# NCERT Solutions Class 12 Maths (Determinants) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-9 (Determinants)Miscellaneous Exercise  This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Q1.

Answer.

Q2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

Answer.   $\begin{array}{rl}& =x\left(a{e}^{x}+b{e}^{-x}+2\right)+2\left(a{e}^{x}-b{e}^{-x}+2x\right)-x\left(a{e}^{x}+b{e}^{-x}+{x}^{2}\right)+{x}^{2}-2\\ & =\left(ax{e}^{x}+bx{e}^{-x}+2x\right)+\left(2a{e}^{x}-2b{e}^{-x}+4x\right)-\left(ax{e}^{x}+bx{e}^{-x}+{x}^{3}\right)+{x}^{2}-2\\ & =2a{e}^{x}-2b{e}^{-x}+{x}^{2}+6x-2\\ & \ne 0\end{array}$ ⇒ L.H.S. ≠ R.H.S. Hence, the given function is not a solution of the corresponding differential equation.  $\begin{array}{l}⇒\frac{{d}^{2}y}{d{x}^{2}}=\left(a+b\right)\cdot \left[{e}^{x}\mathrm{cos}x-{e}^{x}\mathrm{sin}x\right]+\left(b-a\right)\left[{e}^{x}\mathrm{sin}x+{e}^{x}\mathrm{cos}x\right]\\ ⇒\frac{{d}^{2}y}{d{x}^{2}}={e}^{x}\left[\left(a+b\right)\left(\mathrm{cos}x-\mathrm{sin}x\right)+\left(b-a\right)\left(\mathrm{sin}x+\mathrm{cos}x\right)\right]\\ ⇒\frac{{d}^{2}y}{d{x}^{2}}={e}^{x}\left[a\mathrm{cos}x-a\mathrm{sin}x+b\mathrm{cos}x-b\mathrm{sin}x+b\mathrm{sin}x+b\mathrm{cos}x-a\mathrm{sin}x-a\mathrm{cos}x\right]\\ ⇒\frac{{d}^{2}y}{d{x}^{2}}=\left[2{e}^{x}\left(b\mathrm{cos}x-a\mathrm{sin}x\right)\right]\end{array}$  $\begin{array}{rl}& =2{e}^{x}\left(b\mathrm{cos}x-a\mathrm{sin}x\right)-2{e}^{x}\left[\left(a+b\right)\mathrm{cos}x+\left(b-a\right)\mathrm{sin}x\right]+2{e}^{x}\left(a\mathrm{cos}x+b\mathrm{sin}x\right)\\ & ={e}^{x}\left[\left(2b\mathrm{sin}x-2a\mathrm{sin}x\right)-\left(2a\mathrm{cos}x+2b\mathrm{cos}x\right)\\ & ={e}^{x}\left[\left(2b\mathrm{sin}x-2a\mathrm{sin}x\right)+\left(2a\mathrm{cos}x+2b\mathrm{sin}x\right)\right]\\ & ={e}^{x}\left[\left(2b-2a-2b+2a\right)\mathrm{cos}x\right]+{e}^{x}\left[\left(-2a-2b+2a+2b\right)\mathrm{sin}x\right]\\ & =0\end{array}$ Hence, the given function is a solution of the corresponding differential equation.

Q3.