NCERT Solutions for Class 12 Maths Chapter 8 Applications of Integrals Miscellaneous Exercise Solutions

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Class 12 Maths NCERT Solutions Chapter 8 Applications of Integrals Miscellaneous Exercises are provided in the article. Class 12 Chapter 8 Applications of Integrals Miscellaneous Exercises are important for both CBSE Term II exam and for competitive exams. Key topics covered in this chapter are Area Between Two Curves, lines, parabolas; area of circles/ellipses.

Download PDF: NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercises

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CBSE CLASS XII Related Questions

  • 1.
    If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).


      • 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.

              A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length. 


                • 5.

                  At a birthday party, children are being served orange juice in conical cups, as shown in the figure. 


                  Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
                  On the basis of the above information, answer the following questions :


                    • 6.
                      Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).

                        CBSE CLASS XII Previous Year Papers

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