NCERT Solutions for Miscellaneous Exercise Chapter 4 Class 12

NCERT Solutions for Miscellaneous Exercise Chapter 4 Class 12 - Determinants

Edited By Ramraj Saini | Updated on Dec 04, 2023 12:15 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Chapter 4 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 4 class 12 Determinants are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. As the name suggests miscellaneous exercise consists of mixed questions from all other exercises of the chapter. In NCERT solutions for Class 12 Maths chapter 4 miscellaneous exercise, you will get questions like solving determinants using cofactor expansion, solving determinants using properties, solving system of linear equation, checking the consistency of the system of linear equations, etc.

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In this Class 12 Maths chapter 4 miscellaneous solutions, you will get some difficult questions as compared to previous exercises. So, If you are not able to solve these questions at first by yourself, you don't need to be a worry. Over 95% of the questions in the board exams are not asked from Class 12 Maths chapter 4 miscellaneous exercise. You can Check Class 12 Maths chapter 4 miscellaneous exercise solutions in this article. Miscellaneous exercise class 12 chapter 4 are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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Determinants Miscellaneous Exercise

Question:1 Prove that the determinant $\begin{vmatrix} x & \sin \theta &\cos \theta \\ -\sin \theta &-x & 1\\ \cos \theta &1 &x \end{vmatrix}$ is independent of $\theta$ .

Answer:

Calculating the determinant value of $\begin{vmatrix} x & \sin \theta &\cos \theta \\ -\sin \theta &-x & 1\\ \cos \theta &1 &x \end{vmatrix}$ ;

$= x\begin{bmatrix} -x &1 \\ 1& x \end{bmatrix}-\sin \Theta\begin{bmatrix} -\sin \Theta &1 \\ \cos \Theta& x \end{bmatrix} + \cos \Theta \begin{bmatrix} -\sin \Theta &-x \\ \cos \Theta& 1 \end{bmatrix}$

$= x(-x^2-1)-\sin \Theta (-x\sin \Theta-\cos \Theta)+\cos\Theta(-\sin \Theta+x\cos\Theta)$

$= -x^3-x+x\sin^2 \Theta+ \sin \Theta\cos \Theta-\cos\Theta\sin \Theta+x\cos^2\Theta$

$= -x^3-x+x(\sin^2 \Theta+\cos^2\Theta)$

$= -x^3-x+x = -x^3$

Clearly, the determinant is independent of $\Theta$ .

Question:2 Without expanding the determinant, prove that
$\begin{vmatrix} a &a^2 &bc \\ b& b^2 &ca \\ c & c^2 &ab \end{vmatrix}= \begin{vmatrix} 1 &a^2 &a^3 \\ 1 &b^2 &b^3 \\ 1 & c^2 &c^3 \end{vmatrix}$

Answer:

We have the

$L.H.S. = \begin{vmatrix} a &a^2 &bc \\ b& b^2 &ca \\ c & c^2 &ab \end{vmatrix}$

Multiplying rows with a, b, and c respectively.

$R_{1} \rightarrow aR_{1}, R_{2} \rightarrow bR_{2},\ and\ R_{3} \rightarrow cR_{3}$

we get;

$= \frac{1}{abc} \begin{vmatrix} a^2 &a^3 &abc \\ b^2& b^3 &abc \\ c^2& c^3 & abc \end{vmatrix}$

$= \frac{1}{abc}.abc \begin{vmatrix} a^2 &a^3 & 1\\ b^2& b^3 &1 \\ c^2& c^3 & 1 \end{vmatrix}$ $[after\ taking\ out\ abc\ from\ column\ 3].$

$= \begin{vmatrix} a^2 &a^3 & 1\\ b^2& b^3 &1 \\ c^2& c^3 & 1 \end{vmatrix} = \begin{vmatrix} 1&a^2 & a^3\\ 1& b^2 &b^3 \\ 1& c^2 & c^3 \end{vmatrix}$ $[Applying\ C_{1}\leftrightarrow C_{3}\ and\ C_{2} \leftrightarrow C_{3}]$

= R.H.S.

Hence proved. L.H.S. =R.H.S.

Question:3 Evaluate $\begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta &-\sin \alpha \\ -\sin \beta & \cos \beta &0 \\ \sin \alpha \cos \beta &\sin \alpha \sin \beta & \cos \alpha \end{vmatrix}$ .

Answer:

Given determinant $\begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta &-\sin \alpha \\ -\sin \beta & \cos \beta &0 \\ \sin \alpha \cos \beta &\sin \alpha \sin \beta & \cos \alpha \end{vmatrix}$ ;

$= \cos \alpha \cos \beta \begin{vmatrix} \cos \beta &0 \\ \sin \alpha \sin \beta & \cos \alpha \end{vmatrix} - \cos \alpha \sin \beta \begin{vmatrix} -\sin \beta & 0 \\ \sin \alpha \cos \beta & \cos \alpha \end{vmatrix} -\sin \alpha \begin{vmatrix} -\sin \beta &\cos \beta \\ \sin \alpha \cos \beta& \sin \alpha \sin \beta \end{vmatrix}$ $= \cos \alpha \cos \beta (\cos \beta \cos \alpha -0 )- \cos \alpha \sin \beta (-\cos \alpha\sin \beta- 0) -\sin \alpha (-\sin \alpha\sin^2\beta - \sin \alpha \cos^2 \beta)$

$= \cos^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta +\sin^2 \alpha \sin^2\beta + \sin^2 \alpha \cos^2 \beta$

$= \cos^2 \alpha(\cos^2 \beta+\sin^2 \beta) +\sin^2 \alpha(\sin^2\beta+\cos^2 \beta)$

$= \cos^2 \alpha(1) +\sin^2 \alpha(1) = 1$ .

Question:4 If $a,b$ and $c$ are real numbers, and

$\Delta =\begin{vmatrix} b+c & c+a &a+b \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}=0$

Show that either $a+b+c=0$ or $a=b=c$

Answer:

We have given $\Delta =\begin{vmatrix} b+c & c+a &a+b \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}=0$

Applying the row transformations; $R_{1} \rightarrow R_{1} +R_{2} +R_{3}$ we have;

$\Delta =\begin{vmatrix} 2(a+b+c) & 2(a+b+c) &2(a+b+c) \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}$

Taking out common factor 2(a+b+c) from the first row;

$\Delta =2(a+b+c)\begin{vmatrix} 1 & 1 &1 \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}$

Now, applying the column transformations; $C_{1}\rightarrow C_{1} - C_{2}\ and\ C_{2} \rightarrow C_{2}- C_{3}$

we have;

$=2(a+b+c)\begin{vmatrix} 0 & 0 &1 \\ c-b &a-c &b+c \\ a-c & b-a & c+a \end{vmatrix}$

$=2(a+b+c)[(c-b)(b-a)-(a-c)^2]$

$=2(a+b+c)[ab+bc+ca-a^2-b^2-c^2]$

and given that the determinant is equal to zero. i.e., $\triangle = 0$ ;

$(a+b+c)[ab+bc+ca-a^2-b^2-c^2] = 0$

So, either $(a+b+c) = 0$ or $[ab+bc+ca-a^2-b^2-c^2] = 0$ .

we can write $[ab+bc+ca-a^2-b^2-c^2] = 0$ as;

$\Rightarrow -2ab-2bc-2ca+2a^2+2b^2+2c^2 =0$

$\Rightarrow (a-b)^2+(b-c)^2+(c-a)^2 =0$

$\because (a-b)^2,(b-c)^2,\ and\ (c-a)^2$ are non-negative.

Hence $(a-b)^2= (b-c)^2=(c-a)^2 = 0$ .

we get then $a=b=c$

Therefore, if given $\triangle$ = 0 then either $(a+b+c) = 0$ or $a=b=c$ .

Question:5 Solve the equation

$\begin{vmatrix} x+a & x &x \\ x &x+a &x\\ x & x & x+a \end{vmatrix}=0; a\neq 0$

Answer:

Given determinant $\begin{vmatrix} x+a & x &x \\ x &x+a &x\\ x & x & x+a \end{vmatrix}=0; a\neq 0$

Applying the row transformation; $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ we have;

$\begin{vmatrix} 3x+a & 3x+a &3x+a \\ x &x+a &x\\ x & x & x+a \end{vmatrix} =0$

Taking common factor (3x+a) out from first row.

$(3x+a)\begin{vmatrix} 1 & 1 &1 \\ x &x+a &x\\ x & x & x+a \end{vmatrix} =0$

Now applying the column transformations; $C_{1} \rightarrow C_{1}-C_{2}$ and $C_{2} \rightarrow C_{2}-C_{3}$ .

we get;

$(3x+a)\begin{vmatrix} 0 & 0 &1 \\ -a &a &x\\ 0 & -a & x+a \end{vmatrix} =0$

$=(3x+a)(a^2)=0$ as $a^2 \neq 0$ ,

or $3x+a=0$ or $x= -\frac{a}{3}$

Question:6 Prove that $\begin{vmatrix} a^2 &bc &ac+a^2 \\ a^2+ab & b^2 & ac\\ ab &b^2+bc &c^2 \end{vmatrix}=4a^2b^2c^2$ .

Answer:

Given matrix $\begin{vmatrix} a^2 &bc &ac+a^2 \\ a^2+ab & b^2 & ac\\ ab &b^2+bc &c^2 \end{vmatrix}$

Taking common factors a,b and c from the column $C_{1}, C_{2}, and\ C_{3}$ respectively.

we have;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b & a\\ b &b+c &c \end{vmatrix}$

Applying $R_{2} \rightarrow R_{2}-R_{1}\ and\ R_{3} \rightarrow R_{3} - R_{1}$ , we have;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ b & b-c &-c\\ b-a &b &-a \end{vmatrix}$

Then applying $R_{2} \rightarrow R_{2}+R_{1}$ , we get;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ b-a &b &-a \end{vmatrix}$

Applying $R_{3} \rightarrow R_{3}+R_{2}$ , we have;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ 2b &2b &0 \end{vmatrix} = 2ab^2c\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ 1 &1 &0 \end{vmatrix}$

Now, applying column transformation; $C_{2} \rightarrow C_{2 }-C_{1}$ , we have

$\triangle = 2ab^2c\begin{vmatrix} a &c-a &a+c \\ a+b & -a &a\\ 1 &0 &0 \end{vmatrix}$

So we can now expand the remaining determinant along $R_{3}$ we have;

$\triangle = 2ab^2c\left [ a(c-a)+a(a+c) \right ]$

$= 2ab^2c\left [ ac-a^2+a^2+ac) \right ] = 2ab^2c\left [ 2ac \right ]$

$= 4a^2b^2c^2$

Hence proved.

Question:7 If $A^-^1=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$ and $B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$ , find $(AB)^-^1$ .

Answer:

We know from the identity that;

$(AB)^{-1} = B^{-1}A^{-1}$ .

Then we can find easily,

Given $A^-^1=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$ and $B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$

Then we have to basically find the $B^{-1}$ matrix.

So, Given matrix $B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$

$|B| = 1(3-0) -2(-1-0)-2(2-0) = 3+2-4 = 1 \neq 0$

Hence its inverse $B^{-1}$ exists;

Now, as we know that

$B^{-1} = \frac{1}{|B|} adjB$

So, calculating cofactors of B,

$B_{11} = (-1)^{1+1}(3-0) = 3$ $B_{12} = (-1)^{1+2}(-1-0) = 1$

$B_{13} = (-1)^{1+3}(2-0) = 2$ $B_{21} = (-1)^{2+1}(2-4) = 2$

$B_{22} = (-1)^{2+2}(1-0) = 1$ $B_{23} = (-1)^{2+3}(-2-0) = 2$

$B_{31} = (-1)^{3+1}(0+6) = 6$ $B_{32} = (-1)^{3+2}(0-2) = 2$

$B_{33} = (-1)^{3+3}(3+2) = 5$

$adjB = \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}$

$B^{-1} = \frac{1}{|B|} adjB = \frac{1}{1} \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}$

Now, We have both $A^{-1}$ as well as $B^{-1}$ ;

Putting in the relation we know; $(AB)^{-1} = B^{-1}A^{-1}$

$(AB)^{-1}=\frac{1}{1} \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$

$= \begin{bmatrix} 9-30+30 &-3+12-12 &3-10+12 \\ 3-15+10&-1+6-4 &1-5+4 \\ 6-30+25 &-2+12-10 &2-10+10 \end{bmatrix}$

$= \begin{bmatrix} 9 &-3 &5 \\ -2&1 &0 \\ 1 &0 &2\end{bmatrix}$

Question:8(i) Let $A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$ . Verify that,

$\dpi{100} [adj A]^-^1 = adj (A^-^1)$

Answer:

Given that $A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$ ;

So, let us assume that $A^{-1} = B$ matrix and $adjA = C$ then;

$|A| = 1(15-1) -2(10-1) +1(2-3) = 14-18-1 = -5 \neq 0$

Hence its inverse exists;

$A^{-1} = \frac{1}{|A|} adjA$ or $B = \frac{1}{|A|}C$ ;

so, we now calculate the value of $adjA$

Cofactors of A;

$A_{11}= (-1)^{1+1}(15-1) = 14$ $A_{12}= (-1)^{1+2}(10-1) = -9$

$A_{13}= (-1)^{1+3}(2-3) = -1$ $A_{21}= (-1)^{2+1}(10-1) = -9$

$A_{22}= (-1)^{2+2}(5-1) = 4$ $A_{23}= (-1)^{2+3}(1-2) = 1$

$A_{31}= (-1)^{3+1}(2-3) = -1$ $A_{32}= (-1)^{3+2}(1-2) = 1$

$A_{33}= (-1)^{3+3}(3-4) = -1$

$\Rightarrow adjA =C= \begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

$A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

Finding the inverse of C;

$|C| = 14(-4-1)+9(9+1)-1(-9+4) = -70+90+5 = 25 \neq 0$

Hence its inverse exists;

$C^{-1} = \frac{1}{|C|}adj C$

Now, finding the $adjC$ ;

$C_{11}= (-1)^{1+1}(-4-1) = -5$ $C_{12}= (-1)^{1+2}(9+1) = -10$

$C_{13}= (-1)^{1+3}(-9+4) = -5$ $C_{21}= (-1)^{2+1}(9+1) = -10$

$C_{22}= (-1)^{2+2}(-14-1) = -15$ $C_{23}= (-1)^{2+3}(14-9) = -5$

$C_{31}= (-1)^{3+1}(-9+4) = -5$ $C_{32}= (-1)^{3+2}(14-9) = -5$

$C_{33}= (-1)^{3+3}(56-81) = -25$

$adjC = \begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix}$

$C^{-1} = \frac{1}{|C|}adjC = \frac{1}{25}\begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

or $L.H.S. = C^{-1} = [adjA]^{-1} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

Now, finding the R.H.S.

$adj (A^{-1}) = adj B$

$A^{-1} =B= \begin{bmatrix} \frac{-14}{5} &&\frac{9}{5} &&\frac{1}{5} \\ \\ \frac{9}{5}&& \frac{-4}{5}&& \frac{-1}{5}\\ \\ \frac{1}{5}&& \frac{-1}{5} &&\frac{1}{5}\end{bmatrix}$

Cofactors of B;

$B_{11}= (-1)^{1+1}(\frac{-4}{25}-\frac{1}{25}) = \frac{-1}{5}$

$B_{12}= (-1)^{1+2}(\frac{9}{25}+\frac{1}{25}) =- \frac{2}{5}$

$B_{13}= (-1)^{1+3}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{21}= (-1)^{2+1}(\frac{9}{25}+\frac{1}{25}) = -\frac{2}{5}$

$B_{22}= (-1)^{2+2}(\frac{-14}{25}-\frac{1}{25}) = \frac{-3}{5}$

$B_{23}= (-1)^{2+3}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{31}= (-1)^{3+1}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{32}= (-1)^{3+2}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{33}= (-1)^{3+3}(\frac{56}{25}-\frac{81}{25}) = -1$

$R.H.S. = adjB = adj(A^{-1}) =\begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

Hence L.H.S. = R.H.S. proved.

Question:8(ii) Let $A=\begin{bmatrix} 1 &2 &1 \\ 2 & 3 &1 \\ 1 & 1 & 5 \end{bmatrix}$ , Verify that

$(A^-^1)^-^1=A$

Answer:

Given that $A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$ ;

So, let us assume that $A^{-1} = B$

$|A| = 1(15-1) -2(10-1) +1(2-3) = 14-18-1 = -5 \neq 0$

Hence its inverse exists;

$A^{-1} = \frac{1}{|A|} adjA$ or $B = \frac{1}{|A|}C$ ;

so, we now calculate the value of $adjA$

Cofactors of A;

$A_{11}= (-1)^{1+1}(15-1) = 14$ $A_{12}= (-1)^{1+2}(10-1) = -9$

$A_{13}= (-1)^{1+3}(2-3) = -1$ $A_{21}= (-1)^{2+1}(10-1) = -9$

$A_{22}= (-1)^{2+2}(5-1) = 4$ $A_{23}= (-1)^{2+3}(1-2) = 1$

$A_{31}= (-1)^{3+1}(2-3) = -1$ $A_{32}= (-1)^{3+2}(1-2) = 1$

$A_{33}= (-1)^{3+3}(3-4) = -1$

$\Rightarrow adjA =C= \begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

$A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix} = \begin{bmatrix} \frac{-14}{5} &&\frac{9}{5} &&\frac{1}{5} \\ \\ \frac{9}{5} && \frac{-4}{5} &&\frac{-1}{5} \\ \\ \frac{1}{5} &&\frac{-1}{5} &&\frac{1}{5} \end{bmatrix}$

Finding the inverse of B ;

$|B| = \frac{-14}{5}(\frac{-4}{25}-\frac{1}{25})-\frac{9}{5}(\frac{9}{25}+\frac{1}{25})+\frac{1}{5}(\frac{-9}{25}+\frac{4}{25})$

$= \frac{70}{125}-\frac{90}{125}-\frac{5}{125} = \frac{-25}{125} = \frac{-1}{5} \neq 0$

Hence its inverse exists;

$B^{-1} = \frac{1}{|B|}adj B$

Now, finding the $adjB$ ;

$B_{11}= (-1)^{1+1}(\frac{-4}{25}-\frac{1}{25}) = \frac{-1}{5}$ $B_{12}= (-1)^{1+2}(\frac{9}{25}+\frac{1}{25}) = \frac{-2}{5}$

$B_{13}= (-1)^{1+3}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$ $B_{21}= (-1)^{2+1}(\frac{9}{25}+\frac{1}{25}) = -\frac{2}{5}$

$B_{22}= (-1)^{2+2}(\frac{-14}{25}-\frac{1}{25}) = \frac{-3}{5}$ $B_{23}= (-1)^{2+3}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{31}= (-1)^{3+1}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{32}= (-1)^{3+2}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{33}= (-1)^{3+3}(\frac{56}{25}-\frac{81}{25}) = \frac{-25}{25} =-1$

$adjB = \begin{bmatrix} \frac{-1}{5} &&\frac{-2}{5} &&\frac{-1}{5} \\ \\\frac{-2}{5}&& \frac{-3}{5} && \frac{-1}{5}\\ \\ \frac{-1}{5}&& \frac{-1}{5}&& -1 \end{bmatrix}$

$B^{-1} = \frac{1}{|B|}adjB = \frac{-5}{1}\begin{bmatrix} \frac{-1}{5} &&\frac{-2}{5} &&\frac{-1}{5} \\ \\\frac{-2}{5}&& \frac{-3}{5} && \frac{-1}{5}\\ \\ \frac{-1}{5}&& \frac{-1}{5}&& -1 \end{bmatrix}= \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1 & 1 &5 \end{bmatrix}$

$L.H.S. = B^{-1} = (A^{-1})^{-1} = \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1 & 1 &5 \end{bmatrix}$

$R.H.S. = A = \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1& 1 &5 \end{bmatrix}$

Hence proved L.H.S. =R.H.S..

Question:9 Evaluate $\begin{vmatrix} x & y &x+y \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Answer:

We have determinant $\triangle = \begin{vmatrix} x & y &x+y \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Applying row transformations; $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ , we have then;

$\triangle = \begin{vmatrix} 2(x+y) & 2(x+y) &2(x+y) \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Taking out the common factor 2(x+y) from the row first.

$= 2(x+y)\begin{vmatrix} 1 & 1 &1 \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Now, applying the column transformation; $C_{1} \rightarrow C_{1} - C_{2}$ and $C_{2} \rightarrow C_{2} - C_{1}$ we have ;

$= 2(x+y)\begin{vmatrix} 0 & 0 &1 \\ -x & y &x \\ y & x-y & y \end{vmatrix}$

Expanding the remaining determinant;

$= 2(x+y)(-x(x-y)-y^2) = 2(x+y)[-x^2+xy-y^2]$

$= -2(x+y)[x^2-xy+y^2] = -2(x^3+y^3)$ .

Question:10 Evaluate $\begin{vmatrix} 1 & x &y \\ 1 &x+y &y \\ 1 &x &x+y \end{vmatrix}$

Answer:

We have determinant $\triangle = \begin{vmatrix} 1 & x &y \\ 1 &x+y &y \\ 1 &x &x+y \end{vmatrix}$

Applying row transformations; $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{2} \rightarrow R_{2}-R_{3}$ then we have then;

$\triangle = \begin{vmatrix} 0 & -y &0 \\ 0 &y &-x \\ 1 &x &x+y \end{vmatrix}$

Taking out the common factor -y from the row first.

$\triangle = -y\begin{vmatrix} 0 & 1 &0 \\ 0 &y &-x \\ 1 &x &x+y \end{vmatrix}$

Expanding the remaining determinant;

$-y[1(-x-o)] = xy$

Question:11 Using properties of determinants, prove that

$\begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta & \beta ^2 &\gamma +\alpha \\ \gamma &\gamma ^2 &\alpha +\beta \end{vmatrix}=(\beta -\gamma )(\gamma -\alpha )(\alpha -\beta )(\alpha +\beta +\gamma )$

Answer:

Given determinant $\triangle = \begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta & \beta ^2 &\gamma +\alpha \\ \gamma &\gamma ^2 &\alpha +\beta \end{vmatrix}$

Applying Row transformations; and $R_{3} \rightarrow R_{3}-R_{1}$ , then we have;

$\triangle = \begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta -\alpha & \beta ^2 - \alpha^2 &\alpha - \beta \\ \gamma-\alpha &\gamma ^2-\alpha^2 &\alpha -\gamma \end{vmatrix}$

$= (\beta-\alpha)(\gamma-\alpha)\begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ 1 & \beta + \alpha &-1 \\ 0 &\gamma-\beta &0\end{vmatrix}$

Expanding the remaining determinant;

$= (\beta-\alpha)(\gamma-\alpha)[-(\gamma-\beta)(-\alpha-\beta-\gamma)]$

$= (\beta-\alpha)(\gamma-\alpha)(\gamma-\beta)(\alpha+\beta+\gamma)$

$=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)$

hence the given result is proved.

Question:12 Using properties of determinants, prove that

$\begin{vmatrix} x & x^2&1+px^3 \\ y& y^2& 1+py^3\\ z&z^2 & 1+pz^3 \end{vmatrix}=(1+pxyz)(x-y)(y-z)(z-x),$ where p is any scalar.

Answer:

Given the determinant $\triangle = \begin{vmatrix} x & x^2&1+px^3 \\ y& y^2& 1+py^3\\ z&z^2 & 1+pz^3 \end{vmatrix}$

Applying the row transformations; $R_{2} \rightarrow R_{2} - R_{1}$ and $R_{3} \rightarrow R_{3} - R_{1}$ then we have;

$\triangle = \begin{vmatrix} x & x^2&1+px^3 \\ y-x& y^2-x^2& p(y^3-x^3)\\ z-x&z^2-x^2 & p(z^3-x^3) \end{vmatrix}$

Applying row transformation $R_{3} \rightarrow R_{3} - R_{2}$ we have then;

$\triangle =(y-x )(z-x)(z-y)\begin{vmatrix} x & x^2&1+px^3 \\ 1& y+x& p(y^2+x^2+xy)\\ 0&1 & p(x+y+z) \end{vmatrix}$

Now we can expand the remaining determinant to get the result;

$\triangle =(y-x )(z-x)(z-y)[(-1)(p)(xy^2+x^3+x^2y)+1+px^3+p(x+y+z)(xy)]$

$=(x-y)(y-z)(z-x)[-pxy^2-px^3-px^2y+1+px^3+px^2y+pxy^2+pxyz]$

$=(x-y)(y-z)(z-x)(1+pxyz)$

hence the given result is proved.

Question:13 Using properties of determinants, prove that

$\begin{vmatrix} 3a &-a+b &-a+c \\ -b+c &3b &-b+c \\ -c+a &-c+b &3c \end{vmatrix}=3(a+b+c)(ab+bc+ca)$

Answer:

Given determinant $\triangle = \begin{vmatrix} 3a &-a+b &-a+c \\ -b+c &3b &-b+c \\ -c+a &-c+b &3c \end{vmatrix}$

Applying the column transformation, $C_{1} \rightarrow C_{1} +C_{2}+C_{3}$ we have then;

$\triangle = \begin{vmatrix} a+b+c &-a+b &-a+c \\ a+b+c &3b &-b+c \\ a+b+c &-c+b &3c \end{vmatrix}$

Taking common factor (a+b+c) out from the column first;

$=(a+b+c) \begin{vmatrix} 1 &-a+b &-a+c \\ 1 &3b &-b+c \\1 &-c+b &3c \end{vmatrix}$

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$ , we have then;

$\triangle=(a+b+c) \begin{vmatrix} 1 &-a+b &-a+c \\ 0 &2b+a &a-b \\0 &a-c &2c+a \end{vmatrix}$

Now we can expand the remaining determinant along $C_{1}$ we have;

$\triangle=(a+b+c) [(2b+a)(2c+a)-(a-b)(a-c)]$

$=(a+b+c) [4bc+2ab+2ac+a^2-a^2+ac+ba-bc]$

$=(a+b+c)(3ab+3bc+3ac)$

$=3(a+b+c)(ab+bc+ac)$

Hence proved.

Question:14 Using properties of determinants, prove that

$\begin{vmatrix} 1 &1+p &1+p+q \\ 2&3+2p &4+3p+2q \\ 3&6+3p & 10+6p+3q \end{vmatrix}=1$

Answer:

Given determinant $\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 2&3+2p &4+3p+2q \\ 3&6+3p & 10+6p+3q \end{vmatrix}$

Applying the row transformation; $R_{2} \rightarrow R_{2}-2R_{1}$ and $R_{3} \rightarrow R_{3}-3R_{1}$ we have then;

$\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 0&1 &2+p \\ 0&3 & 7+3p \end{vmatrix}$

Now, applying another row transformation $R_{3}\rightarrow R_{3}-3R_{2}$ we have;

$\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 0&1 &2+p \\ 0&0 & 1 \end{vmatrix}$

We can expand the remaining determinant along $C_{1}$ , we have;

$\triangle = 1\begin{vmatrix} 1 & 2+p\\ 0 &1 \end{vmatrix} = 1(1-0) =1$

Hence the result is proved.

Question:15 Using properties of determinants, prove that

$\begin{vmatrix} \sin \alpha &\cos \alpha &\cos (\alpha +\delta ) \\ \sin \beta & \cos \beta & \cos (\beta +\delta )\\ \sin \gamma &\cos \gamma & \cos (\gamma +\delta ) \end{vmatrix}=0$

Answer:

Given determinant $\triangle = \begin{vmatrix} \sin \alpha &\cos \alpha &\cos (\alpha +\delta ) \\ \sin \beta & \cos \beta & \cos (\beta +\delta )\\ \sin \gamma &\cos \gamma & \cos (\gamma +\delta ) \end{vmatrix}$

Multiplying the first column by $\sin \delta$ and the second column by $\cos \delta$ , and expanding the third column, we get

$\triangle =\frac{1}{\sin \delta \cos \delta} \begin{vmatrix} \sin \alpha \sin \delta &\cos \alpha\cos \delta &\cos \alpha \cos \delta -\sin \alpha \sin \delta \\ \sin \beta\sin \delta & \cos \beta\cos \delta & \cos \beta \cos \delta - \sin \beta\sin \delta \\ \sin \gamma\sin \delta &\cos \gamma\cos \delta & \cos \gamma \cos\delta- \sin \gamma\sin \delta \end{vmatrix}$

Applying column transformation, $C_{1} \rightarrow C_{1}+C_{3}$ we have then;

$\triangle =\frac{1}{\sin \delta \cos \delta} \begin{vmatrix} \cos \alpha \cos \delta &\cos \alpha\cos \delta &\cos \alpha \cos \delta -\sin \alpha \sin \delta \\ \cos \beta\cos \delta & \cos \beta\cos \delta & \cos \beta \cos \delta - \sin \beta\sin \delta \\ \cos \gamma\cos \delta &\cos \gamma\cos \delta & \cos \gamma \cos\delta- \sin \gamma\sin \delta \end{vmatrix}$

Here we can see that two columns $C_{1}\ and\ C_{2}$ are identical.

The determinant value is equal to zero. $\therefore \triangle = 0$

Hence proved.

Question:16 Solve the system of equations

$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

Answer:

We have a system of equations;

$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

So, we will convert the given system of equations in a simple form to solve the problem by the matrix method;

Let us take, $\frac{1}{x} = a$ , $\frac{1}{y} = b\ and\ \frac{1}{z} = c$

Then we have the equations;

$2a +3b+10c = 4$

$4a-6b+5c =1$

$6a+9b-20c = 2$

We can write it in the matrix form as $AX =B$ , where

$A= \begin{bmatrix} 2 &3 &10 \\ 4& -6 & 5\\ 6 & 9 & -20 \end{bmatrix} , X = \begin{bmatrix} a\\b \\c \end{bmatrix}\ and\ B = \begin{bmatrix} 4\\1 \\ 2 \end{bmatrix}.$

Now, Finding the determinant value of A;

$|A| = 2(120-45)-3(-80-30)+10(36+36)$

$=150+330+720$

$=1200 \neq 0$

Hence we can say that A is non-singular $\therefore$ its invers exists;

Finding cofactors of A;

$A_{11} = 75$ , $A_{12} = 110$ , $A_{13} = 72$

$A_{21} = 150$ , $A_{22} = -100$ , $A_{23} = 0$

$A_{31} =75$ , $A_{31} =30$ , $A_{33} =-24$

$\therefore$ as we know $A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{1200} \begin{bmatrix} 75 &150 &75 \\ 110& -100& 30\\ 72&0 &-24 \end{bmatrix}$

Now we will find the solutions by relation $X = A^{-1}B$ .

$\Rightarrow \begin{bmatrix} a\\b \\ c \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 75 &150 &75 \\ 110& -100& 30\\ 72&0 &-24 \end{bmatrix}\begin{bmatrix} 4\\1 \\ 2 \end{bmatrix}$

$= \frac{1}{1200}\begin{bmatrix} 300+150+150\\440-100+60 \\ 288+0-48 \end{bmatrix}$

$= \frac{1}{1200}\begin{bmatrix} 600\\400\\ 240 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}\\ \\ \frac{1}{3} \\ \\ \frac{1}{5} \end{bmatrix}$

Therefore we have the solutions $a = \frac{1}{2},\ b= \frac{1}{3},\ and\ c = \frac{1}{5}.$

Or in terms of x, y, and z;

$x =2,\ y =3,\ and\ z = 5$

Question:17 Choose the correct answer.

If $a,b,c,$ are in A.P, then the determinant
$\dpi{100} \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 & x+4 & x+2b\\ x+4 & x+5 &x+2c \end{vmatrix}$ is

(A) $0$ (B) $1$ (C) $x$ (D) $2x$

Answer:

Given determinant $\triangle = \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 & x+4 & x+2b\\ x+4 & x+5 &x+2c \end{vmatrix}$ and given that a, b, c are in A.P.

That means , 2b =a+c

$\triangle = \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 & x+4 & x+(a+c)\\ x+4 & x+5 &x+2c \end{vmatrix}$

Applying the row transformations, $R_{1} \rightarrow R_{1} -R_{2}$ and then $R_{3} \rightarrow R_{3} -R_{2}$ we have;

$\triangle = \begin{vmatrix} -1 &-1 &a-c \\ x+3 & x+4 & x+(a+c)\\ 1 & 1 &c-a \end{vmatrix}$

Now, applying another row transformation, $R_{1} \rightarrow R_{1} + R_{3}$ , we have

$\triangle = \begin{vmatrix} 0 &0 &0 \\ x+3 & x+4 & x+(a+c)\\ 1 & 1 &c-a \end{vmatrix}$

Clearly we have the determinant value equal to zero;

Hence the option (A) is correct.

Question:18 Choose the correct answer.

If x, y, z are nonzero real numbers, then the inverse of matrix $A=\begin{bmatrix} x &0 &0 \\ 0 &y &0 \\ 0 & 0 & z \end{bmatrix}$ is

$(A)\begin{bmatrix} x^-^1 &0 &0 \\ 0 &y^-^1 &0 \\ 0 & 0 & z^-^1 \end{bmatrix}$ $(B)xyz\begin{bmatrix} x^-^1 &0 &0 \\ 0 &y^-^1 &0 \\ 0 & 0 & z^-^1 \end{bmatrix}$

$(C)\frac{1}{xyz}\begin{bmatrix} x &0 &0 \\ 0 &y &0 \\ 0 & 0 & z \end{bmatrix}$ $(D)\frac{1}{xyz}\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

Answer:

Given Matrix $A=\begin{bmatrix} x &0 &0 \\ 0 &y &0 \\ 0 & 0 & z \end{bmatrix}$ ,

$|A| = x(yz-0) =xyz$

As we know,

$A^{-1} = \frac{1}{|A|}adjA$

So, we will find the $adjA$ ,

Determining its cofactor first,

$A_{11} = yz$ $A_{12} = 0$ $A_{13} = 0$

$A_{21} = 0$ $A_{22} = xz$ $A_{23} = 0$

$A_{31} = 0$ $A_{32} = 0$ $A_{33} = xy$

Hence $A^{-1} = \frac{1}{|A|}adjA = \frac{1}{xyz}\begin{bmatrix} yz &0 &0 \\ 0& xz & 0\\ 0& 0& xy \end{bmatrix}$

$A^{-1} = \begin{bmatrix} \frac{1}{x} &&0 &&0 \\ 0&& \frac{1}{y} && 0\\ 0&& 0&& \frac{1}{z} \end{bmatrix}$

Therefore the correct answer is (A)

Question:19 Choose the correct answer.

Let $A=\begin{vmatrix} 1 &\sin \theta &1 \\ -\sin \theta & 1 & \sin \theta \\ -1&-\sin \theta &1 \end{vmatrix},$ where $0\leq \theta \leq 2\pi$ . Then

(A) $Det(A)=0$ nbsp; (B) $Det(A)\in (2,\infty)$

Answer:

Given determinant $A=\begin{vmatrix} 1 &\sin \theta &1 \\ -\sin \theta & 1 & \sin \theta \\ -1&-\sin \theta &1 \end{vmatrix}$

$|A| = 1(1+\sin^2 \Theta) -\sin \Theta(-\sin \Theta+\sin \Theta)+1(\sin^2 \Theta +1)$

$= 1+ \sin ^2 \Theta + \sin ^2 \Theta +1$

$= 2+2\sin ^2 \Theta = 2(1+\sin^2 \Theta)$

Now, given the range of $\Theta$ from $0\leq \Theta \leq 2\pi$

$\Rightarrow 0 \leq \sin \Theta \leq 1$

$\Rightarrow 0 \leq \sin^2 \Theta \leq 1$

$\Rightarrow 1 \leq 1+\sin^2 \Theta \leq 2$

$\Rightarrow 2 \leq 2(1+\sin^2 \Theta) \leq 4$

Therefore the $|A|\ \epsilon\ [2,4]$ .

Hence the correct answer is D.

Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Miscellaneous Exercise:-

Class 12 Maths chapter 4 miscellaneous solutions are designed in a very detailed manner which could be understood by an average student also.
As miscellaneous exercise questions are difficult as compared to the previous exercise, you may not be able to these questions.
You can take NCERT solutions for Class 12 Maths chapter 4 miscellaneous exercise for reference while solving miscellaneous questions.
Miscellaneous exercise chapter 4 Class 12 will check your understanding of this chapter.

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Key Features Of NCERT Solutions For Class 12 Chapter 4 Miscellaneous Exercise

Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 4, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 chapter 4 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 4 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this class 12 maths ch 4 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for class 12 chapter 4 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 4 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. What is the value of determinant, If any two rows of a determinant are identical ?

The value of the determinant is zero if any two rows of a determinant are identical.

2. What is the value of determinant if any two rows of a determinant are proportional ?

The value of the determinant is zero If any two rows of a determinant are proportional.

3. If |A| = 3 and the first row matrix A is multiplied by 3 then what is the value of |A|?

If first row of matrix A is multiplied by 3 then the new value of |A| = 3x3 = 9.

4. Can you find the inverse of a singular matrix ?

No, singular matrices are non-invertible.

5. If transpose of matrix A and matrix A are equal then such matrix is called ?

If the transpose of matrix A and matrix A are equal then such matrix is called symmetric matrix.

6. What is the definition of diagonal matrix ?

The matrix has all the non-diagonal elements zero is called a diagonal matrix.

7. Do all the identity matrices are diagonal matrices ?

Yes, all the identity matrices are diagonal matrices.

8. Do CBSE provides solutions for miscellaneous exercise ?

No, CBSE doesn't provide NCERT solutions for miscellaneous exercises.

Also Read

NCERT Class 12 Maths Notes - Download Chapter wise Free PDFs

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

NCERT Class 12 Physics Chapter 8 Notes Electromagnetic Waves - Download PDF

NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

NCERT Class 12 Physics Chapter 6 Notes Electromagnetic Induction - Download PDF

NCERT Class 12 Physics Chapter 5 Notes Magnetism and Matter - Download PDF

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

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Questions related to CBSE Class 12th

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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Miscellaneous Exercise Chapter 4 Class 12 - Determinants

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Determinants Miscellaneous Exercise

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