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10 Question MCQ (including Assertion)

Chapter 4 Class 12 Determinants | 10 questions | about 8 minutes

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Question 1 of 10
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Question 1 of 10
NCERT Exemplar
Let \(f(t)=\left|\begin{array}{ccc}\cos t & t & 1 \\ 2 \sin t & t & 2 t \\ \sin t & t & t\end{array}\right|\), then \(\lim _{t \rightarrow 0} \frac{f(t)}{t^2}\) is equal to
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Question 2 of 10
NCERT Exemplar
The number of distinct real roots of \(\left|\begin{array}{ccc}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0\) in the interval \(-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}\) is
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Question 3 of 10
If \(A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\), then \(14 A^{-1}\) is given by:
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Question 4 of 10
NCERT Exemplar
If \(\left|\begin{array}{cc}2 x & 5 \\ 8 & x\end{array}\right|=\left|\begin{array}{cc}6 & -2 \\ 7 & 3\end{array}\right|\), then value of \(x\) is
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Question 5 of 10
CBSE Board Exam 2024
Let \(\mathrm{A}=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right]\) be a square matrix such that adj \(\mathrm{A}=\mathrm{A}\). Then, \((\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})\) is equal to :
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Question 6 of 10
CBSE Board Exam 2026
If \(\mathrm{A}(\operatorname{adj} \mathrm{A})=\left[\begin{array}{ccc}2026 & 0 & 0 \\ 0 & 2026 & 0 \\ 0 & 0 & 2026\end{array}\right]\), then the value of \(|\operatorname{adj} \mathrm{A}|\) is equal to :
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Question 7 of 10
CBSE Board Exam 2024
If \(\left|\begin{array}{ccc}1 & 3 & 1 \\ \mathrm{k} & 0 & 1 \\ 0 & 0 & 1\end{array}\right|= \pm 6\), then the value of k is :
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Question 8 of 10
NCERT Exemplar
If A, B and C are angles of a triangle, then the determinant \(\left|\begin{array}{ccc}-1 & \cos \mathrm{C} & \cos \mathrm{B} \\ \cos \mathrm{C} & -1 & \cos \mathrm{~A} \\ \cos \mathrm{~B} & \cos \mathrm{~A} & -1\end{array}\right|\) is equal to
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Question 9 of 10
NCERT Exemplar
If \(f(x)=\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|\), then
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Question 10 of 10
NCERT Exemplar
The value of determinant \(\left|\begin{array}{lll}a-b & b+c & a \\ b-a & c+a & b \\ c-a & a+b & c\end{array}\right|\)
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