JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Determinants

1, lo g_{10} (4^{x} - 2)

and

lo g_{10} (4^{x} + \frac{18}{5})

are in arithmetic progression for a real number x , then the value of the determinant

2 (x - \frac{1}{2}) 1 x x - 1 01 x^{2} x 0

is equal to:

Answer: 2

Solution

Three terms in arithmetic progression satisfy

2 lo g_{10} (4^{x} - 2) = 1 + lo g_{10} (4^{x} + \frac{18}{5})

, so

(4^{x} - 2)^{2} = 10 (4^{x} + \frac{18}{5})

With

t = 4^{x}

this becomes

t^{2} - 4 t + 4 = 10 t + 36

, that is

t^{2} - 14 t - 32 = 0

, whose roots are

t = 16

and

t = - 2

. Only

t = 16

keeps

4^{x} - 2

positive, so

4^{x} = 16

and

x = 2

.
Substituting

x = 2

in the determinant,

312101420 = 3 (0 \cdot 0 - 2 \cdot 1) - 1 (1 \cdot 0 - 2 \cdot 2) + 4 (1 \cdot 1 - 0 \cdot 2) = - 6 + 4 + 4

so the determinant equals

2

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer