NDA 2023
Exam NDA
Year 2023
Subject Mathematics
Chapter Limits NDA 2023 · Mathematics · Limits
Let f(x) a^ x-1 b^ x-1 and g(x)=x-1 . What is _ x arrow 1 f(x)^...
Let
f ( x ) a x − 1 b x − 1 f(x) \quad \frac{a^{x-1} b^{x-1}}{} f ( x ) a x − 1 b x − 1 and
g ( x ) = x − 1 g(x)=x-1 g ( x ) = x − 1 .
What is
lim x → 1 f ( x ) 1 g ( x ) \lim _{x \rightarrow 1} f(x)^{\frac{1}{g(x)}} lim x → 1 f ( x ) g ( x ) 1 equal to?
Options A a b \sqrt{a b} ab B a b a b ab C 2 a b 2 a b 2 ab D a b 2 \frac{\sqrt{a b}}{2} 2 ab Correct answer A. a b \sqrt{a b} ab
Step-by-step solution
Since,
Since, lim x → 1 f ( x ) 1 g ( x ) = lim x → 1 ( a x − 1 + b x − 1 2 ) 1 x − 1 \begin{aligned}
& \text{ Since, } \lim _{x \rightarrow 1} f(x) \frac{1}{g(x)} \\
& =\lim _{x \rightarrow 1}\left(\frac{a^{x-1}+b^{x-1}}{2}\right)^{\frac{1}{x-1}}
\end{aligned} Since, x → 1 lim f ( x ) g ( x ) 1 = x → 1 lim ( 2 a x − 1 + b x − 1 ) x − 1 1
(it is form
1 ∞ 1^{\infty} 1 ∞ )
= e lim x → 1 ( a a − 1 + b x − 1 2 − 1 ) 1 x − 1 = e lim x → m ( a x − 1 − 1 2 ( x − 1 ) + b x − 1 − 1 2 ( x − 1 ) ) = e ( 1 2 ln a + 1 2 mb b ) = e 1 2 ln ( a b ) = a b \begin{aligned}
& =\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 1}\left(\frac{\mathrm{a}^{\mathrm{a}-1}+\mathrm{b}^{\mathrm{x}-1}}{2}-1\right) \frac{1}{\mathrm{x}-1}}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow \mathrm{ m}}\left(\frac{\mathrm{a}^{\mathrm{x}-1}-1}{2(\mathrm{x}-1)}+\frac{\mathrm{b}^{\mathrm{x}-1}-1}{2(\mathrm{x}-1)}\right)} \\
& =e^{\left(\frac{1}{2} \ln a+\frac{1}{2} \operatorname{mb} b\right)}=e^{\frac{1}{2} \ln (a b)}=\sqrt{a b}
\end{aligned} = e l i m x → 1 ( 2 a a − 1 + b x − 1 − 1 ) x − 1 1 = e l i m x → m ( 2 ( x − 1 ) a x − 1 − 1 + 2 ( x − 1 ) b x − 1 − 1 ) = e ( 2 1 l n a + 2 1 mb b ) = e 2 1 l n ( ab ) = ab
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