu = 12

Therefore,

λ+μ=18\lambda + \mu = 18

The correct option is A.

Coefficient Comparison

Given: the numerator is divided by (x−1)3(x-1)^3 and the limit equals −1-1.

Find: values of λ\lambda and μ\mu sufficient to compute λ+μ\lambda+\mu.

After putting h=x−1h=x-1, write

N(h)=h(6+λcos⁡h)−μsin⁡hN(h)=h(6+\lambda \cos h)-\mu \sin h

Using

cos⁡h≈1−h22,sin⁡h≈h−h36\cos h \approx 1-\frac{h^2}{2}, \qquad \sin h \approx h-\frac{h^3}{6}

we get

N(h)=h(6+λ(1−h22))−μ(h−h36)N(h)=h\left(6+\lambda\left(1-\frac{h^2}{2}\right)\right)-\mu\left(h-\frac{h^3}{6}\right)

Expanding and collecting powers of hh,

N(h)=h(6+λ−μ)+h3(−λ2+μ6)N(h)=h(6+\lambda-\mu)+h^3\left(-\frac{\lambda}{2}+\frac{\mu}{6}\right)

Since the denominator is h3h^3, the coefficient of hh must be zero. Hence,

6+λ−μ=06+\lambda-\mu=0

Also, the coefficient of h3h^3 must equal the limit −1-1, so

−λ2+μ6=−1-\frac{\lambda}{2}+\frac{\mu}{6}=-1

Solving,

μ=6+λ\mu=6+\lambda −λ2+6+λ6=−1-\frac{\lambda}{2}+\frac{6+\lambda}{6}=-1 λ=6,μ=12\lambda=6, \qquad \mu=12

Therefore, λ+μ=18\lambda+\mu=18, so the correct option is A.

Common mistakes

  • Ignoring that sin⁡(1−x)=−sin⁡(x−1)\sin(1-x)= -\sin(x-1) is incorrect because the sign changes when the argument is negated. Use sin⁡(−h)=−sin⁡h\sin(-h)=-\sin h after substituting h=x−1h=x-1.

  • Keeping only the first-order terms and stopping too early is wrong because the denominator is h3h^3. You must expand up to the h3h^3 terms in cos⁡h\cos h and sin⁡h\sin h.

  • Not forcing the coefficient of hh in the numerator to vanish is a conceptual error. If that term remains, then dividing by h3h^3 makes the limit blow up like 1h2\frac{1}{h^2} instead of giving a finite value.

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