NCERT Solutions for Class 12 Maths Chapter 6 Applications of Derivatives Exercise 6.3

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Jasmine Grover

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NCERT Solutions for Class 12 Maths Chapter 6 Applications of Derivatives Exercise 6.3 is provided in this article. Chapter 6 Exercise 6.3 includes questions that deal with concepts of tangents and normals. The exercise includes a total of 27 questions.

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Class 12 Chapter 6 Applications of Derivatives Topics:

CBSE Class 12 Mathematics Study Guides:

CBSE CLASS XII Related Questions

  • 1.
    Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).


      • 2.
        Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]


          • 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.
                  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 \(k\).

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

                  • 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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