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If $ax^2-bx+5=0$ does not have two distinct real roots,then minimum value of $5a+b$ is

$\begin{array}{1 1}(A)\;-1&(B)\;0\\(C)\;1&(D)\;5\end{array}$

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Let $f(x)=ax^2-bx+5$
Since $f(x)=0$ does not have two distinct real roots,either $f(x)\geq 0$ or $f(x) \leq 0$
But $f(x)=5>0$
$\Rightarrow f(x) \geq 0 \forall x\in R$
$f(-5)=25a+5b+5 \geq 0$
$\Rightarrow 5a+b\geq -1$
$\Rightarrow$Minimum value of $5a+b$ is $-1$
Hence (A) is the correct answer.
answered Apr 14, 2014

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