NCERT Solutions for Class 12 Chapter 9 Differential Equations Exercise 9.1 Solutions

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Class 12 Maths NCERT Solutions Chapter 9 Differential Equations Exercise 9.1 is provided in the article. Class 12 Chapter 9 Differential Equations Exercises include questions on Order and Degree of Differential Equations, Formation of Differential Equation, Methods of Solving First Order, First Degree Differential Equations.

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CBSE CLASS XII Related Questions

  • 1.
    Find:

    If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

      • \(0\)
      • \(-2\)
      • \(-1\)
      • \(2\)

    • 2.
      Find:

      The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]


        • 3.
          Find:

          If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

            • \(p = -\frac{2}{3}, \, q = \frac{5}{3}\)
            • \(p = -\frac{8}{3}, \, q = \frac{20}{3}\)
            • \(p = \frac{20}{3}, \, q = -\frac{8}{3}\)
            • \(p = 0, \, q = 0\)

          • 4.

            Find:
            Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

              • \(\begin{bmatrix} \frac{1}{3} & 3 \\ 0 & \frac{4}{3} \end{bmatrix}\)
              • \(\begin{bmatrix} \frac{1}{3} & 0 \\ 3 & \frac{4}{3} \end{bmatrix}\)
              • \(\begin{bmatrix} \frac{4}{3} & 3 \\ 1 & 0 \end{bmatrix}\)
              • \(\begin{bmatrix} \frac{4}{3} & 0 1 & \frac{3}{3} \end{bmatrix}\)

            • 5.
              Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).


                • 6.
                  Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

                    CBSE CLASS XII Previous Year Papers

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