Let the function f(x)=(x²-1)|x²-a x+2|+cos |x| be not differentiable at the two points x=α=2 and x=β. Then the distance of the point (α, β) from the line 12 x+5 y+10=0 is equal to :

Solution

Maths sir · 4 steps

The corners of f come from the modulus, so first use x = 2 to find a, then factor what sits inside it.

1. Smooth parts. cos|x| = cos x because cosine is even, so it is differentiable everywhere. The factor x² - 1 is a polynomial. Any corner must come from |x² - ax + 2|, at a root of x² - ax + 2.
2. Find a. Since x = α = 2 is a corner, it must be a root: 4 - 2a + 2 = 0, so a = 3. Then x² - 3x + 2 = (x - 1)(x - 2), and the two candidate corners are x = 1 and x = 2.
3. Pick beta. At x = 2 the factor x² - 1 = 3 is not zero, so the corner of the modulus survives. The question intends β to be the other root, β = 1. Strictly, x² - 1 also vanishes at x = 1, which smooths that corner, but β = 1 is the only value that leads to any of the options.

   d = (|12(2) + 5(1) + 10|)/√(12² + 5²) = 39/13

4. So. The point is (α, β) = (2, 1), and its distance from the line is 39/13 = 3. The answer is (d).

The idea to keep. For g(x) |h(x)|, the corners sit at the simple roots of h. A corner disappears if g is also zero at that root, so check the multiplying factor at each candidate point.

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