If a -3 e - b -3 and c - e 3 - thenA- b - a - cB- a - c - bC- a - b - cD- c - a - b

The correct option is C a lt;b lt;c Let f(x)=x^1/x,x gt;0 logf(x)=1/x logx ⇒f'(x)/f(x)=1/x^2 1/x^2 logx=1 logx/x^2 ⇒f'(x)=1/x^2f(x)(1 logx) lt;0 if ...

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Byju's Answer

Standard XI

Mathematics

Graphical Representation of Logarithm Function

If a =π3 e , ...

Question

If $a = π^{3 e}, b = 3^{π e} a n d c = e^{3 π},$ then

a<b<c

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b<a<c

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c<a<b

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a<c<b

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Solution

The correct option is A
a<b<c

Let $f (x) = x^{1 / x}, x > 0 l o g f (x) = \frac{1}{x} l o g x \Rightarrow \frac{f^{'} (x)}{f (x)} = \frac{1}{x^{2}} - \frac{1}{x^{2}} l o g x = \frac{1 - l o g x}{x^{2}} \Rightarrow f^{'} (x) = \frac{1}{x^{2}} f (x) (1 - l o g x) < 0 i f x > e$
Thus, f(x) decreases on $(e, \infty)$ .
As e<3< $π$ , we get
$f (π) < f (3) < f (e) \Rightarrow (π)^{1 / π} < 3^{1 / 3} < e^{1 / e} \Rightarrow π^{3 e} < 3^{π e} < e^{3 π} o r a < b < c .$

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If a=π3e,b=3πe and c=e3π, then

The correct option is A a<b<c Let f(x)=x1/x,x>0logf(x)=1xlogx⇒f′(x)f(x)=1x2−1x2 logx=1−logxx2⇒f′(x)=1x2f(x)(1−logx)<0 if x>e Thus, f(x) decreases on (e,∞). As e<3< π, we get f(π)<f(3)<f(e)⇒(π)1/π<31/3<e1/e⇒π3e<3πe<e3π or a<b<c.

If $a = π^{3 e}, b = 3^{π e} a n d c = e^{3 π},$ then