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Functions question

2024 · 8 Apr · Shift 1 · Q52
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Functions question

2024 · 8 Apr · Shift 1 · Q52

JEE MainMathematicsFunctionsNumerical+4 / −1
If the range of f(θ)=sin⁡4θ+3cos⁡2θsin⁡4θ+cos⁡2θ,θ∈Rf(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta}, \theta \in \mathbb{R}f(θ)=sin4θ+cos2θsin4θ+3cos2θ​,θ∈R is [α,β][\alpha, \beta][α,β], then the sum of the infinite G.P., whose first term is 64 and the common ratio is αβ\frac{\alpha}{\beta}βα​, is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 96

  1. Rewrite the function in one variable

Let x=cos⁡2θ.x=\cos^2\theta.x=cos2θ. Since θ∈R\theta\in\mathbb Rθ∈R, we have 0≤x≤1.0\le x\le 1.0≤x≤1. Also, sin⁡2θ=1−x⇒sin⁡4θ=(1−x)2.\sin^2\theta=1-x \quad\Rightarrow\quad \sin^4\theta=(1-x)^2.sin2θ=1−x⇒sin4θ=(1−x)2.

So

=\frac{(1-x)^2+3x}{(1-x)^2+x}.$$ Simplify: $$f(x)=\frac{1-2x+x^2+3x}{1-2x+x^2+x} =\frac{x^2+x+1}{x^2-x+1}, \quad 0\le x\le 1.$$ 2. **Find the range of** $f(x)$ We need the maximum and minimum of $$f(x)=\frac{x^2+x+1}{x^2-x+1}.$$ Differentiate: $$f'(x)=\frac{(2x+1)(x^2-x+1)-(x^2+x+1)(2x-1)}{(x^2-x+1)^2}.$$ Now simplify the numerator:

(2x+1)(x^2-x+1)=2x^3-x^2+x+1,

(x^2+x+1)(2x-1)=2x^3+x^2+x-1.

Therefore,Therefore,Therefore,

(2x+1)(x^2-x+1)-(x^2+x+1)(2x-1) =(2x^3-x^2+x+1)-(2x^3+x^2+x-1) =2-2x^2=2(1-x^2).

Hence $$f'(x)=\frac{2(1-x^2)}{(x^2-x+1)^2}.$$ For $0\le x\le 1$, the denominator is always positive, and $$1-x^2\ge 0,$$ with equality only at $x=1$. So $f(x)$ is increasing on $[0,1]$. Thus, - minimum at $x=0$, - maximum at $x=1$. Compute: $$f(0)=\frac{1}{1}=1,$$ $$f(1)=\frac{3}{1}=3.$$ So the range is $$[\alpha,\beta]=[1,3].$$ Thus, $$\alpha=1,\quad \beta=3.$$ 3. **Find the sum of the infinite G.P.** First term: $$a=64$$ Common ratio: $$r=\frac{\alpha}{\beta}=\frac{1}{3}.$$ Since $|r|<1$, the sum to infinity is $$S_\infty=\frac{a}{1-r}= rac{64}{1-\frac13} =\frac{64}{\frac23}=64\cdot\frac32=96.$$ 4. **Final answer** $$\boxed{96}$$ This matches the stored correct answer.
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