Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

Answer:

Option 2

Step-by-step explanation:

At first,we manipulate the denominator and eliminate sine function on basis of a standard formula.

We then apply L-Hospital rule as the expression is indeterminate.

To differentiate the numerator which is a definite integral,apply Leibnitz rule(Rule can be found from your book easily).

Finally,substituting 1,we get our answer.

Answer 2

we have to find the value of $\displaystyle\lim_{x\to 1}\frac{\int\limits^{(x-1)^2}_0{t cost}\,dt}{(x-1)sin(x-1)}$

solution : This type of questions can be solved easily with the help of L - Hospital rule,

applying L - Hospital rule,

differentiating both sides,

differentiation of $\int\limits^{(x-1)^2}_0{tcost^2}\,dt=(x-1)^2cos(x-1)^4\frac{d(x-1)^2}{dx}$

$=\quad\displaystyle\lim{x\to 1}\frac{2(x-1)(x-1)^2cos(x-1)^4-0}{(x-1)cos(x-1)+sin(x-1)}$

$=\displaystyle\lim_{x\to 1}\frac{2(x-1)^3cos(x-1)^4}{(x-1)cos(x-1)+\frac{sin(x-1)}{(x-1)}}$

$=\displaystyle\lim_{x\to 1}\frac{2(x-1)^2cos(x-1)^4}{cos(x-1)+\frac{sin(x-1)}{(x-1)}}$

$=\frac{\displaystyle\lim_{x\to 1}\left[2(x-1)^2cos(x-1)^4\right]}{\displaystyle\lim_{x\to 1}\left[cos(x-1)+\frac{sin(x-1)}{(x-1)}\right]}$

on taking limit we get,

= 0/(1 + 1) = 0

Therefore the value of given limit must be zero. so the correct option is (2)