Question

If f(x) = (x - 1) (x - 3)(x - 4)(x - 6) + 19 for all real value of x is

• A. positive
• B. negative
• C. zero
• D. none of these

Answer 1

Answer: (A)

Note: Discarte's Rule of Signs
No equation can have more +ve real roots than it has changes of sign from +ve to -ve, and from -ve to + ve.
And similarly, number of -ve roots of f(x) can not more than the number of changes of sign in f (- x).
Now given
$f(x)=(x-1)(x-3)(x-4)(x-6)+19$
$\Rightarrow f(x)=x^4-14x^3+67x^2-126x+91$
Here, f(x) has four changes of sign. So the equation can not have more than four positive roots. And f(-x) has no changes of sign, so by Discarte rule, the equation have no relative roots. As
$f(-x)=x^4+14x^3+67x^2+126x+91$
Thus the equation have four positive roots.