If $q^2-4pr=0,\,p>0$ then the domain of the function $f(x) = \log\{px^3 + (p + q)x^2 + (q + r)x + r\}$ is
Answer & explanation
Correct answer: option 2
Given, $q^2-4pr=0$ and p > 0
For f(x) to be defined $px^3 + (p + q)x^2 + (q + r)x + r >0$
$⇒ px^2(x +1) + qx (x +1) + r (x +1)>0⇒ (x+ 1)(px^2+qx+r ) >0$
⇒ x > −1 and $x≠-\frac{q}{2p}$
[Since $q^2-4pq =0$, ∴ at $x=-\frac{q}{2p},px^2+qx+r=0$ and if $x≠-\frac{q}{2p},x+ 1)(px^2+qx+r >0$
∴ Domain = $R-\left[(-∞,-1]∪\{-\frac{q}{2p}\}\right]$