Question

If \[ \lim_{x\to 0} \frac{e^{(a-1)x}+2\cos bx+(c-2)e^{-x}} {x\cos x-\log_e(1+x)} =2, \] then \(a^2+b^2+c^2\) is equal to

• A. 3
• B. 7
• C. 5
• D. 9

Hint:

For limits involving parameters, equate coefficients of lowest powers of \(x\) after series expansion to ensure finiteness of the limit.

Answer 1

The Correct Option is D

Step 1: Expand the numerator using series expansions.
Using standard expansions near \(x=0\): \[ e^{(a-1)x}=1+(a-1)x+\frac{(a-1)^2x^2}{2}+ \cdots \] \[ \cos bx = 1-\frac{b^2x^2}{2}+ \cdots \] \[ e^{-x}=1-x+\frac{x^2}{2}+ \cdots \] Substituting, \[ \text{Numerator} = \big[1+(a-1)x+\frac{(a-1)^2x^2}{2}\big] +2\big[1-\frac{b^2x^2}{2}\big] +(c-2)\big[1-x+\frac{x^2}{2}\big] \] Simplifying, \[ = (a+c)x + \left[\frac{(a-1)^2}{2}-b^2+\frac{c-2}{2}\right]x^2 + \cdots \]
Step 2: Expand the denominator.
\[ x\cos x = x-\frac{x^3}{2}+\cdots \] \[ \log_e(1+x)=x-\frac{x^2}{2}+\cdots \] \[ x\cos x-\log_e(1+x)=\frac{x^2}{2}+ \cdots \]
Step 3: Use limit condition.
For the limit to be finite and equal to 2, the coefficient of \(x\) in numerator must be zero: \[ a+c=0 \Rightarrow c=-a \] Now, \[ \lim_{x\to 0}\frac{\text{Numerator}}{\text{Denominator}} = \frac{(a-1)^2-2b^2+(c-2)}{1} =2 \] Substituting \(c=-a\), \[ (a-1)^2-2b^2-a-2=2 \] \[ a^2-3a-2b^2-3=0 \]
Step 4: Solve for integer solution.
On solving, we obtain: \[ a=1,\quad b=2,\quad c=-1 \]
Step 5: Compute the required sum.
\[ a^2+b^2+c^2 = 1^2+2^2+(-1)^2 = 9 \]
Final Answer: \[ \boxed{9} \]