If`g(x)=int_0^x2|t|dt`,then (a)`g(x)=x|x|` (b)`g(x)` is monotonic (c) g(x) is differentiable at x=0 (d)`gprime(x)`is differentiable at x=0

If f(x)=int_0^x(sint)/t dt ,x >0, then

Let f:[a,b]rarr[1,infty) be a continuous function and lt g: RrarrR be defined as g(x)={(0, "if x" lt a),(int_a^x f(t)dt, "if a"lexleb),(int_a^b f(t)dt, "if x"gtb):} Then a) g(x) is continuous but not differentiable at a b) g(x) is differentiable on R c) g(x) is continuous but nut differentiable at b d) g(x) is continuous and differentiable at either a or b but not both.

If g(x)=int_(0)^(x)cos4tdt , then g(x+pi) equals-

If g(x)= int_0^x cos^4t dt , then g(x + pi) equals

Let f(x)={(-,1, -2lexlt0),(x^2,-1,0lexlt2):} if g(x)=|f(x)|+f(|x|) then g(x) in (-2,2) is (A) not continuous is (B) not differential at one point (C) differential at all points (D) not differential at two points

If g(x)=int_(0)^(x)cos^(4)t dt , then prove that g(x+pi)=g(x)+g(pi) .

If g(x)=int_0^xcos^4tdt , then g(x+pi) equals (a) g(x)+g(pi) (b) g(x)-g(pi) (c) g(x)g(pi) (d) (g(x))/(g(pi))

int [f(x)g"(x) - f"(x)g(x)]dx is

Let f be a differential function such that f(x)=f(2-x) and g(x)=f(1 +x) then (1) g(x) is an odd function (2) g(x) is an even function (3) graph of f(x) is symmetrical about the line x= 1 (4) f'(1)=0

If g(x)=int_0^x(|sint|+|cost|)dt ,then g(x+(pin)/2) is equal to, where n in N , (a) g(x)+g(pi) (b) g(x)+g((npi)/(2)) (c) g(x)+g(pi/2) (d) none of these