NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.3

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Class 11 Maths NCERT Solutions Chapter 14 Mathematical Reasoning Exercise 14.3 are based on the following concepts: 

  • New statements from old statements
  • Implications
  • Validating Statements
  • Conjunction

Download PDF NCERT Solutions for Class 11 Chapter 14 Mathematical Reasoning Exercise 14.3

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Chapter Related Topics
Irrational Number Prime Number Real Number

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CBSE CLASS XII Related Questions

  • 1.

    Evaluate:
    \[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]


      • 2.
        Find:

        The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

          • \(-\frac{\pi}{2}\)
          • \(-\frac{\pi}{4}\)
          • \(\frac{\pi}{4}\)
          • \(\frac{\pi}{2}\)

        • 3.

          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. 


            • 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 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 \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).


                  • 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 \( \cos^{-1}(x^2) \).

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