If `g(x)=f(x)+f(1-x) and f prime prime (x)< 0; 0 <= x <= 1.` show that g(x) increases in `[0,1/2)` and decreases in `(1/2,1]`

Given H(x)=f(x).phi(x) and f prime (x).phi prime(x)=c then prove that (F prime prime prime)/F=(f prime prime prime)/f+(phi prime prime prime)/phi, where c in constant, when f(x),phi(x),F(x) are differentiable

If F(x)=f(x)dotg(x) and f^(prime)(x)dotg^(prime)(x)=c , prove that (F")/F=f^(primeprime)/f+g^(primeprime)/g+(2c)/(fg)&F^(primeprimeprime)/F=f^(primeprimeprime)/f+g^(primeprimeprime)/g

f(x) is a twice differentiable function and g(x) is defined as g(x)=(fprime(x))^2+f prime prime(x)*f(x) on [x_1,x_2]. If x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7,f(x_1)=0,f(x_2)=2,f(x_3)=-3,f(x_4)=4,f(x_5)=-5,f(x_6) and f(x_7)=0, then the minimum number of real roots of g(x)=0 can be

Suppose that f,f^prime and f prime prime are continuous on [0, ln 2] and that f (0) = 0, f prime(0) = 3, f(In 2) = 6,f prime(ln 2) = 4 and int_0^(ln 2) e^(-2x)*f(x)dx=3. Find the value of int_0^(ln2) e^(-2x)*f prime prime(x)dx.

If f(x),g(x)a n dh(x) are three polynomials of degree 2, then prove that varphi(x)=|f(x)g(x)h(x)f^(prime)(x)g^(prime)(x)h^(prime)(x)f^(x)g^(x)h^(x)|i sacon s t a n tpol y nom i a l

If f is twice differentiable such that f^(')(x)=-f(x) and f^(prime)(x)=g(x)dot If h(x) is twice differentiable function such that h^(prime)(x)=(f(x))^2+(g(x))^2dot If h(0)=2,h(1)=4, then the equation y=h(x) represents (a)a curve of degree 2 (b)a curve passing through the origin (c)a straight line with slope 2 (d)a straight line with y intercept equal to 2.

Let f(x)=(2x+7)/(9) and g(x)=(9x+5)/(7) and we define h(x)=f^(-1)(x)+g^(-1)(x)-5x . If (h^(-1)(x)=(ax+b)/(c)) where c is a prime then the value of 5a-2c-b is equal to

Let f(x) be a polynomial of degree 3 having a local maximum at x=-1. If f(-1)=2,\ f(3)=18 , and f^(prime)(x) has a local minimum at x=0, then distance between (-1,\ 2)a n d\ (a ,\ f(a)), which are the points of local maximum and local minimum on the curve y=f(x) is 2sqrt(5) f(x) is a decreasing function for 1lt=xlt=2sqrt(5) f^(prime)(x) has a local maximum at x=2sqrt(5) f(x) has a local minimum at x=1

Let the function f satisfies f(x)*f^(prime)(-x)=f(-x)*f^(prime)(x) foe all x and f(0)=3 The value of f(x).f^(prime)(-x)=f(-x).f^(prime)(x) for all x, is