If `("lim")_(tvecx)(e^tf(x)-e^xf(t))/((t-x)(f(x))^2)=2a n df(0)=1/2,` then find the value of `f^(prime)(0)dot` 4 (b) 2 (c) 0 (d) 1

f(0)=1 , f(2)=e^2 , f'(x)=f'(2-x) , then find the value of int_0^(2)f(x)dx

If F(t)=int(x+y)dt then the value of F(pi/2)-F(0) is 1 (b) -1 (c) e^(pi//2) (d) 0

If lim_(t rarr x)(x^2f^2(t)-t^2f^2(x))/(t-x)=0 and f(1)=e then solution of f(x)=1 is

f(x)=int_(0)^( pi)f(t)dt=x+int_(x)^(1)tf(t)dt, then the value of f(1) is (1)/(2)

Let a function f(x) be such that f"(x)=f'(x)+e^(x)&f(0)=0,f'(0)=1 then find the value of ln(((f(2))^(2))/(4))

Instead of the usual definition of derivative Df(x), if we define a new kind of derivative D^*F(x) by the formula D*f(x)=lim_(h->0)(f^2(x+h)-f^2(x))/h ,w h e r ef^2(x) mean [f(x)]^2 and if f(x)=xlogx ,then D^*f(x)|_(x=e) has the value (A)e (B) 2e (c) 4e (d) none of these

Let f(x)=int_(0)^(x)e^(t)(t-1)(t-2)dt. Then, f decreases in the interval

people living at Mars ,instead of the usual definition of derivative Df(x) define a new kind of derivative Df(x) by the formula Df(x)=lim_(h->0) (f^2(x+h)-f^2(x))/h where f^x means [f(x)]^2. If f(x)=xInx then Df(x)|_(x=e) has the value (a) e (b) 2e (c) 4e (d) non

If f(x)=lim_(n->oo)((x^2+a x+1)+x^(2n)(2x^2+x+b))/(1+x^(2n))and lim_(x->+-1)f(x) exist, then The value of b is (a)-1 (b). 1 ( c.) 0 (d).2

if f(x)=3e^(x^(2)) then f'(x)-2xf(x)+(1)/(3)f(0)-f'(0)