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

Let f(x)a n dg(x) be differentiable for 0lt=xlt=2 such that f(0)=2,g(0)=1,a n df(2)=8. Let there exist a real number c in [0,2] such that f^(prime)(c)=3g^(prime)(c)dot Then find the value of g(2)dot

Let f(x)a n dg(x) be differentiable for 0lt=xlt=2 such that f(0)=2,g(0)=1,a n df(2)=8. Let there exist a real number c in [0,2] such that f^(prime)(c)=3g^(prime)(c)dot Then find the value of g(2)dot

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)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

Leg f be continuous on the interval [0,1] to R such that f(0)=f(1)dot Prove that there exists a point c in [(0,1)/2] such that f(c)=f(c+1/2)dot

Let f(x)a n dg(x) be two differentiable functions in R a n d f(2)=8,g(2)=0,f(4)=10 ,a n dg(4)=8. Then prove that g^(prime)(x)=4f^(prime)(x) for at least one x in (2,4)dot

If f is a continuous function on [0,1], differentiable in (0, 1) such that f(1)=0, then there exists some c in (0,1) such that cf^(prime)(c)-f(c)=0 cf^(prime)(c)+cf(c)=0 f^(prime)(c)-cf(c)=0 cf^(prime)(c)+f(c)=0