# Prob 1If f (x) =intigralupper limit is x power 2and lower limit is 1/x squareroot of sin( t)dt, then find the value of f (1).Prob 2Let a = min x2 + 4x + 5, x ? R and b = lim??0 1-cos 2?/?squarethen find the value ofa-ba+b .Prob 3The value of the Integralupper limit is p/2lower limit is -p/2(ax^3cos x + b sin x + c)dx, depends on how many parameters among (a , b , c).Prob 4Find the value of the Integralupper limit is p/2lower limit is -p/2root of ( cos x - cos^3 x)dx.Prob 5Let f (x) =integralupper limit is xlower limit is 0( 2u+1/u^2-2u+2 )du then find the least value of f (x) in [-1,1].Prob 6Find the values of a for which the function f (x) = cos x - sin x + ax + b is decreasing.Prob 7Find the area bounded by the curves y=x|x| , x - axis , x = 1 & x = -1.Prob 8For x > 0 let f (x) =integralupper limit is x lower limit is 1ln(t)/1+t,Find the function f (x) + f (1/x) and show that f (e) + f (1/e) = 1/2.Prob 9Is the function f (x) = |x^2 - x| differentiable at x = 2. If yes find it’s derivative.Prob 10Check the monotonicity of f (x) =x/sinxin (0, p/2).

Arun Kumar IIT Delhi
8 years ago
Ans 1
$f(x)=\int_{1/x}^{x}\sqrt{sin(t)}dt \\=>f(1)=0;$
because limits are same.

We are told to answer only 1 question.