If limx→0αex+βe−x+γsinxxsin2x=23, where α, β, γ∈R, then which of the following is NOT correct?

limx→0 αex+βe−x+γsin⁡xx3sin⁡xx2=23⇒limx→0 αex+βe−x+γsin⁡xx3=23∴α+β=0……(1)00f0rmlimx→0 αex−βe−x+γcos⁡x3x2=23∴α−β+γ=0… (2) 00 form limx→0 αex+βe−x−γsin⁡x6x=23∴α+β=000 form limx→0 αex−βe−x−γcos⁡x6=23⇒α−β−γ6=23⇒α−β−γ=4. Solving (1), (2),(3) γ=−2,α=1,β=−1.

If limx→0αex+βe−x+γsinxxsin2x=23, where α, β, γ∈R, then which of the following is NOT correct?

limx→0 αex+βe−x+γsin⁡xx3sin⁡xx2=23⇒limx→0 αex+βe−x+γsin⁡xx3=23∴α+β=0……(1)00f0rmlimx→0 αex−βe−x+γcos⁡x3x2=23∴α−β+γ=0… (2) 00 form limx→0 αex+βe−x−γsin⁡x6x=23∴α+β=000 form limx→0 αex−βe−x−γcos⁡x6=23⇒α−β−γ6=23⇒α−β−γ=4. Solving (1), (2),(3) γ=−2,α=1,β=−1.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $\lim_{x \to 0} \frac{{αe}^{x} + {βe}^{- x} + γsinx}{{xsin}^{2} x} = \frac{2}{3},$ where $α, β, γ \in R,$ then which of the following is NOT correct?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$α^{2} + β^{2} + γ^{2} = 6$

$αβ + βγ + γα + 1 = 0$

${αβ}^{2} + {βγ}^{2} + {γα}^{2} + 3 = 0$

$α^{2} - β^{2} + γ^{2} = 4$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} lim_{x \to 0} \frac{{αe}^{x} + {βe}^{- x} + γsin x}{x^{3} {(\frac{\sin x}{x})}^{2}} = \frac{2}{3} \\ \Rightarrow lim_{x \to 0} \frac{{αe}^{x} + {βe}^{- x} + γsin x}{x^{3}} = \frac{2}{3} \\ ∴ α + β = 0 \dots \dots (1) (\frac{0}{0} f 0 rm) \\ lim_{x \to 0} \frac{{αe}^{x} - {βe}^{- x} + γcos x}{3 x^{2}} = \frac{2}{3} \\ ∴ α - β + γ = 0 \dots (2) (\frac{0}{0} form) \\ lim_{x \to 0} \frac{{αe}^{x} + {βe}^{- x} - γsin x}{6 x} = \frac{2}{3} \\ ∴ α + β = 0 (\frac{0}{0} form) \\ lim_{x \to 0} \frac{{αe}^{x} - {βe}^{- x} - γcos x}{6} = \frac{2}{3} \\ \Rightarrow \frac{α - β - γ}{6} = \frac{2}{3} \Rightarrow α - β - γ = 4 . \\ Solving (1), (2),(3) \\ γ = - 2, α = 1, β = - 1 . \end{array}$