Trigonometric elements MCQ Quiz in मराठी - Objective Question with Answer for Trigonometric elements - मोफत PDF डाउनलोड करा

Last updated on Mar 22, 2025

पाईये Trigonometric elements उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Trigonometric elements एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Trigonometric elements MCQ Objective Questions

Top Trigonometric elements MCQ Objective Questions

If \(\rm m = \left[ \begin{array}{cc} 2 &0\\ 0&1\end{array}\right] and \ n= \left[ \begin{array}{cc} 0 &1\\ -2&0\end{array}\right]\) then what is the value of the determinant of m sinθ + n cosθ 

  1. 1
  2. -1
  3. 3
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Trigonometric elements Question 1 Detailed Solution

Download Solution PDF

गणना: 

m sinθ + n cosθ = sin θ \(\rm \left[ \begin{array}{cc} 2 &0\\ 0&1\end{array}\right]\) + cos θ \(\rm \left[ \begin{array}{cc} 0 &1\\ -2&0\end{array}\right]\)

\(\rm \left[ \begin{array}{cc} 2\sin θ &0\\ 0&\sin θ\end{array}\right]\)+\(\rm \left[ \begin{array}{cc} 0 &\cosθ\\ -2\cosθ&0\end{array}\right]\)

\(\rm \left[\begin{array}{cc} 2\sin θ &\cos θ\\ -2\cos θ&\sin θ\end{array}\right]\)

आता,

|m sinθ + n cosθ| = \(\rm \left| \begin{array}{cc} 2\sin θ &\cos θ\\ -2\cos θ&\sin θ\end{array}\right|\)

\(2\sin^2\theta \) + \(\rm 2 cos^2\theta \)

= 2(\(\sin ^2 \theta +\cos^2\theta \))

= 2

म्हणून, पर्याय (4) योग्य आहे.

Trigonometric elements Question 2:

If \(\rm m = \left[ \begin{array}{cc} 2 &0\\ 0&1\end{array}\right] and \ n= \left[ \begin{array}{cc} 0 &1\\ -2&0\end{array}\right]\) then what is the value of the determinant of m sinθ + n cosθ 

  1. 1
  2. -1
  3. 3
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Trigonometric elements Question 2 Detailed Solution

गणना: 

m sinθ + n cosθ = sin θ \(\rm \left[ \begin{array}{cc} 2 &0\\ 0&1\end{array}\right]\) + cos θ \(\rm \left[ \begin{array}{cc} 0 &1\\ -2&0\end{array}\right]\)

\(\rm \left[ \begin{array}{cc} 2\sin θ &0\\ 0&\sin θ\end{array}\right]\)+\(\rm \left[ \begin{array}{cc} 0 &\cosθ\\ -2\cosθ&0\end{array}\right]\)

\(\rm \left[\begin{array}{cc} 2\sin θ &\cos θ\\ -2\cos θ&\sin θ\end{array}\right]\)

आता,

|m sinθ + n cosθ| = \(\rm \left| \begin{array}{cc} 2\sin θ &\cos θ\\ -2\cos θ&\sin θ\end{array}\right|\)

\(2\sin^2\theta \) + \(\rm 2 cos^2\theta \)

= 2(\(\sin ^2 \theta +\cos^2\theta \))

= 2

म्हणून, पर्याय (4) योग्य आहे.

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