Let x₀ be the point of Local maxima of f(x) = a⃗ .( b⃗ × c⃗ ), where
a⃗ = xî - 2ĵ + 3k̂, b⃗ = - 2î + xĵ - k̂, c⃗ = 7î - 2ĵ + xk̂. Then the value of
a⃗ .b⃗ + b⃗ .c⃗ + c⃗ .a⃗ at x = x₀ is :

Solution

Maths sir · 4 steps

Write the triple product as a 3 by 3 determinant, get a cubic in x, and pick the root where it peaks.

1. Set up f(x). The scalar triple product a⃗ · (b⃗ × c⃗) is the determinant whose rows are the components of a⃗, b⃗ and c⃗.

   f(x) = |x, -2, 3; -2, x, -1; 7, -2, x| = x(x²-2) + 2(7-2x) + 3(4-7x) = x³ - 27x + 26

2. Find the maximum. f'(x) = 3x² - 27 = 0 gives x = ± 3. Since f''(x) = 6x, f''(-3) = -18 < 0, so the local maximum is at x₀ = -3.
3. Substitute x0. At x = -3: a⃗ = (-3, -2, 3), b⃗ = (-2, -3, -1), c⃗ = (7, -2, -3). Then a⃗ · b⃗ = 6 + 6 - 3 = 9, b⃗ · c⃗ = -14 + 6 + 3 = -5 and c⃗ · a⃗ = -21 + 4 - 9 = -26.
4. So. The sum is 9 - 5 - 26 = -22, which is option (d).

The idea to keep. A scalar triple product with variable components is just a determinant, so it becomes an ordinary polynomial. Then use f'' < 0 to separate the maximum from the minimum before you substitute.

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