If 1, log₁₀(4ˣ - 2) and log₁₀( 4ˣ + 18/5 ) are in arithmetic progression for a real number x, then the value of the determinant | 2( x - 1/2 ), x - 1, x²; 1, 0, x; x, 1, 0 | is equal to :

Solution

Maths sir · 4 steps

Turn the AP into a quadratic in 4ˣ, get x = 2, and the determinant becomes plain numbers.

1. Use the AP. For three terms in AP, twice the middle term equals the sum of the outer two. Write 1 = log₁₀ 10, so 2 log₁₀(4ˣ - 2) = log₁₀ 10 + log₁₀(4ˣ + 18/5), which gives (4ˣ - 2)² = 10 · 4ˣ + 36.

   t = 4ˣ: t² - 4t + 4 = 10t + 36 ⇒ t² - 14t - 32 = 0 ⇒ (t - 16)(t + 2) = 0

2. Pick the valid root. The log needs 4ˣ - 2 > 0, so t > 2. This rejects t = -2 and leaves 4ˣ = 16, so x = 2.
3. Fill the determinant. With x = 2 the rows are (3, 1, 4), (1, 0, 2) and (2, 1, 0). Expanding along the first row gives 3(0 - 2) - 1(0 - 4) + 4(1 - 0) = -6 + 4 + 4.
4. So. The value of the determinant is 2.

The common slip: -6. Forgetting the minus sign on the middle cofactor gives -6 - 4 + 4 = -6, not 2. The signs along a row go +, -, +.

The idea to keep. When logs are in AP, use 2b = a + c, write any constant as a log, and drop the logs. Always check that the arguments stay positive before you accept a root.

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