JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Application +4 | -1

If the equation

\left(\log _{16} x\right)^2-\left(\log _{16} x\right)+\log _{16} k=0

has a unique solution (where

k

is constant), then the value of

k

View answer and explanation

Correct option: B

The correct answer is B (2) because setting the discriminant of the quadratic equation to zero and solving for k gives k=2.

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\alpha ,\beta

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\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

{x^2} + 4ax + 2 > 0

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

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9{x^2} - 18\left| x \right| + 5 = 0

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)