 # Doubt from Definite Integration

#1
#2

Observe that
(1+x^2)^2 = (1+x^4)+2x^2
For all x \geq 0 we can say
(1+x^2)^2 \geq 1+x^4
\cfrac{1}{(1+x^2)^2} \leq \cfrac{1}{1+x^4}
\Rightarrow \cfrac{1}{1+x^2} \leq \cfrac{1}{\sqrt{1+x^4}}
Integrating preserves the inequality hence integrate both sides with limits from 0 to 1
\displaystyle \int_\limits{0}^1 \cfrac{1}{1+x^2} \leq \int_\limits{0}^1 \cfrac{1}{\sqrt{1+x^4}}
\Rightarrow I \geq \tan^{-1} (1)
\Rightarrow I \geq \cfrac{\pi}{4}

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#3
#4
1 Like
#5
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#6
#7

One can easily see that for the first condition integral of sinx, a and b will have a difference of 4π. And from the second condition we can say that 0 and (a+b) will have a difference of 4.5π.
So
b-a=4π
a+b = 4.5π

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#8

How, I really don't getting a clue from the question?

#9
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#10
#11

apply lebniz but first apply definate integral property
fx=f(a+b-x) then apply product rule of differentiation
. @Gaurav_2020_3

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#12
#13

apply fenymen technique @Gaurav_2020_3

#14

Sorry, but i don't know about that.

#15

1 solution is fenymen technique (integration under diffrenciation).

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#16

I have solved but answer not matching pls check my answer

#17
#18

Sir, you have got correct function then what's wrong?
Ans - (d)

#19
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#20

I was in confusion with the interval of x given in option in answer
for increasing and decreasing function and the interval obtained
By me...that mens i am correct in finding interval for increasing
And decresing pls reply

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