If lim x 0 e ( a - 1 ) x + 2 cos b x + ( c - 2 ) e - x x cos x - log e ( 1 + x ) = 2 , then a 2 + b 2 + c 2 is equal to: [2026]

Question

If $\lim_{x \to 0} \frac{e^{(a - 1) x} + 2 \cos b x + (c - 2) e^{- x}}{x \cos x - \log_{e} (1 + x)} = 2,$ then $a^{2} + b^{2} + c^{2}$ is equal to: [2026]

Smart Achievers · Accepted Answer

(1)

$\lim_{x \to 0} \frac{(1 + (a - 1) x + \frac{{(a - 1)}^{2} x^{2}}{2!}) + 2 (1 - \frac{b^{2} x^{2}}{2!}) + (c - 2) (1 - x + \frac{x^{2}}{2!})}{x (1 - \frac{x^{2}}{2!}) - (x - \frac{x^{2}}{2} . . .)} = 2$

$\lim_{x \to 0} \frac{(1 + 2 + c - 2) + x (a - 1 - c + 2) + x^{2} (\frac{{(a - 1)}^{2}}{2} - b^{2} + (\frac{c - 2}{2}))}{\frac{x^{2}}{2!} - \frac{x^{3}}{2!} + \dots} = 2$

$For which$

$∵$ $c + 1 = 0 \Rightarrow c = - 1$

$∵$ $a - c = - 1 \Rightarrow a = - 2$

$∵$ $\frac{{(a - 1)}^{2}}{2} - b^{2} + (\frac{c - 2}{2}) = 1$

$\frac{9}{2} - b^{2} - \frac{3}{2} = 1 \Rightarrow b^{2} = 2$

$a^{2} + b^{2} + c^{2} = 4 + 2 + 1 = 7$

Solution

Q.
If $\lim_{x \to 0} \frac{e^{(a - 1) x} + 2 \cos b x + (c - 2) e^{- x}}{x \cos x - \log_{e} (1 + x)} = 2,$ then $a^{2} + b^{2} + c^{2}$ is equal to: [2026]

1 7

2 3

3 5

4 9

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Solution

Q. If limx→0e(a-1)x+2cosbx+(c-2)e-xxcosx-loge(1+x)=2, then a2+b2+c2 is equal to: [2026]

1 7

2 3

3 5

4 9

Ans. (1) limx→0(1+(a-1)x+(a-1)2x22!)+2(1-b2x22!)+(c-2)(1-x+x22!)x(1-x22!)-(x-x22...)=2 limx→0(1+2+c-2)+x(a-1-c+2)+x2((a-1)22-b2+(c-22))x22!-x32!+⋯=2 For which ∵ c+1=0⇒c=-1 ∵ a-c=-1⇒a=-2 ∵ (a-1)22-b2+c-22=1 92-b2-32=1⇒b2=2 a2+b2+c2=4+2+1=7

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Q.
If $\lim_{x \to 0} \frac{e^{(a - 1) x} + 2 \cos b x + (c - 2) e^{- x}}{x \cos x - \log_{e} (1 + x)} = 2,$ then $a^{2} + b^{2} + c^{2}$ is equal to: [2026]