If the functions f(x)=x³/3+2 b x+(a x²)/2
and g(x)=x³/3+a x+b x², a ≠ 2 b
have a common extreme point, then a+2 b+7 is equal to :

Solution

Maths sir · 4 steps

Both derivatives vanish at the same x, so subtract them and the common point drops out as x = 1.

1. Differentiate both. At an extreme point the derivative is zero. Here f'(x) = x² + ax + 2b and g'(x) = x² + 2bx + a.
2. Subtract them. If x is a common extreme point, then f'(x) = g'(x) = 0, so f'(x) - g'(x) = 0. The x² terms cancel.

   (a - 2b)x - (a - 2b) = 0 ⇒ (a - 2b)(x - 1) = 0

3. Use the condition. Since a ≠ 2b, the factor a - 2b is not zero, so the common point is x = 1. Putting x = 1 in f'(x) = 0 gives 1 + a + 2b = 0, that is a + 2b = -1.
4. So. a + 2b + 7 = -1 + 7 = 6, which is option (a).

The idea to keep. For a common extreme point, set both derivatives to zero and subtract. The highest power cancels and the common root usually falls out at once.

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