If a-b-c are the integral values of x a b c satisfying -x2-10x-16 x-2- then the value of 2a-3b-4c is

√( x^2+10x 16) x 2 is meaningful when nbsp; x^2+10x 16 ≥ 0 ⇒ x^2 10x+16 ≤ 0 ⇒ (x 2)(x 8) ≤ 0 ⇒ x ∈ [2,8] ⋯ (1) nbsp;Now, √( x^2+10x 16) x 2 ⇒ x^2+10 ...

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Standard XII

Mathematics

Domain

If a,b,c are ...

Question

If $a, b, c$ are the integral values of $x (a < b < c)$ satisfying $\sqrt{- x^{2} + 10 x - 16} < x - 2,$ then the value of $2 a + 3 b - 4 c$ is

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Solution

The correct option is B $1$
$\sqrt{- x^{2} + 10 x - 16} < x - 2$ is meaningful when
$- x^{2} + 10 x - 16 \geq 0$
$\Rightarrow x^{2} - 10 x + 16 \leq 0 \Rightarrow (x - 2) (x - 8) \leq 0 \Rightarrow x \in [2, 8] \dots (1)$

Now, $\sqrt{- x^{2} + 10 x - 16} < x - 2$
$\Rightarrow - x^{2} + 10 x - 16 < (x - 2)^{2} \Rightarrow 2 x^{2} - 14 x + 20 > 0 \Rightarrow x^{2} - 7 x + 10 > 0$
$\Rightarrow (x - 2) (x - 5) > 0$
$\Rightarrow x \in (- \infty, 2) \cup (5, \infty) \dots (2)$

From $(1)$ and $(2)$ ,
$x \in (5, 8]$
$\Rightarrow a = 6, b = 7, c = 8$
$∴ 2 a + 3 b - 4 c = 12 + 21 - 32 = 1$
Hence, the value of $2 a + 3 b - 4 c$ is $1.$

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If a,b,c are the integral values of x (a<b<c) satisfying √−x2+10x−16<x−2, then the value of 2a+3b−4c is

If $a, b, c$ are the integral values of $x (a < b < c)$ satisfying $\sqrt{- x^{2} + 10 x - 16} < x - 2,$ then the value of $2 a + 3 b - 4 c$ is