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Question: The smallest interval \([ a , b ]\) such that \(\int _ { 0 } ^ { 1 } \frac { d x } { \sqrt { 1 + x ^...

The smallest interval [a,b][ a , b ] such that ∫01dx1+x4∈[a,b]\int _ { 0 } ^ { 1 } \frac { d x } { \sqrt { 1 + x ^ { 4 } } } \in [ a , b ] is given by

A

[12,1]\left[ \frac { 1 } { \sqrt { 2 } } , 1 \right]

B

[0,1][ 0,1 ]

C

[12,2]\left[ \frac { 1 } { 2 } , 2 \right]

D

[34,1]\left[ \frac { 3 } { 4 } , 1 \right]

Answer

[12,1]\left[ \frac { 1 } { \sqrt { 2 } } , 1 \right]

Explanation

Solution

Let I=∫01dx1+x4I = \int _ { 0 } ^ { 1 } \frac { d x } { \sqrt { 1 + x ^ { 4 } } }

Here, 0≤x≤1⇒1≤(1+x4)≤20 \leq x \leq 1 \Rightarrow 1 \leq \left( 1 + x ^ { 4 } \right) \leq 2

⇒ 1≤1+x4≤2⇒12≤11+x4≤11 \leq \sqrt { 1 + x ^ { 4 } } \leq \sqrt { 2 } \Rightarrow \frac { 1 } { \sqrt { 2 } } \leq \frac { 1 } { \sqrt { 1 + x ^ { 4 } } } \leq 1

⇒ 12≤∫01dx1+x4≤1\frac { 1 } { \sqrt { 2 } } \leq \int _ { 0 } ^ { 1 } \frac { d x } { \sqrt { 1 + x ^ { 4 } } } \leq 1

Hence [12,1]\left[ \frac { 1 } { \sqrt { 2 } } , 1 \right] is the smallest interval, such that I∈[12,1]I \in \left[ \frac { 1 } { \sqrt { 2 } } , 1 \right]

Note: If m(b−a)≤∫abf(x)dx≤M(b−a)m ( b - a ) \leq \int _ { a } ^ { b } f ( x ) d x \leq M ( b - a ).