In the expansion of xcos- - 1x sin-16- if l1 is the least value of the term independent of x when -8-4 and l2 is the least value of the term independent of x when -16-8- then the ratio l2 - l1 is equal to -

T_r+1 = ^16C_r (xcosθ)^16 r(1x sinθ)^r For term independent of x, nbsp; 16 2r = 0 ⇒ r = 8 T_9 = ^16C_8 ( 1sinθcosθ)^8 = 16C_8 2^8(1sin 2θ)^8 l_1 = ^16C_ ...

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Byju's Answer

Standard XII

Mathematics

Term Independent of x

In the expans...

Question

In the expansion of ${(\frac{x}{cos θ} + \frac{1}{x sin θ})}^{16}$ , if $l_{1}$ is the least value of the term independent of $x$ when $\frac{π}{8} \leq θ \leq \frac{π}{4}$ and $l_{2}$ is the least value of the term independent of $x$ when $\frac{π}{16} \leq θ \leq \frac{π}{8},$ then the ratio $l_{2} : l_{1}$ is equal to :

16 : 1

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8 : 1

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1 : 8

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1 : 16

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Solution

The correct option is A $16 : 1$
$T_{r + 1} =^{16} C_{r} {(\frac{x}{cos θ})}^{16 - r} {(\frac{1}{x sin θ})}^{r}$
For term independent of $x$ ,
$16 - 2 r = 0 \Rightarrow r = 8$
$T_{9} =^{16} C_{8} {(\frac{1}{sin θ cos θ})}^{8} = 16 C_{8} 2^{8} {(\frac{1}{sin 2 θ})}^{8}$
$l_{1} =^{16} C_{8} 2^{8} at θ = \frac{π}{4}$
$l_{2} =^{16} C_{8} \frac{2^{8}}{{(\frac{1}{\sqrt{2}})}^{8}} =^{16} C_{8} 2^{12} at θ = \frac{π}{8}$
$\frac{l_{2}}{l_{1}} = 16 : 1$

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In the expansion of (xcosθ+1xsinθ)16, if l1 is the least value of the term independent of x when π8≤θ≤π4 and l2 is the least value of the term independent of x when π16≤θ≤π8, then the ratio l2:l1 is equal to :

The correct option is A 16:1Tr+1= 16Cr(xcosθ)16−r(1xsinθ)r For term independent of x, 16−2r=0⇒r=8 T9= 16C8(1sinθcosθ)8= 16C8 28(1sin2θ)8 l1= 16C8 28 at θ=π4 l2= 16C828(1√2)8= 16C8 212 at θ=π8 l2l1=16:1