Class 12th

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{\pi }{\int }}xlogsinxdx}\dots \left(i\right)\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\pi -x\right)logsin\left(\pi -x\right)dx}\left[{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)dx={\displaystyle \underset{0}{\overset{a}{\int }}f\left(a-x\right)dx}}\right]\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\pi -x\right)logsinxdx}\dots \left(ii\right)\\ Adding\left(i\right)and\left(ii\right)\\ 2I={\displaystyle \underset{0}{\overset{\pi }{\int }}\left[\left(\pi -x\right)logsinx+xlogsinx\right]dx}\\ 2I={\displaystyle \underset{0}{\overset{\pi }{\int }}\pi logsinxdx}\\ 2I=2\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinxdx\left[?{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)dx=2{\displaystyle \underset{0}{\overset{a/2}{\int }}f\left(x\right)dx}}\right]}\\ \therefore I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinxdx}\dots \left(iii\right)\\ I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsin\left(\frac{\pi }{2}-x\right)dx}\\ I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logcosxdx}\dots \left(iv\right)\\ Onadding\left(iii\right)and\left(iv\right),weget\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(logsinx+logcosx\right)dx}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinxcosxdx}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{log2sinxcosx}{2}dx}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsin2xdx}-\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}log2dx}\\ Put2x=t\Rightarrow 2dx=dt\Rightarrow dx=\frac{dt}{2}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\pi }{\int }}logsintdt}-\pi .log2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}1dx}\left[Changingthe\text{\hspace{0.17em}limit}\right]\\ 2I=I-\pi .log2{\left[x\right]}_0^{\frac{\pi }{2}}\text{\&}\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{1}{\int }}\underset{II}{xlog}}\underset{I}{\left|1+2x\right|}dx\\ ={\left[log\left|1+2x\right|.\left(\frac{x^2}{2}\right)\right]}_0^1-{\displaystyle \underset{0}{\overset{1}{\int }}\left(\frac{1.2}{1+2x}.\frac{x^2}{2}\right)}dx\\ =\frac{1}{2}{\left[x^{2}log\left(1+2x\right)\right]}_0^1-{\displaystyle \underset{0}{\overset{1}{\int }}\left(\frac{x^2}{1+2x}\right)}dx\\ =\frac{1}{2}\left[log3-0\right]-{\displaystyle \underset{0}{\overset{1}{\int }}\left(\frac{x}{2}-\frac{x/2}{1+2x}\right)}dx\\ =\frac{1}{2}log3-\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}x}dx+\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{x}{1+2x}}dx\\ =\frac{1}{2}log3-\frac{1}{2}{\left[\frac{x^2}{2}\right]}_0^1+\frac{1}{2}.\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(2x+1-1\right)}{1+2x}}dx\\ =\frac{1}{2}log3-\frac{1}{4}\left[1-0\right]+\frac{1}{4}{\displaystyle \underset{0}{\overset{1}{\int }}1dx}-\frac{1}{4}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{2x+1}dx}\\ =\frac{1}{2}log3-\frac{1}{4}+\frac{1}{4}{\left[x\right]}_0^1-\frac{1}{4}.\frac{1}{2}{\left[log\left|2x+1\right|\right]}_0^1\\ =\frac{1}{2}log3-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}\left[log3-0\right]\\ =\frac{1}{2}log3-\frac{1}{8}log3=\frac{3}{8}log3\\ Hence,I=\frac{3}{8}log3.\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

This is a Long Answer Type Questions as classified in NCERT Exemplar

Some of the most frequently tested topics for high scores are finding the adjoint and inverse of a matrix, solving equations using Cramer's Rule, expansion using cofactors, and properties of determinants. If students practice questions based on these topics, they can get high scores in the CBSE Board exam and entrance exams.

In finding inverses of matrices, solving systems of linear equations, and understanding matrix transformations, the determinants play a crucial role. The Determinants concept is widely used in physics, engineering, computer Science and competitive exams like JEE. Understanding determinants is important for higher mathematical studies.

The exemplar is beyond the NCERT textbook as it is conceptually more deep and challenging than the NCERT textbook exercises. These MCQs, LA, and SA questions help students improve their logical reasoning and help them prepare for competitive exams. It tests students on higher-order and application-based questions and promotes analytical thinking.

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: (i) — (e);  (ii) — (d);  (iii) — (a);  (iv) — (b);  (v) — (f); (vi) — (c)

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: (i) →  (c); (ii) →  (d); (iii) →  (a); (iv) →  (b).

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: (i) →  (b); (ii) →  (e); (iii) →  (d); (iv) →  (a); (v) →  (c)