If ₓ arrow 0}\{1+x (1²)\}¹/x}=2 a ² θ, a>0 and θ R, then _____

Correct Answer is : θ=nπ±π2,(n∈Z)θ=n π ± π/2,(n Z)θ=nπ±2π,(n∈Z)

Correct Answer is : θ=nπ±π2,(n∈Z)θ=n π ± π/2,(n Z)θ=nπ±2π,(n∈Z)

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AP EAPCET 25-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 65 of 160

Marks: +1, -0

\lim\limits_{x \rightarrow 0}\{1+x \log (1^{2})\}^{\frac{1}{x}}=2 a \sin ^{2} \theta, a>0

\theta \in R

, then _____

Solution: 👈: Video Solution

\lim\limits_{x\rightarrow 0}\{1+x \log (1+a^{2})\}^{\frac{1}{x}}

=2 a \sin ^{2} \theta, a>0

and

\theta \in R

LHS

\lim\limits_{x\rightarrow 0}\{1+x \log (1+a^{2})\}^{\frac{1}{x}}

is of the form ∞1

⇒

e^{\lim\limits_{x\rightarrow 0}\frac{1}{x}\{1+x \log (1+a^{2})-1\}}

⇒

\lim\limits_{x\rightarrow 0}\log (1+a^{2})

⇒

1+a^{2}=2 a \sin ^{2} \theta

\Rightarrow a^{2}-2 a \sin ^{2} \theta+1=0

⇒

a=\frac{2 \sin ^{2} \theta \pm \sqrt{4 \sin ^{4} \theta-4}}{2}

a=\frac{2 \sin ^{2} \theta \pm \sqrt{4(\sin ^{4} \theta-1)}}{2}

a=\sin ^{2} \theta \pm \sqrt{\sin ^{4} \theta-1}

∵ a > 0

⇒

\sin ^{4} \theta=1

\sin ^{2} \theta=1=\sin ^{2} \frac{\pi}{2}

⇒

\theta=n \pi \pm \frac{\pi}{2}

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