If `varphi(x)` is a polynomial function and `varphi^(prime)(x)>varphi(x)AAxgeq1a n dvarphi(1)=0,` then `varphi(x)geq0AAxgeq1` `varphi(x)

If x=varphi(t), y=psi(t),t h e n(d^2 y)/(dx^2) is (a) (varphi^(prime)psi^('')-psi'varphi' ')/((varphi^(prime))^2) (b) (varphi^(prime)psi^('')-psi'varphi' ')/((varphi^(prime))^3) (c) varphi^('')/psi^('') (d) psi^('')/varphi^('')

If the function f(x) increases in the interval (a , b),a n dvarphi(x) [f(x)]^2 , then varphi(x) increases in (a , b) varphi(x) decreases in (a , b) we cannot say that varphi(x) increases or decreases in (a , b)dot none of these

Let inte^x{f(x)-f^(prime)(x)}dx=varphi(x)dot Then inte^xf(x)dx is

If f(t) is an odd function, then prove that varphi(x)=int_a^xf(t)dt is an even function.

If f(x)= log_3x and phi(x)=x^2 ,then the value of f{varphi(3)} will be

Let S denote the set of all polynomials P(x) of degree lt=2 such that P(1)=1,P(0)=0a n dP^(prime)(x)>0AAx in [0,1] , then (a)S=varphi (b)S={(1-a)x^2+a x ;0< a <2} (c)S={(1-a)x^2+a x ;0< a <1} (d)S={(1-a)x^2+a x ;0< a < oo}

The second derivative of a single valued function parametrically represented by x=varphi(t)a n dy=psi(t) (where varphi(t)a n dpsi(t) are different function and varphi^(prime)(t)!=0 ) is given by (d^2y)/(dx^2)=(((dx)/(dt))((d^2y)/(dt^2))-((d^2x)/(dt^2))((dy)/(dt))/(((dx)/(dt))^2) (d^2y)/(dx^2)=(((dx)/(dt))((d^2y)/(dt^2))-((d^2x)/(dt^2))((dy)/(dt))/(((dx)/(dt))^3) (d^2y)/(dx^2)=(((d^2x)/(dt))((d^y)/(dt^2))-((d^x)/(dt^))((d^2y)/(dt^2))/(((dx)/(dt))^3) (d^2y)/(dx^2)=(((d^2x)/(dt^2))((d^y)/(dt^))-((d^2x)/(dt^2))((d^y)/(dt^))/(((dx)/(dt))^3)

If varphi(x)=1/(1+e^(-x)) and S=varphi(5)+varphi(4)+varphi(3)+...+varphi(-3)+varphi(-4)+varphi(-5) , then the value of S is. a. 5 b. 11/2 c. 6 d. 13/2

Suppose |[f'(x),f(x)],[f''(x),f'(x)]|=0 is continuously differentiable function with f^(prime)(x)!=0 and satisfies f(0)=1 and f'(0)=2 then lim_(x->0)(f(x)-1)/x is a. 1//2 b. 1 c. 2 d. 0

A function y=f(x) satisfies (x+1)f^(prime)(x)-2(x^2+x)f(x)=((e^x)^2)/((x+1)),AAx >-1. If f(0)=5, then f(x) is