Let `f(x)` and `g(x)` be differentiable in `[0,1]` such that `f(0)=g(0)=0,f(1)=6` and `g(1)=3`,then show that `f'(C)=2g'(c)`

Let fa n dg be differentiable on [0,1] such that f(0)=2,g(0)=0,f(1)=6a n dg(1)=2. Show that there exists c in (0,1) such that f^(prime)(c)=2g^(prime)(c)dot

If f(x) and g(x) ar edifferentiable function for 0lex le1 such that f(0)=2,g(0) = 0,f(1)=6,g(1)=2 , then in the interval (0,1)

If f(x) and g(x) are differentiable function for 0 le x le 23 such that f(0) =2, g(0) =0 ,f(23) =22 g (23) =10. Then show that f'(x)=2g'(x) for at least one x in the interval (0,23)

f(x) and g(x) are two differentiable functions in [0,2] such that f"(x)=g"(x)=0, f'(1)=2, g'(1)=4, f(2)=3, g(2)=9 then f(x)-g(x) at x=3/2 is

If f(x)a n dg(x) are differentiable functions for 0lt=xlt=1 such that f(0)=10 ,g(0)=2,f(1)=2,g(1)=4, then in the interval (0,1)dot (a) f^(prime)(x)=0 for a l lx (b) f^(prime)(x)+4g^(prime)(x)=0 for at least one x (c) f(x)=2g'(x) for at most one x (d) none of these

If f(x)a n dg(x) are differentiable functions for 0lt=xlt=1 such that f(0)=10 ,g(0)=2,f(1)=2,g(1)=4, then in the interval (0,1)dot (a) f^(prime)(x)=0 for a l lx (b) f^(prime)(x)+4g^(prime)(x)=0 for at least one x (c) f(x)=2g'(x) for at most one x (d) none of these

Let f(x)a n dg(x) be differentiable for 0lt=xlt=1, such that f(0)=0,g(0)=0,f(1)=6. Let there exists real number c in (0,1) such taht f^(prime)(c)=2g^(prime)(c)dot Then the value of g(1) must be 1 (b) 3 (c) -2 (d) -1

Let f(x)a n dg(x) be differentiable for 0lt=xlt=1, such that f(0)=0,g(0)=0,f(1)=6. Let there exists real number c in (0,1) such that f^(prime)(c)=2g^(prime)(c)dot Then the value of g(1) must be (a) 1 (b) 3 (c) -2 (d) -1

Statement 1: If f(x) is differentiable in [0,1] such that f(0)=f(1)=0, then for any lambda in R , there exists c such that f^prime (c) =lambda f(c), 0ltclt1. statement 2: if g(x) is differentiable in [0,1], where g(0) =g(1), then there exists c such that g^prime (c)=0,

If f(x), g(x) be twice differentiable function on [0,2] satisfying f''(x)=g''(x) , f'(1)=4 and g'(1)=6, f(2)=3, g(2)=9, then f(x)-g(x) at x=4 equals to:- (a) -16 (b) -10 (c) -8