Verified Solution • 2025–26 Session moderate

QJEEM18Ma079b: JEE Main 2018 Mathematics - Class 11 JEEM18Ma079b Solution: Consider the following two statements : Statement p : The value …

Consider the following two statements :

Statement p :
The value of sin 120o can be derived by taking θ=240o\theta = {240^o} in the equation
2sin θ2=1+sin⁡θ−1−sin⁡θ{\theta \over 2} = \sqrt {1 + \sin \theta } - \sqrt {1 - \sin \theta }

Statement q :
The angles A, B, C and D of any quadrilateral ABCD satisfy the equation
cos (12(A+C))+cos⁡(12(B+D))=0\left( {{1 \over 2}\left( {A + C} \right)} \right) + \cos \left( {{1 \over 2}\left( {B + D} \right)} \right) = 0

Then the truth values of p and q are respectively :

Multiple Choice Options

  • A F, T
  • B T, F
  • C T, T
  • D F, F

Step-by-Step Verified Solution

Statement p :
sin 120o = cos 30o = 32{{\sqrt 3 } \over 2} ⇒\Rightarrow 2 sin 120o = 3\sqrt 3

So, 1+sin⁡240o−1−sin⁡240o\sqrt {1 + \sin {{240}^o}} - \sqrt {1 - \sin {{240}^o}}

=1−32−1+32≠3= \sqrt {{{1 - \sqrt 3 } \over 2}} - \sqrt {{{1 + \sqrt 3 } \over 2}} \ne \sqrt 3

Statement q :
So,   A + B + C + D = 2 π\pi

⇒\Rightarrow A+C2+B+D2=π{{A + C} \over 2} + {{B + D} \over 2} = \pi

⇒\Rightarrow   cos (A+C2)+cos⁡(B+D2)\left( {{{A + C} \over 2}} \right) + \cos \left( {{{B + D} \over 2}} \right)

= cos (A+C2)\left( {{{A + C} \over 2}} \right) −- cos (A+C2)=0\left( {{{A + C} \over 2}} \right) = 0

Therefore, statement p is false and statement q is true.

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