Let f, g : R → R be two real valued functions defined as f(x) = { - |x + 3|, ,, x < 0; eˣ, ,, x ≥ 0 and g(x) = { x² + k₁x, ,, x < 0; 4x + k₂, ,, x ≥ 0 , where k₁ and k₂ are real constants. If (gof) is differentiable at x = 0, then (gof) (- 4) + (gof) (4) is equal to :

Solution

Maths sir · 5 steps

Find the sign of f on each side of 0 first: that tells you which piece of g each side uses.

1. Build gof near 0. For x < 0 close to 0, x + 3 > 0, so f(x) = -(x+3), which is negative. That uses the x < 0 piece of g: (g° f)(x) = (x+3)² - k₁(x+3). For x ≥ 0, f(x) = eˣ ≥ 1, which uses the other piece: (g° f)(x) = 4eˣ + k₂.
2. Match the slopes. The left derivative is 2(x+3) - k₁, which equals 6 - k₁ at x = 0. The right derivative is 4eˣ, which equals 4 at x = 0. Setting them equal gives k₁ = 2.
3. Match the values. Differentiability needs continuity too. The left value at 0 is 9 - 3k₁ = 3 and the right value is 4 + k₂. So 4 + k₂ = 3, which gives k₂ = -1.
4. Evaluate both. f(-4) = -|-1| = -1, so (g° f)(-4) = g(-1) = 1 - 2 = -1. Next, f(4) = e⁴, so (g° f)(4) = 4e⁴ - 1.
5. So. The sum is 4e⁴ - 2 = 2(2e⁴ - 1), which is option (d).

The idea to keep. For a composite of piecewise functions, find the sign of the inner function on each side first. That picks the piece of the outer function. Then match both values and slopes.

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