If a>0, Ltx→a ax−xaxx−aa=−1 then 'a'=

If a>0, Ltx→a ax−xaxx−aa=−1⇒Ltx→aaxloga−axa−1xx(1+logx)=−1⇒aaloga−aaa−1aa(1+loga)=−1⇒aa(loga−1)aa(loga+1)=−1⇒loga−1=−loga−1⇒2loga=0⇒loga=0⇒a=1

If a>0, Ltx→a ax−xaxx−aa=−1 then 'a'=

If a>0, Ltx→a ax−xaxx−aa=−1⇒Ltx→aaxloga−axa−1xx(1+logx)=−1⇒aaloga−aaa−1aa(1+loga)=−1⇒aa(loga−1)aa(loga+1)=−1⇒loga−1=−loga−1⇒2loga=0⇒loga=0⇒a=1

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If $a > 0, \underset{x \to a}{L t} \frac{a^{x} - x^{a}}{x^{x} - a^{a}} = - 1$ then $' a' =$

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0 $\begin{array}{l} \Rightarrow \underset{x \to a}{L t} \frac{a^{x} \log a - a x^{a - 1}}{x^{x} (1 + \log x)} = - 1 \\ \Rightarrow \frac{a^{a} \log a - a a^{a - 1}}{a^{a} (1 + \log a)} = - 1 \\ \Rightarrow \frac{a^{a} (\log a - 1)}{a^{a} (\log a + 1)} = - 1 \\ \Rightarrow \log a - 1 = - \log a - 1 \\ \Rightarrow 2 \log a = 0 \\ \Rightarrow \log a = 0 \\ \Rightarrow a = 1 \end{array}$

answer is B.

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Detailed Solution

If $a > 0, \underset{x \to a}{L t} \frac{a^{x} - x^{a}}{x^{x} - a^{a}} = - 1$

$\begin{array}{l} \Rightarrow \underset{x \to a}{L t} \frac{a^{x} \log a - a x^{a - 1}}{x^{x} (1 + \log x)} = - 1 \\ \Rightarrow \frac{a^{a} \log a - a a^{a - 1}}{a^{a} (1 + \log a)} = - 1 \\ \Rightarrow \frac{a^{a} (\log a - 1)}{a^{a} (\log a + 1)} = - 1 \\ \Rightarrow \log a - 1 = - \log a - 1 \\ \Rightarrow 2 \log a = 0 \\ \Rightarrow \log a = 0 \\ \Rightarrow a = 1 \end{array}$