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Byju's Answer

Standard XII

Mathematics

Distance Formula

Let f x = ax ...

Question

Let f(x) = $a x^{2} + b x + c, a > 0$ such that $f (- 1 - x) = f (- 1 + x) \forall x ε R .$ Also given that f(x) = 0 has no real roots and b > O
Let p = b - 4a, q = 2a + b, then pq is

negative

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positive

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Cannot be determined.

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Solution

The correct option is A negative
p = b - 4a = - 2a (using (i))
q = 2a + b = 4a
$p q = - 8 a^{2} < 0$ (as a > 0)

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Similar questions

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Consider two quadratic expressions f(x)= $a x^{2} + b x + c$ and g(x)= $a x^{2} + p x + q$ (a,b,c,p,q $\epsilon$ R, b≠p)such that their discriminents are equal. If f(x)=g(x) has a root x=a then.

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Let f(x) = ax2+bx+c,a>0 such that f(−1−x)=f(−1+x)∀ x ε R. Also given that f(x) = 0 has no real roots and b > O Let p = b - 4a, q = 2a + b, then pq is

The correct option is A negativep = b - 4a = - 2a (using (i)) q = 2a + b = 4a pq=−8a2<0 (as a > 0)

Let f(x) = $a x^{2} + b x + c, a > 0$ such that $f (- 1 - x) = f (- 1 + x) \forall x ε R .$ Also given that f(x) = 0 has no real roots and b > O
Let p = b - 4a, q = 2a + b, then pq is

The correct option is A negative
p = b - 4a = - 2a (using (i))
q = 2a + b = 4a
$p q = - 8 a^{2} < 0$ (as a > 0)