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JEE Mathematics 2026 Question with Solution

If lim⁡x→0e(a−1)x+2cos⁡bx+(c−2)e−xxcos⁡x−log⁡e(1+x)=2\lim_{x\to 0} \frac{e^{(a-1)x}+2\cos bx+(c-2)e^{-x}} {x\cos x-\log_e(1+x)} =2, then a2+b2+c2a^2+b^2+c^2 is equal to

  • A

    33

  • B

    77

  • C

    55

  • D

    99

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

lim⁡x→0e(a−1)x+2cos⁡bx+(c−2)e−xxcos⁡x−log⁡e(1+x)=2\lim_{x\to 0} \frac{e^{(a-1)x}+2\cos bx+(c-2)e^{-x}} {x\cos x-\log_e(1+x)} =2

Find: a2+b2+c2a^2+b^2+c^2

Expand the numerator near x=0x=0 using standard series:

e(a−1)x=1+(a−1)x+(a−1)2x22+⋯e^{(a-1)x}=1+(a-1)x+\frac{(a-1)^2x^2}{2}+ \cdots cos⁡bx=1−b2x22+⋯\cos bx = 1-\frac{b^2x^2}{2}+ \cdots e−x=1−x+x22+⋯e^{-x}=1-x+\frac{x^2}{2}+ \cdots

Substituting these into the numerator,

Numerator=[1+(a−1)x+(a−1)2x22]+2[1−b2x22]+(c−2)[1−x+x22]\text{Numerator} = \big[1+(a-1)x+\frac{(a-1)^2x^2}{2}\big] +2\big[1-\frac{b^2x^2}{2}\big] +(c-2)\big[1-x+\frac{x^2}{2}\big]

Simplifying,

=(a+c)x+[(a−1)22−b2+c−22]x2+⋯= (a+c)x + \left[\frac{(a-1)^2}{2}-b^2+\frac{c-2}{2}\right]x^2 + \cdots

Now expand the denominator:

xcos⁡x=x−x32+⋯x\cos x = x-\frac{x^3}{2}+\cdots log⁡e(1+x)=x−x22+⋯\log_e(1+x)=x-\frac{x^2}{2}+\cdots

Therefore,

xcos⁡x−log⁡e(1+x)=x22+⋯x\cos x-\log_e(1+x)=\frac{x^2}{2}+ \cdots

For the limit to be finite and equal to 22, the coefficient of xx in the numerator must be zero:

a+c=0⇒c=−aa+c=0 \Rightarrow c=-a

Using the quadratic term,

lim⁡x→0NumeratorDenominator=(a−1)2−2b2+(c−2)1=2\lim_{x\to 0}\frac{\text{Numerator}}{\text{Denominator}} = \frac{(a-1)^2-2b^2+(c-2)}{1} =2

Substituting c=−ac=-a,

(a−1)2−2b2−a−2=2(a-1)^2-2b^2-a-2=2 a2−3a−2b2−3=0a^2-3a-2b^2-3=0

On solving, we obtain:

a=1,b=2,c=−1a=1,\quad b=2,\quad c=-1

Hence,

a2+b2+c2=12+22+(−1)2=9a^2+b^2+c^2 = 1^2+2^2+(-1)^2 = 9

Therefore, the correct option is D.

Series Coefficient Matching

Given: the limit equals 22. Find: the value of a2+b2+c2a^2+b^2+c^2.

The denominator starts from order x2x^2, since

xcos⁡x−log⁡e(1+x)=x22+⋯x\cos x-\log_e(1+x)=\frac{x^2}{2}+\cdots

So the numerator must not contain a nonzero linear term; otherwise the quotient would diverge like 1x\frac{1}{x}.

From the numerator expansion, the linear coefficient is a+ca+c. Therefore,

a+c=0a+c=0

which gives

c=−ac=-a

Next compare the coefficients of x2x^2 because that determines the finite limit:

(a−1)22−b2+c−2212=2\frac{\frac{(a-1)^2}{2}-b^2+\frac{c-2}{2}}{\frac{1}{2}}=2

Multiplying by 22,

(a−1)2−2b2+(c−2)=2(a-1)^2-2b^2+(c-2)=2

Substitute c=−ac=-a:

(a−1)2−2b2−a−2=2(a-1)^2-2b^2-a-2=2 a2−3a−2b2−3=0a^2-3a-2b^2-3=0

Using the solution obtained in the working,

a=1, b=2, c=−1a=1,\ b=2,\ c=-1

Finally,

a2+b2+c2=1+4+1=9a^2+b^2+c^2=1+4+1=9

Therefore, the required value is 99.

Common mistakes

  • Students may ignore the linear term in the numerator. That is incorrect because the denominator begins with x2x^2, so a nonzero xx term would make the limit unbounded. Set the coefficient of xx equal to zero first.

  • Students may expand cos⁡bx\cos bx incorrectly as 1−bx221-\frac{bx^2}{2}. This is wrong because the correct expansion is 1−b2x22+⋯1-\frac{b^2x^2}{2}+\cdots. Always square the argument coefficient in the cosine series.

  • Students may forget that dividing by x22\frac{x^2}{2} doubles the numerator's x2x^2 coefficient. This leads to an incorrect equation for the parameters. Compare coefficients carefully after forming the quotient.

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