JEE maths#2

Calculus Level 4

If $\displaystyle \lim_{x \to 0} \int_0^x \frac {t^2 \ dt}{(x-\sin x)\sqrt{a+t}} = 1$ , what is the value of $a$ ?

1 solution

Prakhar Bindal
Mar 29, 2017

Firstly we observe that its 0/0 form (taking x-sinx out of integral as its not a function of t)

Differentiate up and down using Newton Leibniz Rule

To get

Limit as (x^2)/(1-cosx)sqrt(a)

Use standard limit x approaching zero (1-cosx)/x^2 = 1/2 which can be derived by using L Hospital's rule

So 2 = root a

a = 4

over rated, how ids mains prep ?

A Former Brilliant Member - 4 years, 2 months ago

Well going good . WBU?

Prakhar Bindal - 4 years, 2 months ago

i'm trying to studying a lot but no use...... in short : mri pdi h , CBSE ki chemistry ka ppr acha ni hua ta uski tension h .

A Former Brilliant Member - 4 years, 2 months ago

Your "in short " was amazing @shubham dhull

Aakash Khandelwal - 4 years, 2 months ago

LOL :P but what's true is true :| WBU ?

A Former Brilliant Member - 4 years, 2 months ago

What do we differentiate up and down ? After taking $x - sin x$ common do we actually need to do the integral, then differentiate or something else ?

Vishal Yadav - 4 years, 2 months ago

Using L'Hopital Rule.

Akshat Sharda - 3 years, 4 months ago

@Prakhar Bindal I have some silly doubts..

When x tends to zero, the upper and lower limit are very close and hence the area under the curve should be negligible (or zero). how can it be 1?
To apply Newton Leibnitz and L'Hopital rule, the integral sign should be separately in the numerator and denominator.

Would be glad if you could take some time to reply.

Arpan Sarangi - 3 years, 1 month ago