JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 19)
If 1, log10(4x $$-$$ 2) and log10$$\left( {{4^x} + {{18} \over 5}} \right)$$ are in arithmetic progression for a real number x, then the value of the determinant $$\left| {\matrix{
{2\left( {x - {1 \over 2}} \right)} & {x - 1} & {{x^2}} \cr
1 & 0 & x \cr
x & 1 & 0 \cr
} } \right|$$ is equal to :
Answer
2
Explanation
1, $$lo{g_{10}}({4^x} - 2),\,lo{g_{10}}\left( {{4^x} + {{18} \over 5}} \right)$$ in AP.
$$ \therefore $$ 2$$ \times $$$$lo{g_{10}}({4^x} - 2) = 1 + \,lo{g_{10}}\left( {{4^x} + {{18} \over 5}} \right)$$
$$lo{g_{10}}{({4^x} - 2)^2} = \,lo{g_{10}}\left( {10.\left( {{4^x} + {{18} \over 5}} \right)} \right)$$
$${({4^x} - 2)^2} = 10.\left( {{4^x} + {{18} \over 5}} \right)$$
$${({4^x})^2} + 4 - {4.4^x} = {10.4^x} + 36$$
$${({4^x})^2} - {14.4^x} - 32 = 0$$
$${({4^x})^2} + {2.4^x} - {16.4^x} - 32 = 0$$
$${4^x}({4^x} + 2) - 16.({4^x} + 2) = 0$$
$$({4^x} + 2)({4^x} - 16) = 0$$
4x = -2 (Not Possible)
Or 4x = 16
$$ \Rightarrow $$ x = 2
Therefore $$\left| {\matrix{ {2(x - 1/2)} & {x - 1} & {{x^2}} \cr 1 & 0 & x \cr x & 1 & 0 \cr } } \right|$$
$$ = \left| {\matrix{ 3 & 1 & 4 \cr 1 & 0 & 2 \cr 2 & 1 & 0 \cr } } \right|$$
$$ = 3( - 2) - 1(0 - 4) + 4(1 - 0)$$
$$ = - 6 + 4 + 4 = 2$$
$$ \therefore $$ 2$$ \times $$$$lo{g_{10}}({4^x} - 2) = 1 + \,lo{g_{10}}\left( {{4^x} + {{18} \over 5}} \right)$$
$$lo{g_{10}}{({4^x} - 2)^2} = \,lo{g_{10}}\left( {10.\left( {{4^x} + {{18} \over 5}} \right)} \right)$$
$${({4^x} - 2)^2} = 10.\left( {{4^x} + {{18} \over 5}} \right)$$
$${({4^x})^2} + 4 - {4.4^x} = {10.4^x} + 36$$
$${({4^x})^2} - {14.4^x} - 32 = 0$$
$${({4^x})^2} + {2.4^x} - {16.4^x} - 32 = 0$$
$${4^x}({4^x} + 2) - 16.({4^x} + 2) = 0$$
$$({4^x} + 2)({4^x} - 16) = 0$$
4x = -2 (Not Possible)
Or 4x = 16
$$ \Rightarrow $$ x = 2
Therefore $$\left| {\matrix{ {2(x - 1/2)} & {x - 1} & {{x^2}} \cr 1 & 0 & x \cr x & 1 & 0 \cr } } \right|$$
$$ = \left| {\matrix{ 3 & 1 & 4 \cr 1 & 0 & 2 \cr 2 & 1 & 0 \cr } } \right|$$
$$ = 3( - 2) - 1(0 - 4) + 4(1 - 0)$$
$$ = - 6 + 4 + 4 = 2$$
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