Question

jee mains related questions...plz give the correct ans...if u dont know plz dont reply just for the sake of points​

Answer 1

Step-by-step explanation:

We have,

$f(x) = \log_{2}( {x}^{2} - 6x + 10 )$

Argument of log must be positive,

${x}^{2} - 6x + 10> 0$

Now, determinant of the above quadratic equation is $(-6)^{2}-4(10) = - 4$ which is negative

Since, coefficient of x² is positive and D is negative, so, the quadratic equation is positive.

So,

$\implies {x}^{2} - 6x + 10> 0 \: \: \forall \: x \in \: R$