x-1" varepsilonरेई "[xi_(4)d" भाวे "" विधेय "f(x)=(1)/(5^(x))" थो "...

Check whether the Lagrange formula is applicable to following functions : f(x) =x^(2) "on" [3,4] f(x) = root5 (x^(4)(x-1)) " on " [-1//2,1//2] If it is , find the values of xi appearing in this formula .

If f(x)=cos(log x), then value of f(x)f(4)-(1)/(2){f((x)/(4))+f(4x)} is (a) 1(b)-1(c)0(d)+-1

Let phi(x) be the inverse of the function f(x) and f'=(1)/(1+x^(5)), then (d)/(dx)phi(x) is

If f(x)=(x-1)/(x+1) , then f(f(a x)) in terms of f(x) is equal to (a)(f(x)-1)/(a(f(x)-1)) (b) (f(x)+1)/(a(f(x)-1)) (f(x)-1)/(a(f(x)+1)) (d) (f(x)+1)/(a(f(x)+1))

If f(x)=(x-1)/(x+1) , then f(f(a x)) in terms of f(x) is equal to (a)(f(x)-1)/(a(f(x)-1)) (b) (f(x)+1)/(a(f(x)-1)) (f(x)-1)/(a(f(x)+1)) (d) (f(x)+1)/(a(f(x)+1))

If f(x)=(x-1)/(x+1), then f(f(ax)) in terms of f(x) is equal to (a) (f(x)-1)/(a(f(x)-1)) (b) (f(x)+1)/(a(f(x)-1))(c)(f(x)-1)/(a(f(x)+1))(d)(f(x)+1)/(a(f(x)+1))

If f(x) is an invertible function and g(x)=2f(x)+5, then the value of g^(-1)(x)i s (a) 2f^(-1)(x)-5 (b) 1/(2f^(-1)(x)+5) 1/2f^(-1)(x)+5 (d) f^(-1)((x-5)/2)

If f(x) is an invertible function and g(x)=2f(x)+5, then the value of g^(-1)(x)i s (a) 2f^(-1)(x)-5 (b) 1/(2f^(-1)(x)+5) 1/2f^(-1)(x)+5 (d) f^(-1)((x-5)/2)