jee mains related questions...plz give the correct ans...if u dont know plz dont reply just for the sake of points
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jee mains related questions...plz give the correct ans...if u dont know plz dont reply just for the sake of points
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Step-by-step explanation:
We have,
[tex]f(x) = \log_{2}( {x}^{2} - 6x + 10 ) [/tex]
Argument of log must be positive,
[tex] {x}^{2} - 6x + 10> 0 [/tex]
Now, determinant of the above quadratic equation is [tex] (-6)^{2}-4(10) = - 4[/tex] which is negative
Since, coefficient of x² is positive and D is negative, so, the quadratic equation is positive.
So,
[tex] \implies {x}^{2} - 6x + 10> 0 \: \: \forall \: x \in \: R \\ [/tex]