The integral values ofx satisfying -x2-10x-16-x-2 is-are

√( x^2+10x 16) lt;x 2 we must have x^2+10x 16≥ 0 ⇒ x^2 10x+16≤0 ⇒ (x 2)(x 8)≤0 ⇒ 2 ≤ x ≤ 8⋯(1) Now sqauring both sides x^2+10x 16 lt;x^2 4x+4 ⇒ 2x^2 14x ...

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Standard XII

Mathematics

Inequality

The integral ...

Question

The integral value(s) of
$x$ satisfying $\sqrt{- x^{2} + 10 x - 16} < x - 2$ is/are

5

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6

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7

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8

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Solution

The correct option is D $8$
$\sqrt{- x^{2} + 10 x - 16} < x - 2$
we must have
$- x^{2} + 10 x - 16 \geq 0 \Rightarrow x^{2} - 10 x + 16 \leq 0 \Rightarrow (x - 2) (x - 8) \leq 0$

$\Rightarrow 2 \leq x \leq 8 \dots (1)$

Now sqauring both sides
$- x^{2} + 10 x - 16 < x^{2} - 4 x + 4 \Rightarrow 2 x^{2} - 14 x + 20 > 0 \Rightarrow x^{2} - 7 x + 10 > 0 \Rightarrow (x - 2) (x - 5) > 0$

$\Rightarrow x > 5 or x < 2 \dots (2)$

From equation $(1) and (2)$ , we get
$5 < x \leq 8 \Rightarrow x = 6, 7, 8$

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The integral value(s) of x satisfying √−x2+10x−16<x−2 is/are

The integral value(s) of
$x$ satisfying $\sqrt{- x^{2} + 10 x - 16} < x - 2$ is/are