NCERT Solutions for Differential Equations Exercise 9.2 Solutions

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Class 12 Maths NCERT Solutions Chapter 9 Differential Equations Exercise 9.2 is provided in the article. Class 12 Chapter 9 Differential Equations Exercises include questions on Order and Degree of Differential Equations.

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CBSE CLASS XII Related Questions

  • 1.
    Find the domain of \(p(x)=\sin^{-1}(1-2x^2)\). Hence, find the value of \(x\) for which \(p(x)=\frac{\pi}{6}\). Also, write the range of \(2p(x)+\frac{\pi}{2}\).


      • 2.
        Obtain the value of \[ \Delta = \begin{vmatrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{vmatrix} \] in terms of \(x, y, z\). Further, if \(\Delta = 0\) and \(x, y, z\) are non–zero real numbers, prove that \[ x^{-1} + y^{-1} + z^{-1} = -1 \]


          • 3.
            Find the general solution of the differential equation \[ y\log y\,\frac{dx}{dy}+x=\frac{2}{y}. \]


              • 4.
                Evaluate : \[ \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}}(\sin|x|+\cos|x|)\,dx \]


                  • 5.

                    Smoking increases the risk of lung problems. A study revealed that 170 in 1000 males who smoke develop lung complications, while 120 out of 1000 females who smoke develop lung related problems. In a colony, 50 people were found to be smokers of which 30 are males. A person is selected at random from these 50 people and tested for lung related problems. Based on the given information answer the following questions: 

                    (i) What is the probability that selected person is a female? 
                    (ii) If a male person is selected, what is the probability that he will not be suffering from lung problems? 
                    (iii)(a) A person selected at random is detected with lung complications. Find the probability that selected person is a female. 
                    OR 
                    (iii)(b) A person selected at random is not having lung problems. Find the probability that the person is a male. 
                     


                      • 6.
                        If \[ P = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \quad \text{and} \quad Q = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 1 & -1 & 5 \end{bmatrix} \] find \( QP \) and hence solve the following system of equations using matrix method:
                        \[ x - y = 3,\quad 2x + 3y + 4z = 13,\quad y + 2z = 7 \]

                          CBSE CLASS XII Previous Year Papers

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