If a function f(x) defined by
f( x ) = { aeˣ + be⁻ˣ,, - 1 ≤ x < 1; cx²,, 1 ≤ x ≤ 3; ax² + 2cx,, 3 < x ≤ 4
be continuous for some a, b, c ∈ R and f'(0) + f'(2) = e, then the value of of a is :

Solution

Maths sir · 4 steps

Use continuity at x=1 and x=3 to write b and c in terms of a, then the derivative condition gives a.

1. Continuity at 3. Both pieces must give the same value at x=3: 9c = 9a + 6c. So 3c = 9a, which gives c = 3a.
2. Continuity at 1. At x=1 you need ae + b/e = c = 3a. Multiply by e to get b = 3ae - ae².
3. Find the derivatives. For -1 ≤ x < 1, f'(x) = aeˣ - be⁻ˣ, so f'(0) = a - b. For 1 ≤ x ≤ 3, f'(x) = 2cx, so f'(2) = 4c = 12a.

   (a - b) + 12a = e ⇒ 13a - (3ae - ae²) = e ⇒ a(e² - 3e + 13) = e

4. So. a = e/(e² - 3e + 13), which is option (c).

The idea to keep. With a piecewise function and unknown constants, use one continuity equation at each joining point to write every constant in terms of one of them. Then apply the extra condition.

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