Let fx - ax2 - 2a-1 x-9a-4x2-8x-32 and gx - x-x2 -1- If fx is less than M - 34 for all x - where M is the maximum value of gx - then set of values of a lies in

M = (1 4( 1)( 1))4( 1)= 34 ∴ f(x ) lt;0 ∀ x ∈ℝ ⇒ ax^2 + 2 (a+1 ) x+9a+4 lt; 0 ∀ x ∴ a lt;0 and D lt;0 D = 4 (a+1 )^2 4a (9a+4 ) lt;0 ⇒ ...

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Byju's Answer

Standard XII

Mathematics

Test for Monotonicity about a Point

Let fx = ax2...

Question

Let $f (x) = \frac{a x^{2} + 2 (a + 1) x + 9 a + 4}{x^{2} - 8 x + 32}$ and $g (x) = x - x^{2} - 1.$ If $f (x)$ is less than $M + \frac{3}{4}$ for all $x \in R,$ where $M$ is the maximum value of $g (x),$ then set of values of $a$ lies in

(- \infty, \frac{- 1}{2}]

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(- \infty, \frac{- 1}{2})

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(- \infty, - 5)

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(- \infty, 0)

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Solution

The correct option is D $(- \infty, 0)$
$M = \frac{- (1 - 4 (- 1) (- 1))}{4 (- 1)} = \frac{- 3}{4}$
$∴ f (x) < 0 \forall x \in R$
$\Rightarrow a x^{2} + 2 (a + 1) x + 9 a + 4 < 0 \forall x$

$∴ a < 0$ and $D < 0$
$D = 4 {(a + 1)}^{2} - 4 a (9 a + 4) < 0 \Rightarrow a^{2} + 2 a + 1 - 9 a^{2} - 4 a < 0 \Rightarrow - 8 a^{2} - 2 a + 1 < 0 \Rightarrow 8 a^{2} + 2 a - 1 > 0 \Rightarrow 8 a^{2} + 4 a - 2 a - 1 > 0 \Rightarrow (2 a + 1) (4 a - 1) > 0$

But $a < 0$
$∴ a \in (- \infty, \frac{- 1}{2})$

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Let f(x)=ax2+2(a+1)x+9a+4x2−8x+32 and g(x)=x−x2−1. If f(x) is less than M+34 for all x∈R, where M is the maximum value of g(x), then set of values of a lies in

The correct option is D (−∞,0)M=−(1−4(−1)(−1))4(−1)=−34 ∴f(x)<0 ∀ x∈R ⇒ax2+2(a+1)x+9a+4<0 ∀ x ∴a<0 and D<0 D=4(a+1)2−4a(9a+4)<0⇒a2+2a+1−9a2−4a<0⇒−8a2−2a+1<0⇒8a2+2a−1>0⇒8a2+4a−2a−1>0⇒(2a+1)(4a−1)>0 But a<0 ∴a∈(−∞,−12)

Let $f (x) = \frac{a x^{2} + 2 (a + 1) x + 9 a + 4}{x^{2} - 8 x + 32}$ and $g (x) = x - x^{2} - 1.$ If $f (x)$ is less than $M + \frac{3}{4}$ for all $x \in R,$ where $M$ is the maximum value of $g (x),$ then set of values of $a$ lies in