Let f x - ax 2- bx - c - a -0 such that f -1- x - f -1- x - x - R - Also given that f x -0 has no real roots and b - OLet -4 a -2 b - c - -9 a -3 b - c - y -9 a -3 a - c- Then which of the following is correct-A- -B- -C- -D- - Y

The correct option is C α lt; γ lt; β f ( 1 x) = f( 1 + x) ⇒ f(x) is symmetric about x = 1 So vertex of parabola must be at x = –1 and it is concave up. ...

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Byju's Answer

Standard XII

Mathematics

LaGrange's Mean Value theorem

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 $α = 4 a - 2 b + c, β = 9 a + 3 b + c, γ = 9 a - 3 a + c .$ Then which of the following is correct?

β < α < γ

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γ < α < β

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α < γ < β

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α < β < γ

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Solution

The correct option is C $α < γ < β$

$f (- 1 - x) = f (- 1 + x) \Rightarrow f (x)$ is symmetric about x = -1
So vertex of parabola must be at x = –1 and it is concave up.
$\Rightarrow - \frac{b}{2 a} = - 1 \Rightarrow b = 2 a . . . . . . . . (i) α = f (- 2) = 4 a - 2 b + c β = f (3) = 9 a + 3 b + c γ = f (- 3) = 9 a - 3 b + c$
As f(x) has no real roots so values of f(x) increase as we move further away from x = –1 on both sides
Hence $f (3) > f (- 3) > f (- 2) \Rightarrow β > γ > α$ .

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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 α=4a−2b+c,β=9a+3b+c,γ=9a−3a+c. Then which of the following is correct?

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 $α = 4 a - 2 b + c, β = 9 a + 3 b + c, γ = 9 a - 3 a + c .$ Then which of the following is correct?