Verified Solution • 2025–26 Session moderate

QJEEM19Made0b: JEE Main 2019 Mathematics - Class 12 JEEM19Made0b Solution: If Δ 1 =

If Δ1=|xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x| and
Δ2=|xsin⁡2θcos⁡2θ−sin⁡2θ−x1cos⁡2θ1x|, x≠0x \ne 0 ;

then for all θ∈(0,π2)\theta \in \left( {0,{\pi \over 2}} \right) :

Multiple Choice Options

  • A Δ1−Δ2{\Delta _1} - {\Delta _2} = x (cos 2 θ\theta – cos 4 θ\theta )
  • B Δ1+Δ2{\Delta _1} + {\Delta _2} = - 2x3
  • C Δ1+Δ2{\Delta _1} + {\Delta _2} = – 2(x3 + x –1)
  • D Δ1−Δ2{\Delta _1} - {\Delta _2} = - 2x3

Step-by-Step Verified Solution

Δ1=|xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x|

= x(–x2 –1) – sin θ\theta (–xsin θ\theta – cos θ\theta ) + cos θ\theta (–sin θ\theta + xcos θ\theta )

= –x3 – x + xsin2 θ\theta + sin θ\theta cos θ\theta – cos θ\theta sin θ\theta + xcos2 θ\theta

= –x3 – x + x = –x3

Similarly Δ2=−x3{\Delta _2} = - {x^3}

Δ1+Δ2=−2x3{\Delta _1} + {\Delta _2} = - 2{x^3}

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