If the middle term in the expansion of 1- x - x sin x 10 is 63-8- then the value of 6 sin 2 x -sin x -2 isA- B- 3-2D- 61-8D- 5

The correct option is A 0Middle Term = T_6 = ^10C_5(1x)^5(xsin x)^5 = 252(sin x)^5 ∴ 252(sin x)^5 = 638 ⇒sin x = 12 ⇒ 2sin x 1 = 0 ∴ 6sin^2 x + sin x 2 = ...

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Standard XII

Mathematics

Middle Terms

If the middle...

Question

If the middle term in the expansion of ${(\frac{1}{x} + x sin x)}^{10}$ is $\frac{63}{8},$ then the value of $6 {sin}^{2} x + sin x - 2$ is

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\frac{61}{8}

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Solution

The correct option is A $0$
Middle Term $= T_{6} =^{10} C_{5} {(\frac{1}{x})}^{5} (x sin x)^{5} = 252 (sin x)^{5}$
$∴ 252 (sin x)^{5} = \frac{63}{8}$
$\Rightarrow sin x = \frac{1}{2}$
$\Rightarrow 2 sin x - 1 = 0$
$∴ 6 {sin}^{2} x + sin x - 2$
$= (2 sin x - 1) (3 sin x + 2) = 0$

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If the middle term in the expansion of (1x+xsinx)10 is 638, then the value of 6sin2x+sinx−2 is

The correct option is A 0Middle Term =T6=10C5(1x)5(xsinx)5=252(sinx)5 ∴252(sinx)5=638 ⇒sinx=12 ⇒2sinx−1=0 ∴6sin2x+sinx−2 =(2sinx−1)(3sinx+2)=0

If the middle term in the expansion of ${(\frac{1}{x} + x sin x)}^{10}$ is $\frac{63}{8},$ then the value of $6 {sin}^{2} x + sin x - 2$ is

The correct option is A $0$
Middle Term $= T_{6} =^{10} C_{5} {(\frac{1}{x})}^{5} (x sin x)^{5} = 252 (sin x)^{5}$
$∴ 252 (sin x)^{5} = \frac{63}{8}$
$\Rightarrow sin x = \frac{1}{2}$
$\Rightarrow 2 sin x - 1 = 0$
$∴ 6 {sin}^{2} x + sin x - 2$
$= (2 sin x - 1) (3 sin x + 2) = 0$