Assertion (A): The maximum value of log13(x2−4x+5) is zeroReason (R): logax≤0 for x≥1 and 0<a<1

Let x2−4x+5= x2−4x+4+1= (x−2)2+1≥1also 0< a=13<1⇒ log13 (x2−4x+5)≤0as we know that logax ≤0 for x≥1 and 0<a<1Assertion (A) is trueReason is also True

Assertion (A): The maximum value of log13(x2−4x+5) is zeroReason (R): logax≤0 for x≥1 and 0<a<1

Let x2−4x+5= x2−4x+4+1= (x−2)2+1≥1also 0< a=13<1⇒ log13 (x2−4x+5)≤0as we know that logax ≤0 for x≥1 and 0<a<1Assertion (A) is trueReason is also True

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$\begin{array}{l} A s s e r t i o n (A) : T h e \max i m u m v a l u e o f \log_{\frac{1}{3}} (x^{2} - 4 x + 5) i s z e r o \\ Re a s o n (R) : \log_{a} x \leq 0 f o r x \geq 1 a n d 0 < a < 1 \end{array}$

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$B o t h (A) (R) a r e t r u e a n d (R) i s t h e c o r r e c t \exp l a n a t i o n o f (A)$

$B o t h (A) a n d (R) a r e f a l s e$

$(A) i s t r u e b u t (R) i s f a l s e$

$B o t h (A) a n d (R) a r e t r u e b u t (R) i s n o t c o r r e c t \exp l a n a t i o n o f (A)$

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Detailed Solution

$\begin{array}{l} L e t x^{2} - 4 x + 5 = x^{2} - 4 x + 4 + 1 \\ = {(x - 2)}^{2} + 1 \geq 1 \\ a l s o 0 < a = \frac{1}{3} < 1 \\ \Rightarrow \log_{\frac{1}{3}} (x^{2} - 4 x + 5) \leq 0 \\ a s w e k n o w t h a t \log_{a}^{x} \leq 0 f o r x \geq 1 a n d 0 < a < 1 \\ A s s e r t i o n (A) i s t r u e \\ Re a s o n i s a l s o T r u e \end{array}$