The number of real roots of $x^{10} - x^{7} + x^{5} + 2 x^{4} + 6 = 0$

atmost five

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atmost four

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atmost three

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atmost two

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Solution

The correct option is A atmost four
$x^{10} - x^{7} + x^{5} + 2 x^{4} + 6 = 0$
$f (x) = x^{10} - x^{7} + x^{5} + 2 x^{4} + 6$
$+ - + + +$
Maximum number of positive roots (sign changes)are $2$ .
Now, $f (- x) = x^{10} + x^{7} - x^{5} + 2 x^{4} + 6$
$+ + - + +$
Maximum number of negative roots (sign changes) are $2$
So, maximum number of real roots $= 4$

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