Let f(x)=x⁵+2 x³+3 x+1, x ∈ R, and g(x) be a function such that g(f(x))=x for all x ∈ R. Then g(7)/g′(7) is equal to :

Solution

Maths sir · 4 steps

Find the x that f sends to 7. Then g(7) is that x, and g'(7) is one over f' there.

1. g undoes f. Since g(f(x)) = x, g is the inverse of f. This works because f'(x) = 5x⁴ + 6x² + 3 > 0 for every x, so f is strictly increasing and one-one.
2. Find g(7). Look for the x with f(x) = 7. Try x = 1: f(1) = 1 + 2 + 3 + 1 = 7. So g(7) = 1.

   g'(f(x))· f'(x) = 1 ⇒ g'(7) = 1/f'(1)

3. Find g'(7). Differentiating g(f(x)) = x by the chain rule gives the line above. Here f'(1) = 5 + 6 + 3 = 14, so g'(7) = 1/14.
4. So. g(7)/g'(7) = 1/(1/14) = 14, which is option (d).

The idea to keep. For an inverse, never try to find g's formula. Solve f(a) = b to get g(b) = a, then use g'(b) = 1/f'(a), with f' taken at a, not at b.

Answer from the NTA final answer key. Solution by PrepWiser.