Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Trigonometric Functions and Equations question

2024 · 4 Apr · Shift 2 · Q51
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Trigonometric Functions and Equations
  5. /2024 · 4 Apr · Shift 2 · Q51

Trigonometric Functions and Equations question

2024 · 4 Apr · Shift 2 · Q51

JEE MainMathematicsTrigonometric Functions and EquationsNumerical+4 / −1
Let S={sin⁡22θ:(sin⁡4θ+cos⁡4θ)x2+(sin⁡2θ)x+(sin⁡6θ+cos⁡6θ)=0S=\left\{\sin ^2 2 \theta:\left(\sin ^4 \theta+\cos ^4 \theta\right) x^2+(\sin 2 \theta) x+\left(\sin ^6 \theta+\cos ^6 \theta\right)=0\right.S={sin22θ:(sin4θ+cos4θ)x2+(sin2θ)x+(sin6θ+cos6θ)=0 has real roots }\}}. If α\alphaα and β\betaβ be the smallest and largest elements of the set SSS, respectively, then 3((α−2)2+(β−1)2)3\left((\alpha-2)^2+(\beta-1)^2\right)3((α−2)2+(β−1)2) equals ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 4

Let A= rac{}{ }\sin^4\theta+\cos^4\theta,\qquad B=\sin 2\theta,\qquad C=\sin^6\theta+\cos^6\theta. We are given the quadratic Ax2+Bx+C=0Ax^2+Bx+C=0Ax2+Bx+C=0 and want the set S={sin⁡22θ:the quadratic has real roots}.S=\{\sin^2 2\theta : \text{the quadratic has real roots}\}.S={sin22θ:the quadratic has real roots}.

We must find the smallest and largest possible values of t=sin⁡22θt=\sin^2 2\thetat=sin22θ for which the quadratic has real roots.

1. Rewrite everything in terms of t=sin⁡22θt=\sin^2 2\thetat=sin22θ

Use standard identities: sin⁡4θ+cos⁡4θ=(sin⁡2θ+cos⁡2θ)2−2sin⁡2θcos⁡2θ=1−2sin⁡2θcos⁡2θ.\sin^4\theta+\cos^4\theta=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta=1-2\sin^2\theta\cos^2\theta.sin4θ+cos4θ=(sin2θ+cos2θ)2−2sin2θcos2θ=1−2sin2θcos2θ. Since sin⁡22θ=4sin⁡2θcos⁡2θ=t,\sin^2 2\theta=4\sin^2\theta\cos^2\theta=t,sin22θ=4sin2θcos2θ=t, we get sin⁡2θcos⁡2θ=t4,\sin^2\theta\cos^2\theta=\frac t4,sin2θcos2θ=4t​, so A=1−2⋅t4=1−t2.A=1-2\cdot \frac t4=1-\frac t2.A=1−2⋅4t​=1−2t​.

Now, sin⁡6θ+cos⁡6θ=(sin⁡2θ)3+(cos⁡2θ)3.\sin^6\theta+\cos^6\theta=(\sin^2\theta)^3+(\cos^2\theta)^3.sin6θ+cos6θ=(sin2θ)3+(cos2θ)3. Using a3+b3=(a+b)3−3ab(a+b),a^3+b^3=(a+b)^3-3ab(a+b),a3+b3=(a+b)3−3ab(a+b), with a=sin⁡2θa=\sin^2\thetaa=sin2θ, b=cos⁡2θb=\cos^2\thetab=cos2θ, and a+b=1a+b=1a+b=1, we get C=1−3sin⁡2θcos⁡2θ=1−3⋅t4=1−3t4.C=1-3\sin^2\theta\cos^2\theta=1-3\cdot \frac t4=1-\frac{3t}{4}.C=1−3sin2θcos2θ=1−3⋅4t​=1−43t​. Also, B=sin⁡2θ,B=\sin 2\theta,B=sin2θ, so B2=t.B^2=t.B2=t.

Thus the quadratic has coefficients A=1−t2,B=sin⁡2θ,C=1−3t4,A=1-\frac t2,\qquad B=\sin 2\theta,\qquad C=1-\frac{3t}{4},A=1−2t​,B=sin2θ,C=1−43t​, where 0≤t≤10\le t\le 10≤t≤1.

2. Condition for real roots

A quadratic has real roots when its discriminant is nonnegative: B2−4AC≥0.B^2-4AC\ge 0.B2−4AC≥0. Substitute: t−4(1−t2)(1−3t4)≥0.t-4\left(1-\frac t2\right)\left(1-\frac{3t}{4}\right)\ge 0.t−4(1−2t​)(1−43t​)≥0. Now simplify the product: (1−t2)(1−3t4)=1−3t4−t2+3t28=1−5t4+3t28.\left(1-\frac t2\right)\left(1-\frac{3t}{4}\right)=1-\frac{3t}{4}-\frac t2+\frac{3t^2}{8}=1-\frac{5t}{4}+\frac{3t^2}{8}.(1−2t​)(1−43t​)=1−43t​−2t​+83t2​=1−45t​+83t2​. Multiply by 4: 4AC=4−5t+3t22.4AC=4-5t+\frac{3t^2}{2}.4AC=4−5t+23t2​. Hence t−(4−5t+3t22)≥0,t-\left(4-5t+\frac{3t^2}{2}\right)\ge 0,t−(4−5t+23t2​)≥0, 6t−4−3t22≥0.6t-4-\frac{3t^2}{2}\ge 0.6t−4−23t2​≥0. Multiply by 2: 12t−8−3t2≥0,12t-8-3t^2\ge 0,12t−8−3t2≥0, 3t2−12t+8≤0.3t^2-12t+8\le 0.3t2−12t+8≤0.

Solve the quadratic equation 3t2−12t+8=0.3t^2-12t+8=0.3t2−12t+8=0. Its roots are t=12±144−966=12±436=2±233.t=\frac{12\pm\sqrt{144-96}}{6}=\frac{12\pm 4\sqrt3}{6}=2\pm \frac{2\sqrt3}{3}.t=612±144−96​​=612±43​​=2±323​​. So the inequality gives 2−233≤t≤2+233.2-\frac{2\sqrt3}{3}\le t\le 2+\frac{2\sqrt3}{3}.2−323​​≤t≤2+323​​. But also t=sin⁡22θ∈[0,1]t=\sin^2 2\theta\in[0,1]t=sin22θ∈[0,1]. Therefore, S=[2−233, 1].S=\left[2-\frac{2\sqrt3}{3},\,1\right].S=[2−323​​,1]. So α=2−233,β=1.\alpha=2-\frac{2\sqrt3}{3},\qquad \beta=1.α=2−323​​,β=1.

3. Compute the required expression

We need 3((α−2)2+(β−1)2).3\left((\alpha-2)^2+(\beta-1)^2\right).3((α−2)2+(β−1)2). Now, α−2=−233  ⟹  (α−2)2=43,\alpha-2=-\frac{2\sqrt3}{3}\implies (\alpha-2)^2=\frac{4}{3},α−2=−323​​⟹(α−2)2=34​, and (β−1)2=(1−1)2=0.(\beta-1)^2=(1-1)^2=0.(β−1)2=(1−1)2=0. Thus 3(43+0)=4.3\left(\frac{4}{3}+0\right)=4.3(34​+0)=4.

4. Final answer

4\boxed{4}4​

This matches the stored correct answer.

PreviousNext

More from Trigonometric Functions and Equations

  • The number of solutions of sin2x+(2+2x−x2)sinx−3(x−1)2=0, where −π≤x≤π, is ​.2024 · Numerical
  • Let ∣cosθcos(60−θ)cos(60+θ)∣≤81​,θϵ[0,2π]. Then, the sum of all θ∈[0,2π], where cos3θ attains its maximum value, is :2024 · MCQ
  • If 2tan2θ−5secθ=1 has exactly 7 solutions in the interval [0,2nπ​], for the least value of n∈N, then ∑k=1n​2kk​ is equal to:2024 · MCQ
  • If α,−2π​<α<2π​ is the solution of 4cosθ+5sinθ=1, then the value of tanα is2024 · MCQ
  • The sum of the solutions x∈R of the equation cos6x−sin6x3cos2x+cos32x​=x3−x2+6 is2024 · MCQ
  • If 2sin3x+sin2xcosx+4sinx−4=0 has exactly 3 solutions in the interval [0,2nπ​],n∈N, then the roots of the equation x2+nx+(n−3)=0 belong to :2024 · MCQ
  • If θ∈[−2π,2π], then the number of solutions of 22​cos2θ+(2−6​)cosθ−3​=0, is equal to:2025 · MCQ
  • If θϵ[−67π​,34π​], then the number of solutions of 3​cosec2θ−2(3​−1)cosecθ−4=0, is equal to :2025 · MCQ