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 Ratio and Identites question

2025 · 23 Jan · Shift 2 · Q34
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 Ratio and Identites
  5. /2025 · 23 Jan · Shift 2 · Q34

Trigonometric Ratio and Identites question

2025 · 23 Jan · Shift 2 · Q34

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
Let the range of the function f(x)=6+16cos⁡x⋅cos⁡(π3−x)⋅cos⁡(π3+x)⋅sin⁡3x⋅cos⁡6x,x∈Rf(x)=6+16 \cos x \cdot \cos \left(\frac{\pi}{3}-x\right) \cdot \cos \left(\frac{\pi}{3}+x\right) \cdot \sin 3 x \cdot \cos 6 x, x \in \mathbf{R}f(x)=6+16cosx⋅cos(3π​−x)⋅cos(3π​+x)⋅sin3x⋅cos6x,x∈R be [α,β][\alpha, \beta][α,β]. Then the distance of the point (α,β)(\alpha, \beta)(α,β) from the line 3x+4y+12=03 x+4 y+12=03x+4y+12=0 is :
  1. A
    11
  2. B
    10
  3. C
    8
  4. D
    9
View written solutionFree

Correct answer: A

  1. Given function
f(x)=6+16cos⁡x⋅cos⁡(π3−x)⋅cos⁡(π3+x)⋅sin⁡3x⋅cos⁡6xf(x)=6+16\cos x\cdot \cos\left(\frac{\pi}{3}-x\right)\cdot \cos\left(\frac{\pi}{3}+x\right)\cdot \sin 3x\cdot \cos 6xf(x)=6+16cosx⋅cos(3π​−x)⋅cos(3π​+x)⋅sin3x⋅cos6x

We need the range [α,β][\alpha,\beta][α,β] of f(x)f(x)f(x), then the distance of point (α,β)(\alpha,\beta)(α,β) from the line

3x+4y+12=0.3x+4y+12=0.3x+4y+12=0.
  1. Simplify the product of cosines

Use

cos⁡(A−B)cos⁡(A+B)=cos⁡2A−sin⁡2B.\cos(A-B)\cos(A+B)=\cos^2 A-\sin^2 B.cos(A−B)cos(A+B)=cos2A−sin2B.

Here, A=π3A=\frac{\pi}{3}A=3π​ and B=xB=xB=x, so

cos⁡(π3−x)cos⁡(π3+x)=cos⁡2π3−sin⁡2x=14−sin⁡2x.\cos\left(\frac{\pi}{3}-x\right)\cos\left(\frac{\pi}{3}+x\right) =\cos^2\frac{\pi}{3}-\sin^2 x =\frac14-\sin^2 x.cos(3π​−x)cos(3π​+x)=cos23π​−sin2x=41​−sin2x.

Thus,

cos⁡x⋅cos⁡(π3−x)⋅cos⁡(π3+x)=cos⁡x(14−sin⁡2x).\cos x\cdot \cos\left(\frac{\pi}{3}-x\right)\cdot \cos\left(\frac{\pi}{3}+x\right) =\cos x\left(\frac14-\sin^2 x\right).cosx⋅cos(3π​−x)⋅cos(3π​+x)=cosx(41​−sin2x).

Now,

14−sin⁡2x=14−(1−cos⁡2x)=cos⁡2x−34.\frac14-\sin^2 x=\frac14-(1-\cos^2 x)=\cos^2 x-\frac34.41​−sin2x=41​−(1−cos2x)=cos2x−43​.

So,

cos⁡x(cos⁡2x−34)=14(4cos⁡3x−3cos⁡x)=14cos⁡3x\cos x\left(\cos^2 x-\frac34\right)=\frac14(4\cos^3 x-3\cos x)=\frac14\cos 3xcosx(cos2x−43​)=41​(4cos3x−3cosx)=41​cos3x

using

cos⁡3x=4cos⁡3x−3cos⁡x.\cos 3x=4\cos^3 x-3\cos x.cos3x=4cos3x−3cosx.

Hence the function becomes

f(x)=6+16⋅14cos⁡3x⋅sin⁡3x⋅cos⁡6x.f(x)=6+16\cdot \frac14\cos 3x\cdot \sin 3x\cdot \cos 6x.f(x)=6+16⋅41​cos3x⋅sin3x⋅cos6x.

That is,

f(x)=6+4cos⁡3xsin⁡3xcos⁡6x.f(x)=6+4\cos 3x\sin 3x\cos 6x.f(x)=6+4cos3xsin3xcos6x.
  1. Use product-to-sum identities

Since

2sin⁡3xcos⁡3x=sin⁡6x,2\sin 3x\cos 3x=\sin 6x,2sin3xcos3x=sin6x,

we get

4cos⁡3xsin⁡3xcos⁡6x=2sin⁡6xcos⁡6x=sin⁡12x.4\cos 3x\sin 3x\cos 6x=2\sin 6x\cos 6x=\sin 12x.4cos3xsin3xcos6x=2sin6xcos6x=sin12x.

Therefore,

f(x)=6+sin⁡12x.f(x)=6+\sin 12x.f(x)=6+sin12x.
  1. Find the range

Since

−1≤sin⁡12x≤1,-1\le \sin 12x\le 1,−1≤sin12x≤1,

it follows that

5≤f(x)≤7.5\le f(x)\le 7.5≤f(x)≤7.

So,

α=5,β=7.\alpha=5,\qquad \beta=7.α=5,β=7.

Thus the point is

(α,β)=(5,7).(\alpha,\beta)=(5,7).(α,β)=(5,7).
  1. Distance from the line

Distance of point (x1,y1)(x_1,y_1)(x1​,y1​) from line Ax+By+C=0Ax+By+C=0Ax+By+C=0 is

d=∣Ax1+By1+C∣A2+B2.d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.d=A2+B2​∣Ax1​+By1​+C∣​.

Here, A=3A=3A=3, B=4B=4B=4, C=12C=12C=12, and (x1,y1)=(5,7)(x_1,y_1)=(5,7)(x1​,y1​)=(5,7). So

d=∣3⋅5+4⋅7+12∣32+42=∣15+28+12∣5=555=11.d=\frac{|3\cdot 5+4\cdot 7+12|}{\sqrt{3^2+4^2}} =\frac{|15+28+12|}{5} =\frac{55}{5}=11.d=32+42​∣3⋅5+4⋅7+12∣​=5∣15+28+12∣​=555​=11.
  1. Check options
  • A: 111111 ✅
  • B: 101010 ❌
  • C: 888 ❌
  • D: 999 ❌

So the correct option is A.

PreviousNext

More from Trigonometric Ratio and Identites

  • If r=1∑13​{sin(4π​+(r−1)6π​)sin(4π​+6rπ​)1​}=a3​+b,a,b∈Z, then a2+b2 is equal to :2025 · MCQ
  • If sinx+sin2x=1, x∈(0,2π​), then (cos12x+tan12x)+3(cos10x+tan10x+cos8x+tan8x)+(cos6x+tan6x) is equal to:2025 · MCQ
  • If tanA=x(x2+x+1)​1​,tanB=x2+x+1​x​​ and tanC=(x−3+x−2+x−1)1/2,0<A,B,C<2π​, then A+B…2024 · MCQ
  • Suppose θ∈[0,4π​] is a solution of 4cosθ−3sinθ=1. Then cosθ is equal to :2024 · MCQ
  • If sinx=−53​, where π<x<23π​, then 80(tan2x−cosx) is equal to2024 · MCQ
  • If the value of 5cos36∘−3sin18∘3cos36∘+5sin18∘​ is ca5​−b​, where a,b,c are natural numbers and gcd(a,c)=1, then a+b+c is equal to :2024 · MCQ
  • Let the set of all a∈R such that the equation cos2x+asinx=2a−7 has a solution be [p,q] and r=tan9∘−tan27∘−cot63∘1​+tan81∘, then pqr is equal to ​…2024 · Numerical
  • For α,β∈(0,π/2), let 3sin(α+β)=2sin(α−β) and a real number k be such that tanα=ktanβ. Then, the value of k is equal to2024 · MCQ