The smallest interval [a, , ,b] such that _0^1 dx 1 + x^4 [a, , ,… - Vidyadip

MCQ

The smallest interval $[a,\,\,b]$ such that $\int_0^1 {\frac{{dx}}{{\sqrt {1 + {x^4}} }}} \in [a,\,\,b]$ is given by

✓
$\left[ {\frac{1}{{\sqrt 2 }},\,\,1} \right]$
B
$[0,\,\,1]$
C
$\left[ {\frac{1}{2},\,\,2} \right]$
D
$\left[ {\frac{3}{4},\,\,1} \right]$

✓

Answer

Correct option: A.

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

a
(a) Let $I = \int_0^1 {\frac{{dx}}{{\sqrt {1 + {x^4}} }}} $

Here, $0 \le x \le 1 \Rightarrow 1 \le (1 + {x^4}) \le 2$

==> $1 \le \sqrt {1 + {x^4}} \le \sqrt 2 $

$\Rightarrow \frac{1}{{\sqrt 2 }} \le \frac{1}{{\sqrt {1 + {x^4}} }} \le 1$

==> $\frac{1}{{\sqrt 2 }} \le \int_0^1 {\frac{{dx}}{{\sqrt {1 + {x^4}} }} \le 1} $

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

Note: If $m = $ least value of $f(x)$ and $M=$ greatest value of $f(x)$ in $[a, b],$ then

$m(b - a) \le \int_a^b {f(x)dx \le M(b - a)} $.

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