If f : R → R and g : R → R defined by f(x) = 2x + 3 and g(x) = x2 + 7, then the value of x for which f(g(x)) = 25 is ...

Let f : N → R : f(x) = (frac{(2 x-1)}{2}) and g : Q → R : g(x) = x + 2 be two functions. Then, (gof) ((frac{3}{2})) is 1) 3 ...

Let (f(x)=frac{x-1}{x+1}), then f(f(x)) is 1) (frac{1}{x}) 2) (-frac{1}{x}) 3) (frac{1}{x+1}) 4) (frac{1}{x-1})

If f(x) = (1-frac{1}{x}), then f(f((frac{1}{x}))) 1) (frac{1}{x}) 2) (frac{1}{1+x}) 3) (frac{x}{x-1}) 4) (frac{1}{x-1})

If f : R → R, g : R → R and h : R → R are such that f(x) = x2, g(x) = tan x and h(x) = log x, then the value of ...

If f(x) = (frac{3 x+2}{5 x-3}) then (fof)(x) is 1) x 2) -x 3) f(x) 4) -f(x)

If the binary operation * is defind on the set Q+ of all positive rational numbers by a * b = (frac{a b}{4}). Then, (3 *left(frac{1}{5} * frac{1}{2}right)) is equal to 1) (frac{3}{160}) ...

The number of binary operations that can be defined on a set of 2 elements is 1) 8 2) 4 3) 16 4) 64

Let * be a binary operation on set Q of rational numbers defined as a * b = (frac{a b}{5}). Write the identity for *. 1) 5 2) 3 3) 1 ...

For binary operation * defind on R – {1} such that a * b = (frac{a}{b+1}) is 1) not associative 2) not commutative 3) commutative 4) both (a) and (b) ...