NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.4

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CBSE CLASS XII Related Questions

  • 1.

    Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not. 


      • 2.
        A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.


          • 3.

            For two vectors \(\vec{a}\) and \(\vec{b}\):  

            Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

              • Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
              • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
              • Assertion (A) is true, but Reason (R) is false.
              • Assertion (A) is false, but Reason (R) is true.

            • 4.

              Check whether the function \[ f(x)= \begin{cases} \dfrac{|x-3|}{2(x-3)}, & x<3,\\[6pt] \dfrac{x-6}{6}, & x\geq 3 \end{cases} \] is continuous at \(x=3\) or not. 


                • 5.

                  If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is: 

                    • \(\frac{1}{3}\)
                    • \(\frac{1}{9}\)
                    • \(3\)
                    • \(9\)

                  • 6.
                    For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

                      • local maximum value is 2
                      • local minimum value is \( -2 \)
                      • local maximum value is \( -2 \)
                      • local minimum value \( < \) local maximum value
                    CBSE CLASS XII Previous Year Papers

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