Number of integral values of x satisfying the inequality 3-46 x-10-x2-27-64 i-A- 7B- 8C- infiniteD- 6

The correct option is A 7(34)^6x+10 x^22764 ⇒(34)^6x+10 x^2(34)^3 ⇒6x+10 x^2 gt;3 (∵34 lt;1) ⇒ x^2 6x 70 ⇒(x+1)(x 7)0 nbsp; ⇒ x∈( 1,7) Hence, x={0,1, ...

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Standard XII

Mathematics

Inequality

Number of int...

Question

Number of integral values of $x$ satisfying the inequality ${(\frac{3}{4})}^{6 x + 10 - x^{2}} < \frac{27}{64}$ is

6

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7

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8

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infinite

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Solution

The correct option is B $7$
${(\frac{3}{4})}^{6 x + 10 - x^{2}} < \frac{27}{64}$
$\Rightarrow {(\frac{3}{4})}^{6 x + 10 - x^{2}} < {(\frac{3}{4})}^{3}$
$\Rightarrow 6 x + 10 - x^{2} > 3 (∵ \frac{3}{4} < 1)$
$\Rightarrow x^{2} - 6 x - 7 < 0$
$\Rightarrow (x + 1) (x - 7) < 0$
$\Rightarrow x \in (- 1, 7)$
Hence, $x = {0, 1, 2, 3, 4, 5, 6}$

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Number of integral values of x satisfying the inequality (34)6x+10−x2<2764 is

The correct option is B 7(34)6x+10−x2<2764 ⇒(34)6x+10−x2<(34)3 ⇒6x+10−x2>3 (∵34<1) ⇒x2−6x−7<0 ⇒(x+1)(x−7)<0 ⇒x∈(−1,7) Hence, x={0,1,2,3,4,5,6}

Number of integral values of $x$ satisfying the inequality ${(\frac{3}{4})}^{6 x + 10 - x^{2}} < \frac{27}{64}$ is