if vartriangle x 20c 1 and cos x and 1 cos x 1 sin x and

If \[\vartriangle \left( x \right) = \left| {\begin{array}{*{20}{c}} 1&{\cos x}&{1 - \cos x} \\ {1 + \sin x}&{\cos x}&{1 + \sin x - \cos x} \\ {\sin x}&{\sin x}&1 \end{array}} \right|\] then $$\int\limits_0^{\frac{\pi }{2}} {\vartriangle \left( x \right)dx} $$ is equal to

A. $$\frac{1}{4}$$

B. $$\frac{1}{2}$$

C. $$0$$

D. $$ - \frac{1}{2}$$

Answer : $$ - \frac{1}{2}$$

Solution :
\[{C_3} \to {C_3} + {C_2} - {C_1}\,\,{\text{gives }}\vartriangle \left( x \right) = \left| {\begin{array}{*{20}{c}} 1&{\cos x}&0 \\ {1 + \sin x}&{\cos x}&0 \\ {\sin x}&{\sin x}&1 \end{array}} \right| = \cos x - \cos x\left( {1 + \sin x} \right) = - \sin x \cdot \cos x\]
$$\therefore \,\,\int\limits_0^{\frac{\pi }{2}} {\vartriangle \left( x \right)dx = - \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\sin 2x\,dx} } $$
\[\int\limits_0^{\frac{\pi }{2}} {\vartriangle \left( x \right)dx = - \frac{1}{2}\left[ { - \frac{{\cos 2x}}{2}} \right]_0^{\frac{\pi }{2}} = \frac{1}{4}\left( {\cos \pi - \cos 0} \right) = - \frac{1}{2}.} \]

Releted MCQ Question on
Algebra >> Matrices and Determinants

Releted Question 1

Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$ be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$ be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then

A. $$C$$ is empty

B. $$B$$ has as many elements as $$C$$

C. $$A = B \cup C$$

D. $$B$$ has twice as many elements as elements as $$C$$

View Answer

Releted Question 2

If $$\omega \left( { \ne 1} \right)$$ is a cube root of unity, then
\[\left| {\begin{array}{*{20}{c}} 1&{1 + i + {\omega ^2}}&{{\omega ^2}}\\ {1 - i}&{ - 1}&{{\omega ^2} - 1}\\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|=\]

A. 0

B. 1

C. $$i$$

D. $$\omega $$

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Releted Question 3

Let $$a, b, c$$ be the real numbers. Then following system of equations in $$x, y$$ and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$ has

A. no solution

B. unique solution

C. infinitely many solutions

D. finitely many solutions

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Releted Question 4

If $$A$$ and $$B$$ are square matrices of equal degree, then which one is correct among the followings?

A. $$A + B = B + A$$

B. $$A + B = A - B$$

C. $$A - B = B - A$$

D. $$AB=BA$$

View Answer

If \[\vartriangle \left( x \right) = \left| {\begin{array}{*{20}{c}} 1&{\cos x}&{1 - \cos x} \\ {1 + \sin x}&{\cos x}&{1 + \sin x - \cos x} \\ {\sin x}&{\sin x}&1 \end{array}} \right|\] then $$\int\limits_0^{\frac{\pi }{2}} {\vartriangle \left( x \right)dx} $$ is equal to

Releted MCQ Question on
Algebra >> Matrices and Determinants

Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$ be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$ be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then

If $$\omega \left( { \ne 1} \right)$$ is a cube root of unity, then
\[\left| {\begin{array}{*{20}{c}} 1&{1 + i + {\omega ^2}}&{{\omega ^2}}\\ {1 - i}&{ - 1}&{{\omega ^2} - 1}\\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|=\]

If $$A$$ and $$B$$ are square matrices of equal degree, then which one is correct among the followings?

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Matrices and Determinants

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If \[\vartriangle \left( x \right) = \left| {\begin{array}{*{20}{c}} 1&{\cos x}&{1 - \cos x} \\ {1 + \sin x}&{\cos x}&{1 + \sin x - \cos x} \\ {\sin x}&{\sin x}&1 \end{array}} \right|\] then $$\int\limits_0^{\frac{\pi }{2}} {\vartriangle \left( x \right)dx} $$ is equal to

Releted MCQ Question on Algebra >> Matrices and Determinants

Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$ be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$ be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then

If $$\omega \left( { \ne 1} \right)$$ is a cube root of unity, then \[\left| {\begin{array}{*{20}{c}} 1&{1 + i + {\omega ^2}}&{{\omega ^2}}\\ {1 - i}&{ - 1}&{{\omega ^2} - 1}\\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|=\]

If $$A$$ and $$B$$ are square matrices of equal degree, then which one is correct among the followings?

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Releted MCQ Question on
Algebra >> Matrices and Determinants

If $$\omega \left( { \ne 1} \right)$$ is a cube root of unity, then
\[\left| {\begin{array}{*{20}{c}} 1&{1 + i + {\omega ^2}}&{{\omega ^2}}\\ {1 - i}&{ - 1}&{{\omega ^2} - 1}\\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|=\]

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Matrices and Determinants