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QJEEM21Ma1a45: JEE Main 2021 Mathematics - Class 12 JEEM21Ma1a45 Solution: If , then a + b + c is equal to ____________.

If lim⁡x→0aex−bcos⁡x+ce−xxsin⁡x=2\mathop {\lim }\limits_{x \to 0} {{a{e^x} - b\cos x + c{e^{ - x}}} \over {x\sin x}} = 2 , then a + b + c is equal to ____________.

Step-by-Step Verified Solution

lim⁡x→0{a(1+x+x22!+.....)−b(1−x22!+x44!......)+c(1−x+x22!......)}x(x−x33!+.....)=2\mathop {\lim }\limits_{x \to 0} {{\left\{ {a\left( {1 + x + {{{x^2}} \over {2!}} + .....} \right) - b\left( {1 - {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}}......} \right) + c\left( {1 - x + {{{x^2}} \over {2!}}......} \right)} \right\}} \over {x\left( {x - {{{x^3}} \over {3!}} + .....} \right)}} = 2

∴\therefore lim⁡x→0(a−b+c)+x(a−c)+x2(a2+b2+c2)+....x2(1−x26....)=2\mathop {\lim }\limits_{x \to 0} {{(a - b + c) + x(a - c) + {x^2}\left( {{a \over 2} + {b \over 2} + {c \over 2}} \right) + ....} \over {{x^2}\left( {1 - {{{x^2}} \over 6}....} \right)}} = 2

For this limit to exist

a −- b + c = 0 & a −- c = 0

& a2+b2+c2=2{a \over 2} + {b \over 2} + {c \over 2} = 2

⇒\Rightarrow a + b + c = 4

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