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Binomial Theorem question

2020 · 9 Jan · Shift 2 · Q23
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  5. /2020 · 9 Jan · Shift 2 · Q23

Binomial Theorem question

2020 · 9 Jan · Shift 2 · Q23

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
In the expansion of (xcos⁡θ+1xsin⁡θ)16{\left( {{x \over {\cos \theta }} + {1 \over {x\sin \theta }}} \right)^{16}}(cosθx​+xsinθ1​)16, if ℓ1{\ell _1}ℓ1​ is the least value of the term independent of x when π8≤θ≤π4{\pi \over 8} \le \theta \le {\pi \over 4}8π​≤θ≤4π​ and ℓ2{\ell _2}ℓ2​ is the least value of the term independent of x when π16≤θ≤π8{\pi \over {16}} \le \theta \le {\pi \over 8}16π​≤θ≤8π​, then the ratio ℓ2{\ell _2}ℓ2​: ℓ1{\ell _1}ℓ1​ is equal to :
  1. A
    8 : 1
  2. B
    16 : 1
  3. C
    1 : 8
  4. D
    1 : 16
View written solutionFree

Correct answer: B

  1. General term in the binomial expansion

Given (xcos⁡θ+1xsin⁡θ)16.\left(\frac{x}{\cos\theta}+\frac{1}{x\sin\theta}\right)^{16}.(cosθx​+xsinθ1​)16.

The general term is Tr+1=(16r)(xcos⁡θ)16−r(1xsin⁡θ)r.T_{r+1}=\binom{16}{r}\left(\frac{x}{\cos\theta}\right)^{16-r}\left(\frac{1}{x\sin\theta}\right)^r.Tr+1​=(r16​)(cosθx​)16−r(xsinθ1​)r.

So,

=\binom{16}{r}x^{16-2r}\frac{1}{\cos^{16-r}\theta\,\sin^r\theta}.$$ 2. **Term independent of $x$** For the term independent of $x$, power of $x$ must be zero: $$16-2r=0 \implies r=8.$$ Hence the constant term is $$T_9=\binom{16}{8}\frac{1}{\cos^8\theta\sin^8\theta}.$$ Using $\sin\theta\cos\theta=\frac{1}{2}\sin2\theta$, $$T_9=\binom{16}{8}\frac{1}{(\sin\theta\cos\theta)^8} =\binom{16}{8}\frac{1}{\left(\frac{1}{2}\sin2\theta\right)^8} =\binom{16}{8}\frac{2^8}{\sin^8 2\theta}.$$ So minimizing the constant term is equivalent to maximizing $\sin 2\theta$. --- 3. **Find $\ell_1$ for $\frac{\pi}{8}\le \theta\le \frac{\pi}{4}$** Here, $$\frac{\pi}{4}\le 2\theta\le \frac{\pi}{2}.$$ On this interval, $\sin 2\theta$ is maximum at $$2\theta=\frac{\pi}{2} \implies \theta=\frac{\pi}{4},$$ and the maximum value is $1$. Therefore, $$\ell_1=\binom{16}{8}2^8.$$ --- 4. **Find $\ell_2$ for $\frac{\pi}{16}\le \theta\le \frac{\pi}{8}$** Here, $$\frac{\pi}{8}\le 2\theta\le \frac{\pi}{4}.$$ On this interval, $\sin 2\theta$ is increasing, so its maximum occurs at $$2\theta=\frac{\pi}{4} \implies \theta=\frac{\pi}{8}.$$ Thus, $$\sin 2\theta=\sin\frac{\pi}{4}=\frac{1}{\sqrt2}.$$ Hence, $$\ell_2=\binom{16}{8}\frac{2^8}{\left(\frac{1}{\sqrt2}\right)^8}.

Now, (12)8=2−4=116.\left(\frac{1}{\sqrt2}\right)^8=2^{-4}=\frac{1}{16}.(2​1​)8=2−4=161​. So, ℓ2=(168)28⋅16.\ell_2=\binom{16}{8}2^8\cdot 16.ℓ2​=(816​)28⋅16.

  1. Compute the ratio

ℓ2:ℓ1=((168)28⋅16):((168)28)=16:1.\ell_2:\ell_1=\left(\binom{16}{8}2^8\cdot 16\right):\left(\binom{16}{8}2^8\right)=16:1.ℓ2​:ℓ1​=((816​)28⋅16):((816​)28)=16:1.

Therefore the correct option is B: 16:1.\boxed{\text{B: }16:1}. B: 16:1​.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer is also B. So they agree.

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