RD Sharma Solutions Class 12 Mathematics Chapter 5 MCQ

# RD Sharma Solutions Class 12 Mathematics Chapter 5 MCQ

Edited By Kuldeep Maurya | Updated on Jan 25, 2022 05:32 PM IST

School students who will appear for their board exams undergo a lot of stress due to the vast mathematics syllabus and the short time within which they have to complete it. The vast syllabus is why students should use the RD Sharma Class 12th Exercise MCQ for their exam preparations to score well in their board exams.

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

The RD Sharma Class 12 Chapter 5 Exercise MCQ is the top choice for many students who have experienced the benefits of using the book. Chapter 5 of the Class 12 book has concepts on determinants which was again explored in Class 11. RD Sharma Class 12 solutions Chapter 5 ex MCQ book spans the entire Chapter and has 37 questions covering the concepts of area of a triangle, adjoint and inverse of matrices, and applications of Determinants and matrices.

## Determinants Excercise:MCQ

Determinants exercise multiple choise question 1

Correct option : (d)
i.e. A+B=0
Hint:
$I\! f\; \; \; \; det(A+B)=0\: \Rightarrow det[a_{ij}+b_{ij}]=0$
Given:
Here given that,
A and B are square matrices of order 2.
We have to find that det(A+B)=0 is possible in which condition.
Solution:
Determinant A denoted as [aij] and determinant B as [bij]
$\Rightarrow (A+B)=[a_{ij}]+[b_{ij}]$
$\Rightarrow (A+B)=[a_{ij}+b_{ij}]$
$\Rightarrow det(A+B)=det[a_{ij}+b_{ij}]$
Here det(A+B)=0
$\Rightarrow det[a_{ij}+b_{ij}]=0$
$\Rightarrow a_{ij}+b_{ij}=0$
$\Rightarrow A+B=0$
Hence option (d) is correct.

Determinants exercise multiple choise question 2

Correct option (d)
Hint:
Using the properties of determinant.
Given:
$\begin{vmatrix} a+b &c+d \\ e+f &g+h \end{vmatrix}$
Solution:
We know that,
If each element of a row or column of determinant is expressed as a sum of two or more terms, then the determinant can be expressed as the sum of two or more determinant.
$So,\; \; \; \; \begin{vmatrix} a+b &c+d \\ e+f &g+h \end{vmatrix}=\begin{vmatrix} a+b &c \\ e+f &g \end{vmatrix}+\begin{vmatrix} a+b &d \\ e+f &h \end{vmatrix}$
$=\left | a\, c\, e\, g \right |+\left | b\, c\, f\, g \right |+\left | a\, d\, e\, h \right |+\left | b\, d\, f\, h \right |$
Hence,
$So,\; \; \; \; \begin{vmatrix} a+b &c+d \\ e+f &g+h \end{vmatrix}\neq \begin{vmatrix} a &c \\ e &g \end{vmatrix}+\begin{vmatrix} b &d \\ f &h \end{vmatrix}$
Hence option (d) is correct.

Determinants exercise multiple choise question 3

Correct option (d)
Hint:
If A is a square matrix of order n, then
$det(A)=a_{11}c_{11}+a_{21}c_{21}+a_{31}c_{31}$
Given:
$A=\begin{vmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31} &a_{32} &a_{33} \end{vmatrix}\; \; and\; \; c_{ij}\; is\; co\! f\! actor\; o\! f\; a_{ij}\; in\; A$
We have to find the value of |A|.
Solution:
We know that,
If A is a square matrix of order n, then
$det(A)=a_{11}c_{11}+a_{21}c_{21}+a_{31}c_{31}$.

Determinants exercise multiple choise question 4

Correct option (b)

Solution:
We know that,
Minor of an element can never be equal to cofactor of the same element.

$\therefore c_{ij}=(-1)^{i+j}M_{ij}$

Determinants exercise multiple choise question 5

Correct option (d) none of these
Hint:
Simply solve the given determinant.
Given:
$\begin{vmatrix} x &2 &x \\ x^{2} &x &6 \\ x &x &6 \end{vmatrix}=ax^{4}+bx^{3}+cx^{2}+dx+e$
We have to find value of 5a+4b+3c+2d+e.
Solution:
Here, taking L.H.S
$L.H.S=\begin{vmatrix} x &2 &x \\ x^{2} &x &6 \\ x &x &6 \end{vmatrix}$
On solving determinant we have,
$=x(6x-6x)-2(6x^{2}-6x)+x(x^{3}-x^{2})$
$=0-12x^{2}+12x+x^{4}-x^{3}$
$=x^{4}-x^{3}-12x^{2}+12x$
Comparing with R.H.S
$ax^{4}+bx^{3}+cx^{2}+dx+e$
We get a=1, b=-1,c=-12, d=12, e=0
$\Rightarrow$ 5a+4b+3c+2d+e=5(1)+4(-1)+3(-12)+2(12)+0
=-11
Hence, option (d) is correct.

Determinants exercise multiple choise question 6

Correct option (a)
Hint:
$Suppose\: \: nx=u,\: (n+1)x=v,\: (n+2)x=w$
then solve determinant.
Given:
$\begin{vmatrix} a^{2} &a &1 \\ cos\: nx &cos(n+1)x &cos(n+2)x \\ sin\: nx &sin(n+1)x &sin(n+2)x \end{vmatrix}$
Solution:
$\begin{vmatrix} a^{2} &a &1 \\ cos\: nx &cos(n+1)x &cos(n+2)x \\ sin\: nx &sin(n+1)x &sin(n+2)x \end{vmatrix}$
$Putting\: \: nx=u,\: (n+1)x=v,\: (n+2)x=w$,
Then we have,
$\begin{vmatrix} a^{2} &a &1 \\ cos\: u &cos\, v &cos\, w \\ sin\: u &sin\, v&sin\, w \end{vmatrix}$
$=a^{2}(cos\, v\, sin\, w-sin\, v\, cos\, w)-a(cos\, u\, sin\, w-sin\, u\, cos\, w)+1(cos\, u\, sin\, v-cos\, v\, sin\, u)$
$=a^{2}sin(w-v)-a\, sin(w-u)+sin(v-u)$
$=a^{2}sin\, x-a\, sin2x+sin\, x$
Hence, it is independent of n.

Determinants exercise multiple choise question 7

Correct option (a)
Hint:
First solve these determinants, then find conclusion.
Given:
$\Delta _{1}=\begin{vmatrix} 1 &1 &1 \\ a &b &c \\ a^{2} &b^{2} &c^{2} \end{vmatrix},\: \: \Delta _{2}=\begin{vmatrix} 1 &bc &a \\ 1 &ca &b \\ 1 &ab &c \end{vmatrix}$
Solution:
Here,
$\Delta _{1}=\begin{vmatrix} 1 &1 &1 \\ a &b &c \\ a^{2} &b^{2} &c^{2} \end{vmatrix}$
$=bc^{2}-b^{2}c-(ac^{2}-a^{2}c)+ab^{2}-a^{2}b$
$=bc^{2}-b^{2}c-ac^{2}+a^{2}c+ab^{2}-a^{2}b$
$\Delta _{2}=\begin{vmatrix} 1 &bc &a \\ 1 &ca &b \\ 1 &ab &c \end{vmatrix}$
$=c^{2}a-ab^{2}-bc(c-b)+a(ab-ac)$
$=c^{2}a-ab^{2}-bc^{2}+bc^{2}+a^{2}b-a^{2}c$
$=-(bc^{2}-b^{2}c-ac^{2}+a^{2}c+ab^{2}-a^{2}b)$
$\Rightarrow \Delta _{1}=-\Delta _{2}$
$\Rightarrow \Delta _{1}+\Delta _{2}=0$
Hence, option (a) is correct.

Determinants exercise multiple choise question 8

Correct option (a)
Hint:
Solve the given determinant Dk
Given:
$D_{k}=\begin{vmatrix} 1 &n &n \\ 2k &n^{2}+n+2 &n^{2}+n \\ 2k-1 &n^{2} &n^{2}+n+2 \end{vmatrix}\; \; and\; \; \sum_{k=1}^{n}D_{k}=48$
We have to find the value of n.
Solution:
$D_{k}=\begin{vmatrix} 1 &n &n \\ 2k &n^{2}+n+2 &n^{2}+n \\ 2k-1 &n^{2} &n^{2}+n+2 \end{vmatrix}$
Applying row transformation,
R2→R2-R3, We get
$D_{k}=\begin{vmatrix} 1 &n &n \\ 1 &n+2 &-2 \\ 2k-1 &n^{2} &n^{2}+n+2 \end{vmatrix}$
Applying R2→R2-R3, We get
$D_{k}=\begin{vmatrix} 0 &-2 &n+2 \\ 1 &n+2 &-2 \\ 2k-1 &n^{2} &n^{2}+n+2 \end{vmatrix}$
Expanding along Row1, we get
$=2(n^{2}+n+2+4k-2)+(n+2)(n^{2}-(n-2)(2k-1))$
$=2(n^{2}+n+4k)+(n+2)(n^{2}-2nk+n-4k+2)$
$=2n^{2}+2n+8k+n^{3}-2n^{2}k+n^{2}-4nk+2n+2n^{2}-4nk+2n-8k+4$
$D_{k}=n^{3}+5n^{2}-2n^{2}k+6n-8nk+4$
Given:
$\sum_{k=1}^{n}D_{k}=48$
$\Rightarrow \sum_{k=1}^{n}1(n^{3}+5n^{2}+6n+4)-2n^{2}\sum_{k=1}^{n}k-8n\sum_{k=1}^{n}k=48$
$\Rightarrow n(n^{3}+5n^{2}+6n+4)-2n^{2}\frac{n(n+1)}{2}-8n\frac{n(n+1)}{2}=48$
$\Rightarrow n^{4}+5n^{3}+6n^{2}+4n-n^{4}-n^{3}-4n^{3}-4n^{2}=48$
$\Rightarrow 2n^{2}+4n=48$
$\Rightarrow n^{2}+2n-24=0$
$\Rightarrow (n+6)(n-4)=0$
$\Rightarrow n=-6,4$
Hence according to option n=4

Determinants exercise multiple choise question 9

Correct option (b)
Hint:
Simply we will put x=0, then solve determinant.
Given:
Let
$\begin{vmatrix} x^{2}+3x &x-1 &x+3 \\ x+1 &-2x &x-4 \\ x-3 &x+4 &3x \end{vmatrix}=ax^{4}+bx^{3}+cx^{2}+dx+e$
be an identity in x
Solution:
$\begin{vmatrix} x^{2}+3x &x-1 &x+3 \\ x+1 &-2x &x-4 \\ x-3 &x+4 &3x \end{vmatrix}=ax^{4}+bx^{3}+cx^{2}+dx+e$
Putting x=0
$\Rightarrow \begin{vmatrix} 0 &-1 &3 \\ 1 &0 &-4 \\ -3 &4 &0 \end{vmatrix}=e$
$\Rightarrow e=1(0-12)+3(4)$
$\Rightarrow e=0$

Determinants exercise multiple choise question 10

Correct option (a)
Hint:
Solve given determinant
Given:
$Here\; \begin{vmatrix} -2a &a+b &a+c \\ b+a &-2b &b+c \\ c+a &c+b &-2c \end{vmatrix} is \; given$
The three factor are given, we have to find another one factor.
Solution:
$\begin{vmatrix} -2a &a+b &a+c \\ b+a &-2b &b+c \\ c+a &c+b &-2c \end{vmatrix}$
Appling C1→C1+C2 , we get
$=\begin{vmatrix} b-a &a+b &a+c \\ a-b &-2b &b+c \\ a+b+2c &b+c &-2c \end{vmatrix}$
Appling C2→C2+C3 , we get
$=\left | b-a2a+b+ca+ca-bc-bb+ca+b+2cb-c-2c \right |$
Appling R3→R3+R2 , we get
$=\begin{vmatrix} b-a &2a+b+a &a+c \\ a-b &c-b &b+c \\ 2(c+a) &0 &b-c \end{vmatrix}$
Appling R2→R2+R1 , we get
$=\begin{vmatrix} b-a &2a+b+a &a+c \\ 0 &2(a+c) &a+b+2c \\ 2(c+a) &0 &b-c \end{vmatrix}$
Expanding along C1 , we get
$=(b-a)[2(a+c)(b-c)]+2(a+c)[(2a+b+c)(a+b+2c)-2(a+c)^{2}]$
$=2(a+c)[(b-a)(b-c)+(2a+b+c)(a+b+2c)-2(a+c)^{2}]$
$=2(a+c)[b^{2}-bc-ab+ac+2a^{2}+2ab+4ac+ab+b^{2}+2bc+ac+bc+2c^{2}-2a^{2}-2c^{2}-4ac]$
$=2(a+c)[2b^{2}+2ab+2bc+2ac]$
$=4(a+c)[b^{2}+ab+bc+ca]$
$=4(a+c)(a+b)(b+c)$
Hence another factor of the given determinant is 4.

Determinants exercise multiple choise question 11

Correct option (d)
Hint:
Put x=0 in the determinant and then check the skew-symmetric.
Given:
$\begin{vmatrix} 0 &x^{2}-a &x^{3}-b \\ x^{2}+a &0 &x^{2}+c \\ x^{4}+b &x-c &0 \end{vmatrix}=0$
Solution:
$\begin{vmatrix} 0 &x^{2}-a &x^{3}-b \\ x^{2}+a &0 &x^{2}+c \\ x^{4}+b &x-c &0 \end{vmatrix}$
If we put x=0 in the above determinant
$\Rightarrow \begin{vmatrix} 0 &-a &-b \\ a &0 &c \\ b &-c &0 \end{vmatrix}=A$(Let)
$\Rightarrow A^{T}=\begin{vmatrix} 0 &a &b \\ -a &0 &-c \\ -b &c &0 \end{vmatrix}$
$\Rightarrow =\begin{vmatrix} 0 &-a &-b \\ a &0 &c \\ b &-c &0 \end{vmatrix}=-A$
$\Rightarrow A^{T}=-A$
Hence matrix is skew symmetric
Hence value of x is 0.

Determinants exercise multiple choise question 12

Correct option (b)
Hint:
First solve the given determinant, then conclude answer.
Given:
$\left | a\, b\, 2a\alpha +3b\: b\: c\: 2b\alpha +3c\: 2a\alpha +3b\: 2b\alpha +3c\: 0 \right |=0$
We have to find the conclusion of determinant
Solution:
$\begin{vmatrix} a &b &2a\alpha +3b \\ b &c &2b\alpha +3c \\ 2a\alpha +3b &2b\alpha +3c &0 \end{vmatrix}=0$
Expanding along R1, we have
$a\left[-(2 b \alpha+3 c)^{2}\right]-b[-(2 b \alpha+3 c)(2 a \alpha+3 b)]+(2 a \alpha+3 b)[b(2 b \alpha+3 c)-c(2 a \alpha+3 b)]=0$
$\Rightarrow -a(2b\alpha +3c)^{2}+b(2b\alpha +3c)(2a\alpha +3b)+(2a\alpha +3b)(2b^{2}\alpha +3bc-2ac\alpha -3bc)=0$
$\Rightarrow -a(2b\alpha +3c)^{2}+b(2b\alpha +3c)(2a\alpha +3b)+(2a\alpha +3b)(2\alpha (b^{2}-ac))=0$
$\Rightarrow (2b\alpha +3c)[-2ab\alpha -3ac+2ab\alpha +3b^{2}]+(2a\alpha +3b)(2\alpha )(b^{2}-ac)=0$
$\Rightarrow (2b\alpha +3c)[3b^{2}-3ac]+(2a\alpha +3b)(2\alpha )(b^{2}-ac)=0$
$\Rightarrow (b^{2}-ac)[6b\alpha +9c+4a\alpha ^{2}+6b\alpha ]=0$
$\Rightarrow (b^{2}-ac)[4a\alpha ^{2}+12b\alpha +9c ]=0$
$\Rightarrow (b^{2}-ac)=0\; \; or\; \;[4a\alpha ^{2}+12b\alpha +9c ]=0$
$\Rightarrow a,b,c\: are in\; G.P\; or\; [4a\alpha ^{2}+12b\alpha +9c ]=0\; \; whose\; root\; is\; \alpha .$
$Hence\; \alpha \; is\;a\; root\; o\! f\; [4a\alpha ^{2}+12b\alpha +9c ]=0\; \; or\; a,b,c\; are\; in\; G.P .$

Determinants exercise multiple choise question 13

Correct option (a)
Hint:
Simply solve this determinant.
Given:
$\Delta=\left|\begin{array}{ccc} 1 & \omega^{n} & \omega^{2 n} \\ \omega^{2 n} & 1 & \omega^{n} \\ \omega^{n} & \omega^{2 n} & 1 \end{array}\right|$
We have to find $\Delta$
Solution:
$\Delta=\left|\begin{array}{ccc} 1 & \omega^{n} & \omega^{2 n} \\ \omega^{2 n} & 1 & \omega^{n} \\ \omega^{n} & \omega^{2 n} & 1 \end{array}\right|$
Applying C1→C1+C2+C3
$\Rightarrow \Delta=\left|\begin{array}{ccc} 1+\omega^{n}+\omega^{2n} & \omega^{n} & \omega^{2 n} \\ 1+\omega^{n}+\omega^{2n} & 1 & \omega^{n} \\ 1+\omega^{n}+\omega^{2n} & \omega^{2 n} & 1 \end{array}\right|$
$\Rightarrow \Delta=\left|\begin{array}{ccc} 0 & \omega^{n} & \omega^{2 n} \\ 0 & 1 & \omega^{n} \\ 0 & \omega^{2 n} & 1 \end{array}\right|$ $[\because 1+\omega ^{n}+\omega ^{2n}=0]$
$\Rightarrow \Delta=0$

Determinants exercise multiple choise question 14

Correct option (c)
Hint:
Simply solve this determinant
Given:
$A_{r}=\left|\begin{array}{ccc} 1 & r & 2^{r} \\ 2 & n & n^{2} \\ n & \frac{n(n+1)}{2} & 2^{n+1} \end{array}\right|$
We have to find
$\sum_{r=1}^{n}A^{r}$
Solution:
If
$A_{r}=\left|1\, r\, 2^{r}\, 2\, n\, n^{2}\, n\, \frac{n(n+1)}{2}\, 2^{n+1}\right|$
$\Rightarrow \sum_{r=1}^{n}A_{r}=\left| n\frac{n(n+1)}{2} 2^{n+1}-2\; 2\; n\; n^{2}\; n\frac{n(n+1)}{2} 2^{n+1} \right|$
$Since,\; \; \sum_{r=1}^{n}r=1+2+3+.....+n\; \; =\frac{n(n+1)}{2}$
$\sum_{r=1}^{n}2^{r}=2+2^{2}+2^{3}+.....+2^{n}\; \; =\frac{2(2^{n}-1)}{2-1}=2^{n+1}-2$
Now applying, R1→R1-R3
$\Rightarrow \sum_{r=1}^{n}A_{r}=\left | 0-2\; 2\; n\; n^{2}\; n\frac{n(n+1)}{2}2^{n+1} \right |$
$\Rightarrow \sum_{r=1}^{n}A_{r}=-2[n(n+1)-n^{2}]$
$\Rightarrow \sum_{r=1}^{n}A_{r}=-2[n^{2}+n-n^{2}]$
$\Rightarrow \sum_{r=1}^{n}A_{r}=-2n$

Determinants exercise 5 multiple choise question 15

Correct option (c)
Hint:
Simply solve this determinant, then conclude answer
Given:
If $a$ > 0 and discriminant of $ax^{2}+2bx+c$ is negative.

Here we have to find $\Delta$.
Solution:
If $a$ > 0 and discriminant of $ax^{2}+2bx+c$ is negative.
$\Rightarrow (2b)^{2}-4ac< 0$
$\Rightarrow 4b^{2}-4ac< 0$
$\Rightarrow b^{2}-ac< 0\: but\: \: a> 0$ .......(i)
$\Delta =\begin{vmatrix} a &b &ax+b \\ b &c &bx+c \\ ax+b &bx+c &0 \end{vmatrix}$
R1→R1+R2
$\Rightarrow \Delta =\begin{vmatrix} ax+b &bx+c &ax^{2}+2bx+c \\ b &c &bx+c \\ ax+b &bx+c &0 \end{vmatrix}$
R1→R1-R3
$\Rightarrow \Delta =\begin{vmatrix} 0 &0 &ax^{2}+2bx+c \\ b &c &bx+c \\ ax+b &bx+c &0 \end{vmatrix}$
Expanding along R1
$\Rightarrow \Delta =(ax^{2}+2bx+c)[b(bx+c)-c(ax+b)]$
$\Rightarrow \Delta =(ax^{2}+2bx+c)[b^{2}x+bc-acx-bc]$
$\Rightarrow \Delta =(ax^{2}+2bx+c)[b^{2}x-acx]$
$\Rightarrow \Delta =(ax^{2}+2bx+c)x(b^{2}-ac)$
$As\: \: b^{2}-ac< 0,\: \Delta\: \: must\: \: be\: \: negative$.

Determinants exercise multiple choise question 16

Correct option (b)
Hint:
Using properties of determinant
Given:
$\begin{vmatrix} 5^{2} &5^{3} &5^{4} \\ 5^{3} &5^{4} &5^{5} \\ 5^{4} &5^{5} &5^{6} \end{vmatrix}$
Solution:
$\begin{vmatrix} 5^{2} &5^{3} &5^{4} \\ 5^{3} &5^{4} &5^{5} \\ 5^{4} &5^{5} &5^{6} \end{vmatrix}$
Taking 52 and 53 common from R1 and R2 respectively
$\Rightarrow 5^{2}\times 5^{3}\begin{vmatrix} 1 &5 &5^{2} \\ 1 &5 &5^{2} \\ 5^{4} &5^{5} &5^{6} \end{vmatrix}$
Here R1 and R2 are identical.
Hence from the properties of determinant, if any two row or column of the matrix is identical ,then the value of determinant is equal to zero.
Hence,
$\begin{vmatrix} 5^{2} &5^{3} &5^{4} \\ 5^{3} &5^{4} &5^{5} \\ 5^{4} &5^{5} &5^{6} \end{vmatrix}=0$

Determinants exercise multiple choise question 17

Correct option (b)

Hint:
Use logarithm properties

Solution:

$\left|\begin{array}{cc} \log _{3}^{512} & \log _{4}^{3} \\ \log _{3}^{8} & \log _{4}^{9} \end{array}\right| \times\left|\begin{array}{ll} \log _{2}^{3} & \log _{8}^{3} \\ \log _{3}^{4} & \log _{3}^{4} \end{array}\right|$

$=\left|\begin{array}{cc} \log _{3}{ }^{2^{9}} & \log _{4}^{3} \\ \log _{3}^{2^{3}} & \log _{4}^{3^{2}} \end{array}\right| \times\left|\begin{array}{cc} \log _{2}^{3} & \log _{8}^{3} \\ \log _{3}^{2^{2}} & \log _{3}^{2^{2}} \end{array}\right|$

$=\left|\begin{array}{cc} 9\log _{3}{ }^{2} & \log _{4}^{3} \\ 3\log _{3}^{2} & 2\log _{4}^{3} \end{array}\right| \times\left|\begin{array}{cc} \log _{2}^{3} & \log _{8}^{3} \\ 2\log _{3}^{2} & 2\log _{3}^{2} \end{array}\right|$

$=\left[\left(9 \times \frac{\log 2}{\log 3} \times 2 \times \frac{\log 3}{\log 4}\right)-3 \times \frac{\log 2}{\log 3} \times \frac{\log 3}{\log 4}\right] \times\left[\frac{\log 3}{\log 2} \times 2 \times \frac{\log 2}{\log 3}-\frac{\log 3}{\log 8} \times 2 \times \frac{\log 2}{\log 3}\right]$

$=\left(18 \times \frac{\log 2}{2 \log 2}-3 \times \frac{\log 2}{2 \log 2}\right)\left(2-2 \times \frac{1}{3 \log 2} \times \log 2\right)$

$=(9-\frac{3}{2})(2-\frac{2}{3})$

$=\frac{15}{2}\times \frac{4}{3}$

$=10$

Determinants exercise multiple choise question 18

Correct option (a)
Hint:
If $a,b,c$ are in A.P $\Rightarrow 2b=a+c$
Given:
If $a,b,c$ are in A.P
$Let \: \: \Delta= \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 &x+4 &x+2b \\ x+4 &x+5 &x+2c \end{vmatrix}$
We have to find the value of given determinant $\Delta$
Solution:
$\Delta= \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 &x+4 &x+2b \\ x+4 &x+5 &x+2c \end{vmatrix}$
If $a,b,c$ are in A.P $\Rightarrow 2b=a+c$
$\Rightarrow \Delta= \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 &x+4 &x+a+c \\ x+4 &x+5 &x+2c \end{vmatrix}$
Applying R1→R1-R2 ; R3→R3-R2
$\Rightarrow \Delta= \begin{vmatrix} -1 &-1 &a-c \\ x+3 &x+4 &x+a+c \\ 1 &1 &c-a \end{vmatrix}$
Applying R1→R1+R3
$\Rightarrow \Delta= \begin{vmatrix} 0 &0 &0 \\ x+3 &x+4 &x+a+c \\ 1 &1 &c-a \end{vmatrix}$
Here R1 is zero
$\Rightarrow \Delta= 0$
$Hence \: \: \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 &x+4 &x+2b \\ x+4 &x+5 &x+2c \end{vmatrix}=0$

Determinants exercise multiple choise question 19

Correct option
Hint:
Here A+B+C=$\pi$, then we can write sin(A+B+C)=sin$\pi$
Given:
Here given that A+B+C=$\pi$
Here we have to find the value of
$\left|\begin{array}{ccc} \sin (A+B+C) & \sin (A+C) & \cos C \\ -\sin B & 0 & \tan A \\ \cos (A+B) & \tan (B+C) & 0 \end{array}\right|$
Solution:
$Let\: \: \Delta =\left|\begin{array}{ccc} \sin (A+B+C) & \sin (A+C) & \cos C \\ -\sin B & 0 & \tan A \\ \cos (A+B) & \tan (B+C) & 0 \end{array}\right|$
Here A+B+C=$\pi$
$\Delta =\left | sin\, sin\, \pi \: sin\: sin(\pi -B)\: cos\: cos\, C-sin\, sin\, B\, 0\, tan\, tan\, A\, cos\, cos(\pi -C)\, tan\, tan(\pi -A)\, 0 \right |$....(i)
$[\because sin\, sin(\pi -B)=sin\, sin\, B, \, \, cos\, cos(\pi -c)=C,\, tan\, tan(\pi -A)=A]$
$\Delta =\left | 0\: sin\, sin\, B\: cos\, cos\, C-sin\, sin\, B\: 0\: tan\, tan\, A-cos\, cos\, C-tan\, tan\, A\, \, 0 \right |$
Transposing the determinant, we get
$\Delta =\begin{vmatrix} 0 &-sinB &-cosC \\ sinB &0 &-tanA \\ cosC &tanA &0 \end{vmatrix}$ ....(ii)
Adding (i) and (ii) we get
$2\Delta =0\Rightarrow \Delta =0$

Determinants exercise multiple choise question 20

Correct option (b)
Hint:
Simply first solve the determinant.
Given:
Here the distinct real roots of
$\left|\begin{array}{ccc} \cos e c x & \sec x & \sec x \\ \sec x & cos\, ecx & \sec x \\ \sec x & \sec x & \cos e c x \end{array}\right|=0$
lies in the interval
$\frac{-\pi }{4}\leq x\leq \frac{\pi }{4}$
We have to find the number of roots of the given determinant.
Solution:
$Here\; \; \left|\begin{array}{ccc} \cos e c x & \sec x & \sec x \\ \sec x & cos\, ecx & \sec x \\ \sec x & \sec x & \cos e c x \end{array}\right|=0$
Applying C1→C1-C3; C2→C2-C3
Taking common cos sec x-sec sec x from column 1 and 2, we get
$\Rightarrow(cos\, \, ecx-\sec x)^{2}\left|\begin{array}{ccc} 1 & 0 & \sec x \\ 0 & 1 & \sec x \\ -1 & -1 & \cos e c x \end{array}\right|=0$
Expanding along R1 , we get
$\Rightarrow(cos\, \, ecx-\sec x)^{2}[(cos\, ecx+sec\, x)+sec\, x(0+1)]=0$
$\Rightarrow(cos\, \, ecx-\sec x)^{2}(cos\, ecx+2sec\, x)=0$
$\Rightarrow(cos\, \, ecx-\sec x)^{2}=0\; \; or\; \; (cos\, ecx+2sec\, x)=0$
$\Rightarrow cos\, \, ecx-\sec x =0\; \; or\; \; cos\, ecx=-2sec\, x$
$\Rightarrow cos\, \,cos\, x-sin sin\, x =0\; \; or\; \; cos \, cos\, x=-2sin\, sin\, x$
$\Rightarrow sin\, x =cos\, x\; \; or\; \;\frac{cos\, cos\, x}{sin\, sin\, x}=-2$
$\Rightarrow \frac{cos\, cos\, x}{sin\, sin\, x}=1\; \;or\, \, cot\, cot\, x=-2$
$\Rightarrow cot\, cot\, x=1\; \;or\, \, cot\, cot\, x=-2$
$\Rightarrow tan\, tan\, x=1\; \;or\, \, tan\, tan\, x=-\frac{1}{2}$
Hence, there are two solutions.

Determinants exercise multiple choise question 21

Correct option (d)
Hint:
First solve the given determinant then find |A| for $\Theta$ =0 and for
$\theta =\frac{\pi }{2},\frac{3\pi }{2},.......$
Given:
Given that,
$Let\; \; A=\begin{vmatrix} 1 &sin\theta &1 \\ -sin\theta &1 &sin\theta \\ -1 &-sin\theta &1 \end{vmatrix}$
$W\! here\; \; 0\leq \theta \leq 2\pi$
We have to find the |A|
Solution:
$Here\; \; A=\begin{vmatrix} 1 &sin\theta &1 \\ -sin\theta &1 &sin\theta \\ -1 &-sin\theta &1 \end{vmatrix}$
Expanding along R1, we get
$\left | A \right |=1+sin^{2}\theta -sin\theta (-sin\theta +sin\theta )+sin^{2}\theta +1$
$\Rightarrow \left | A \right |=2+2sin^{2}$
$\Rightarrow \left | A \right |=2(1+\theta )$
$Given\; that\; \; 0\leq \theta \leq 2\pi$
For $\theta$=0
$\Rightarrow \left | A \right |=2$
$F\! or\; \theta =\frac{\pi }{2},\frac{3\pi }{2},.......$
$\Rightarrow \left | A \right |=2(1+1)$
$\Rightarrow \left | A \right |=4$
$Hence \left | A \right |\varepsilon [2,4]$

Determinants exercise multiple choise question 22

Correct option (c)
Hint:
Solve both determinants then equating them, we get required value of x
Given:
Here
$\begin{vmatrix} 2x &5 \\ 8 &x \end{vmatrix}=\begin{vmatrix} 6 &-2 \\ 7 &3 \end{vmatrix}$
We have to find the value of x.
Solution:
Here
$\begin{vmatrix} 2x &5 \\ 8 &x \end{vmatrix}=\begin{vmatrix} 6 &-2 \\ 7 &3 \end{vmatrix}$
$\Rightarrow 2x^{2}-40=18+14$
$\Rightarrow 2x^{2}=32+40$
$\Rightarrow 2x^{2}=72$
$\Rightarrow x^{2}=36$
$\Rightarrow x=\pm 6$

Determinants exercise multiple choise question 23

Correct option (c)
Hint:
If all the element in a diagonal of determinant is zero, then determinant must be zero.
Given:
Given that,
$f(x)=\begin{vmatrix} 0 &x-a &x-b \\ x+a &0 &x-c \\ x+b &x+c &0 \end{vmatrix}$
We have to find $f(x)$
Solution:
Here
$f(x)=\begin{vmatrix} 0 &x-a &x-b \\ x+a &0 &x-c \\ x+b &x+c &0 \end{vmatrix}$
Put X=0,
$\Rightarrow f(0)=\begin{vmatrix} 0 &-a &-b \\ a &0 &-c \\ b &c &0 \end{vmatrix}$
Expanding along R1
$\Rightarrow f(0)=a(0+bc)-b(ac-0)$
$\Rightarrow f(0)=abc-abc$
$\Rightarrow f(0)=0$

Determinants exercise multiple choise question 24

Correct option (c)
Hint:
Using properties of determinant.
Given:
$Let\: \:\;\Delta =\begin{vmatrix} a-b &b+c &a \\ b-c &c+a &b \\ c-a &a+b &c \end{vmatrix}$
We have to find the value of $\Delta$.
Solution:
$\Delta =\begin{vmatrix} a-b &b+c &a \\ b-c &c+a &b \\ c-a &a+b &c \end{vmatrix}$
Using properties,
$\Rightarrow \Delta =\begin{vmatrix} a &b &a \\ b &c &b \\ c &a &c \end{vmatrix}+\begin{vmatrix} -b &c &a \\ -c &a &b \\ -a &b &c \end{vmatrix}$
Here two columns are identical
$\Rightarrow \begin{vmatrix} a &b &a \\ b &c &b \\ c &a &c \end{vmatrix}=0$
$\Rightarrow \Delta =0+(-1)\begin{vmatrix} b &c &a \\ c &a &b \\ a &b &c \end{vmatrix}$
Applying R1→R1+R2+R3
$\Rightarrow \Delta =-\begin{vmatrix} a+b+c &a+b+c &a+b+c \\ c &a &b \\ a &b &c \end{vmatrix}$
Taking (a+b+c) common from R1
$\Rightarrow \Delta =-(a+b+c)\begin{vmatrix} 1 &1 &1 \\ c &a &b \\ a &b &c \end{vmatrix}$
Expanding along R1, we have
$\Rightarrow \Delta =-(a+b+c)[(ac-b^{2})-(c^{2}-ab)+(bc-a^{2})]$
$\Rightarrow \Delta =-(a+b+c)[ac-b^{2}-c^{2}+ab+bc-a^{2}]$
$\Rightarrow \Delta =(a+b+c)[a^{2}+b^{2}+c^{2}-ab-bc-ca]$
$[\because a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)]$
$\Rightarrow \Delta =a^{3}+b^{3}+c^{3}-3abc$
Hence, this is the required answer.

Determinants exercise multiple choise question 25

Correct option (d)
Hint:
Simply solve the determinant.
Given:
x, y, z are non-zero value and
$\begin{vmatrix} 1+x &1 &1 \\ 1 &1+y &1 \\ 1 &1 &1+z \end{vmatrix}=0$
We have to find
$x^{-1}+y^{-1}+z^{-1}\; \; i.e.\; \; \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
Solution:
$Here\; \; \; \begin{vmatrix} 1+x &1 &1 \\ 1 &1+y &1 \\ 1 &1 &1+z \end{vmatrix}=0$
Applying C1→C1-C3; C2→C2-C3
$\Rightarrow \begin{vmatrix} x &0 &1 \\ 0 &y &1 \\ -z &-z &1+z \end{vmatrix}=0$
Expanding along R1
$\Rightarrow x[y(1+z)+z]+0+(0+zy)=0$
$\Rightarrow x(y+yz+z)+zy=0$
$\Rightarrow xy+xyz+xz+zy=0$
$\Rightarrow xy+xz+zy=-xyz$
$\Rightarrow \frac{xy}{xyz}+\frac{xz}{xyz}+\frac{zy}{xyz}=\frac{-xyz}{xyz}$
$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{z}{z}=-1$
$\Rightarrow x^{-1}+y^{-1}+z^{-1}=-1$
Hence, the required solution.

Determinants exercise multiple choise question 26

Correct option (d)
Hint:
Evaluate the given determinant by applying row or column operation.
Given:
$Let\; \; \; \; \Delta =\begin{vmatrix} b^{2}-ab &b-c &bc-ac \\ ab-a^{2} &a-b &b^{2}-ab \\ bc-ca &c-a &ab-a^{2} \end{vmatrix}$
We have to find the value of $\Delta$
Solution:
$Here\; \; \; \; \Delta =\begin{vmatrix} b^{2}-ab &b-c &bc-ac \\ ab-a^{2} &a-b &b^{2}-ab \\ bc-ca &c-a &ab-a^{2} \end{vmatrix}$
$\Rightarrow \Delta =\begin{vmatrix} b(b-a) &b-c &c(b-a) \\ a(b-a) &a-b &b(b-a) \\ c(b-a) &c-a &a(b-a) \end{vmatrix}$
Taking common (b-a) from column 1 and 3
$\Rightarrow \Delta=(b-a)^{2}\left|\begin{array}{lll} b & b-c & c \\ a & a-b & b \\ c & c-a & a \end{array}\right|$
Here, C1 and C2 are same and if any two row or column of a matrix is identical then determinant will be zero.
$\Rightarrow \Delta=\left|\begin{array}{lll} b & b & c \\ a & a & b \\ c & c & a \end{array}\right|$
$\Rightarrow \Delta=0$
$Hence,\; \; \; \; \begin{vmatrix} b^{2}-ab &b-c &bc-ac \\ ab-a^{2} &a-b &b^{2}-ab \\ bc-ca &c-a &ab-a^{2} \end{vmatrix}=0$

Determinants exercise multiple choise question 27

Correct option (a)
Hint:
Use formula,
\begin{aligned} &\cos (x+y)=\cos x \cos y-\sin x \sin y \\ &\sin (x+y)=\sin x \cos y+\cos x \sin y \end{aligned}
Given:
Here x, y $\in$ R,
$\Delta=\left|\begin{array}{ccc} \cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ \cos (x+y) & -\sin (x+y) & 0 \end{array}\right|$
We have to find the value of $\Delta$.
Solution:
$Here\; \; \Delta=\left|\begin{array}{ccc} \cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ \cos (x+y) & -\sin (x+y) & 0 \end{array}\right|$
Using formula,
\begin{aligned} &\cos (x+y)=\cos x \cos y-\sin x \sin y \\ &\sin (x+y)=\sin x \cos y+\cos x \sin y \end{aligned}
$\Rightarrow \Delta=\left|\begin{array}{ccc} \cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ \\cos x \cos y-\sin x \sin y & -\sin x \cos y+\cos x \sin y & 0 \end{array}\right|$
Applying R3→R3 - R1 cos y + R2 sin y
$\Rightarrow \Delta=\left|\begin{array}{ccc} \cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ 0 & 0 & -\cos y+\sin y \end{array}\right|$
Expanding along R3
$\begin{array}{ll} \Rightarrow \quad \Delta =(\sin y-\cos y)\left(\sin ^{2} x+\cos ^{2} x\right) \\ \Rightarrow \quad \Delta =(\sin y-\cos y)\end{array}$$[\because x + x=1]$
Again, multiplying and dividing by $\sqrt{2}$, then we have
\begin{aligned} &\Delta=\sqrt{2} \times \frac{1}{\sqrt{2}}(\sin y-\cos y) \\ &\Delta=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin y-\frac{1}{\sqrt{2}} \cos y\right) \\ &\Delta=\sqrt{2}\left(\cos \frac{\pi}{4} \sin y-\sin \frac{\pi}{4} \cos y\right) \\ &\Delta=\sqrt{2} \sin \left(y-\frac{\pi}{4}\right) \end{aligned}
$Since \: \: -1\leq sin\, \theta \leq 1$
$\begin{gathered} -1 \leq \sin \sin \left(y-\frac{\pi}{4}\right) \leq 1 \\ -\sqrt{2} \leq \sqrt{2} \sin \sin \left(y-\frac{\pi}{4}\right) \leq \sqrt{2} \end{gathered}$
$Hence,\: \: \sqrt{2}sin\: sin(y-\frac{\pi }{4})\in [-\sqrt{2}, \sqrt{2}]$

Determinants exercise multiple choise question 28

Correct option (a)
Hint:
First solve the given determinant then find the maximum value of sin function, which is 1.
Given:
Given that,
$\Delta =\begin{vmatrix} 1 &1 &1 \\ 1 &1+sin\, \theta &1 \\ 1+cos\, \theta &1 &1 \end{vmatrix}$
We have to find the maximum value of $\Delta$.
Solution:
$Here\: \: \Delta =\begin{vmatrix} 1 &1 &1 \\ 1 &1+sin\, \theta &1 \\ 1+cos\, \theta &1 &1 \end{vmatrix}$
Applying R1 → R1 - R2
$\Rightarrow \Delta =\begin{vmatrix} 1 &1 &1 \\ 1 &1+sin\, \theta &1 \\ 1+cos\, \theta &1 &1 \end{vmatrix}$
Expanding along R1
\begin{aligned} &\Rightarrow \quad \Delta=\sin \theta[1-1-\cos \theta] \\ &\Rightarrow \quad \Delta=-\sin \theta \cos \theta \\ &\Rightarrow \quad \Delta=\frac{-2 \sin \theta \cos \theta}{2} \\ &\Rightarrow \quad \Delta=\frac{-\sin 2 \theta}{2} \end{aligned} $[\because \sin \sin 2 \theta=2 \sin \sin \theta \cos \cos \theta]$
We know that the maximum value of sin function is 1.
Hence, maximum value of
$\Delta =\frac{1}{2}$

Determinants exercise multiple choise question 29

Correct option (b)
Hint:
Solve determinant by applying row and column operation.
Given:
\begin{aligned} &\text { Let } \Delta=\left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right| \end{aligned}
We have to find the value of $\Delta$
Solution:
\begin{aligned} &\text { Here } \: \: \Delta=\left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right| \end{aligned}
Applying R1 → R1 - R2
$\Rightarrow \Delta =\left|\begin{array}{ccc} -2y & y &y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right|$
Applying R3 → R3 - R2
$\Rightarrow \Delta =\left|\begin{array}{ccc} -2y & y &y \\ x+2 y & x & x+y \\ -y & 2 y & -y \end{array}\right|$
Taking common y from R1 and R3
$\Rightarrow \Delta =y^{2}\left|\begin{array}{ccc} -2 & 1 &1 \\ x+2 y & x & x+y \\ -1 & 2 & -1 \end{array}\right|$
Applying C2 → C2 + 2C1; C3 → C3 -C1
$\Rightarrow \Delta =y^{2}\left|\begin{array}{ccc} -2 & -3 &3 \\ x+2 y & x & x+y \\ -1 & 0 & 0 \end{array}\right|$
Expanding along R3, we get
\begin{aligned} &\Delta=y^{2}[(-1)(3 y-9 x-12 y)] \\ &\Delta=y^{2}[9 x+9 y] \end{aligned}
\begin{aligned} &\Delta=9y^{2}[ x+ y] \end{aligned}
\begin{aligned} &\text { Hence } \left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right| \end{aligned}=9y^{2}[x+y]

Determinants exercise multiple choise question 30

Correct option (a)
Hint:
Solve the determinant by applying row operation then use limit.
Given:
Given that,
$f(x)=\left|\begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x & 2 x \\ \sin x & x & x \end{array}\right|$
We have to find
$\lim_{x\rightarrow 0}\frac{f(x)}{x^{2}}$
Solution:
Here, we have
$f(x)=\left|\begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x & 2 x \\ \sin x & x & x \end{array}\right|$
Applying R2 → R2 - R3
$f(x)=\left|\begin{array}{ccc} \cos x & x & 1 \\ \sin x & 0 & x \\ \sin x-cos\, x & 0 & x-1 \end{array}\right|$
Expanding along column 2
\begin{aligned} &f(x)=-x[\sin \sin x \cdot(x-1)-x(\sin \sin x-\cos \cos x)] \\ &f(x)=-x[x x-\sin \sin x-x x+x x] \\ &f(x)=-x[x \cos x-\sin x] \\ &f(x)=x[\sin x-x \cos x] \end{aligned}
Again,
\begin{aligned} \lim _{x \rightarrow 0} \frac{f(x)}{x^{2}} &=\lim _{x \rightarrow 0} \frac{x(\sin x-x \cos x)}{x^{2}} \\ &=\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}-\cos x\right) \\ \lim _{x \rightarrow 0} \frac{f(x)}{x^{2}} &=\lim _{x \rightarrow 0} \frac{\sin x}{x}-\lim _{x \rightarrow 0} \cos x \\ &=1-1=0 \end{aligned}
$Hence, \: \: \lim_{x\rightarrow 0}\frac{f(x)}{x^{2}} =0$

Determinants exercise multiple choise question 31

Correct option (c)
Hint:
Simply solve the determinant and find $a$.
Given:
Given that there are two values of $a$ which makes the determinant.
$\Delta=\left|\begin{array}{ccc} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a \end{array}\right|=86$
We have to find the sum of these two values of $a$.
Solution:
Here, we have
$\left|\begin{array}{ccc} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a \end{array}\right|=86$
Expanding along R1, we get
\begin{aligned} &\Rightarrow \quad 1\left(2 a^{2}+4\right)+2(4 a)+5(8-0)=86 \\ &\Rightarrow \quad 2 a^{2}+4+8 a+40=86 \\ &\Rightarrow \quad 2 a^{2}+8 a-42=0 \\ &\Rightarrow \quad a^{2}+4 a-21=0 \\ &\Rightarrow \quad(a+7)(a-3)=0 \\ &\Rightarrow \quad a=-7 \text { or } a=3 \end{aligned}
Hence sum of the two values of $a$ = - 7 + 3 = -4.

Determinants exercise multiple choise question 32

Correct option (d)
Hint:
We know that,
If each element of a row or column of determinant is expressed as a sum of two or more terms, then the determinant can be expressed as the sum of two or more determinant.
Given:
Given that
$\begin{vmatrix} a &p &x \\ b &q &y \\ c &r &z \end{vmatrix}=16$
We have to find the value of
$\left|\begin{array}{lll} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{array}\right|$
Solution:
By using properties we can write,
$\left|\begin{array}{lll} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{array}\right|=\left|\begin{array}{lll} p & x & a \\ q & y & b \\ r & z & c \end{array}\right|+\left|\begin{array}{ccc} x & a & p \\ y & b & q \\ z & c & r \end{array}\right|$
Applying C1↔C3 in first determinant and applying C2↔C1 in second determinant.
$=(-)\left|\begin{array}{lll} a & x & p \\ b & y & q \\ c & z & r \end{array}\right|+(-)\left|\begin{array}{lll} a & x & p \\ b & y & q \\ c & z & r \end{array}\right|$
Again applying C2↔C3 in first determinant and C2↔C3 in second determinant.
$=\left | a\: p\: x\: b\: q\: y\: c\: r\: z \right |+\left | a\: p\: x\: b\: q\: y\: c\: r\: z \right |$
$=16+16 = 32$
$\left|\begin{array}{lll} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{array}\right|=32$

Determinants exercise multiple choise question 33

Correct option (c)
Hint:
Using combination properties
$i.e. \: \: nC_{1}=n\: \: and\: \: nC_{2}=\frac{n(n-1)}{2}\: \: and \: \: so\: \: on.$
Given:
We have
$\left | 1\: 1\: 1\: nC_{1}n+ 2C_{1}n+4C_{1}\: nC_{2}n+2C_{2}n+4C_{2} \right |$
We have to find the value of given determinant.
Solution:
We have,
$\left|\begin{array}{ccc} 1 & 1 & 1 \\ n C_{1} & n+2 C_{1} & n+4 C_{1} \\ n C_{2} & n+2 C_{2} & n+4 C_{2} \end{array}\right|$
We know that
\begin{aligned} &n C_{1}=n, n C_{2}=\frac{n(n-1)}{2} n+2 C_{2}=\frac{(n+2)(n+1)}{2} \quad \text { and } \quad n+4 C_{2}=\frac{(n+4)(n+3)}{2}\\ &\left|\begin{array}{ccc} 1 & 1 & 1 \\ n C_{1} & n+2 C_{1} & n+4 C_{1} \\ n C_{2} & n+2 C_{2} & n+4 C_{2} \end{array}\right|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ n & n+2 & n+4 \\ \frac{n(n-1)}{2} & \frac{(n+2)(n+1)}{2} & \frac{(n+4)(n+3)}{2} \end{array}\right| \end{aligned}
Applying C2 → C2 - C1; C3 → C3 - C1
$=\left|1\: 0\: 0 \: n\: 2\: 4 \frac{n(n-1)}{2} 2\: n+1\: 4 n+6\right|$
Expanding along R1
$=2(4n+6)-4(2n+1)$
$=8n+12-8n-4$
$=8$
$Hence,\: \: \left|\begin{array}{ccc} 1 & 1 & 1 \\ n C_{1} & n+2 C_{1} & n+4 C_{1} \\ n C_{2} & n+2 C_{2} & n+4 C_{2} \end{array}\right|=8$

Determinants exercise multiple choise question 34

Correct option (c)
Hint:
Solve the determinant by applying column operations.
Given:
Given that,
$\left|\begin{array}{lll} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0$
We have to find the real root of the given determinant in interval
$[\frac{-\pi }{4},\frac{\pi }{4}]$
Solution:
Here given that,
$\left|\begin{array}{lll} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0$
Applying C1 → C1 - C2; C2 → C2 - C3
$\Rightarrow \quad\left|\begin{array}{ccc} \sin x-\cos x & 0 & \cos x \\ \cos x-\sin x & \sin x-\cos x & \cos x \\ 0 & \cos x-\sin x & \sin x \end{array}\right|=0$
Taking common (sin x- cos x) from column 1 and 2 , we get
\begin{aligned} &\quad(\sin x-\cos x)^{2}\left|\begin{array}{ccc} 1 & 0 & \cos x \\ -1 & 1 & \cos x \\ 0 & -1 & \sin x \end{array}\right|=0 \\ &\Rightarrow (\sin x-\cos x)^{2}(1(\sin x+\cos x)+\cos x(1))=0 \\ &\Rightarrow (\sin x-\cos x)^{2}(\sin x+2 \cos x)=0 \\ &\Rightarrow (\sin x-\cos x)^{2}=0 \text { or } \sin x+2 \cos x=0 \\ &\Rightarrow \, \sin x-\cos x=0 \text { or } \frac{\sin x}{\cos x}=-2 \\ &\Rightarrow \frac{\sin x}{\cos x}=1 \: \text \: or \tan x=-2 \\ &\Rightarrow \tan x=1 \text { or } \tan x=-2 \notin\left[\frac{-\pi}{4}, \frac{\pi}{4}\right] \end{aligned}
$Since, \frac{-\pi }{4}\leq x\leq \frac{\pi }{4}$
Hence tan x=1 is the required answer.

Determinants exercise multiple choise question 35

Correct option (a)
Hint:
A, B and C are angle of a $\Delta$, then ∠A+∠B+∠C=180 and solve determinant.
Given:
Given that,
A, B and C are angle of a triangle
We have to find the value of
$|-1 \cos \cos C \cos \cos B \cos \cos C-1 \cos \cos A \cos \cos B \cos \cos A-1|$
Solution:
We have
$|-1 \cos \cos C \cos \cos B \cos \cos C-1 \cos \cos A \cos \cos B \cos \cos A-1|$
Expanding along R1, we have
\begin{aligned} &=-1\left(1-\cos ^{2} A\right)-\cos C(-\cos C-\cos A \cos B)+\cos B(\cos A \cos C+\cos B) \\ &=-1+\cos ^{2} A+\cos ^{2} C+\cos A \cos B \cos C+\cos A \cos B \cos C+\cos ^{2} B \\ &=\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cos B \cos C-1 \\ &=\left(\frac{1+\cos 2 A}{2}\right)+\left(\frac{1+\cos 2 B}{2}\right)+\cos ^{2} C+2 \cos A \cos B \cos C-1 \quad\left[\because A=\frac{1+\cos \cos 2 A}{2}\right] \\ &=\frac{2+\cos 2 A+\cos 2 B}{2}+\cos ^{2} C+2 \cos A \cos B \cos C-1 \\ &=\frac{2+2 \cos \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)}{2}+\cos ^{2} C+2 \cos A \cos B \cos C-1 \end{aligned}
\begin{aligned} &=1+\cos (A+B) \cos (A-B)+\cos ^{2} C+2 \cos A \cos B \cos C-1 \\ &=1+\cos (180-C) \cos (A-B)+\cos ^{2} C+2 \cos A \cos B \cos C-1 \\ &{[\angle A+\angle B+\angle C=180]} \\ &=1-\cos C \cos (A-B)+\cos ^{2} C+2 \cos A \cos B \cos C-1 \\ &=\cos \cos C[-\cos \cos (A-B)+\cos \cos C]+2 \cos \cos A \cos \cos B \cos \cos C \\ &=\cos \cos C[-\cos \cos (A-B)-\cos \cos (A+B)]+2 \cos \cos A \cos \cos B \cos \cos C \\ &{[\because \cos \cos C=\cos \cos (180-(A+B))=-\cos \cos (A+B)]} \\ &=-\cos \cos C\left[2 \cos \cos \left(\frac{A-B+A+B}{2}\right) \cos \cos \left(\frac{A-B-A-B}{2}\right)\right]+2 \cos \cos A \\ &\cos \cos B \cos \cos C \\ &=-2 \cos C \cos A \cos B+2 \cos A \cos B \cos C \\ &=0 \end{aligned}
Hence 0 is the required answer.

Determinants exercise multiple choise question 36

Correct option(b)
Hint:
Use determinant to solve the area of triangle.
Given:
Area of a triangle = 9sq.units
Vertices of triangle = (-3,0),(3,0) and (0,k)
We have to find the value of k.
Solution:
We have,
Area of a triangle = 9sq.units
$\frac{1}{2}\left|\begin{array}{ccc} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{array}\right|=9$
Solving the determinant, we get
$\left | -k(-6) \right |=18$
$\left | 6k \right |=18$
$k=\pm 3$
Hence k = 3 is required value of k.

Determinants exercise multiple choise question 37

Correct option (d)
Hint:
We use |kA| = kn|A|, where n is order of A.
Given:
|A| = 8 and A is 3x3 matrix.
We have to find |3A|.
Solution:
Here |A| = 8
We know that
|kA| = kn|A|, where n is order of A.
$\therefore$ |3A| = 33 |A|
=27x8 = 216 [Since A is 3x3 matrix]
Hence |3A| = 216

Determinants Exercise MCQs Question 2 (iv)

Answer:$\left|\begin{array}{lll} \frac{1}{a} & a^{2} & b c \\ \frac{1}{b} & b^{2} & a c \\ \frac{1}{c} & c^{2} & a b \end{array}\right|=0$
Hint: We will try to do any two column or row equal
Given: $\left|\begin{array}{lll} \frac{1}{a} & a^{2} & b c \\ \frac{1}{b} & b^{2} & a c \\ \frac{1}{c} & c^{2} & a b \end{array}\right|$
Solution:$\left|\begin{array}{lll} \frac{1}{a} & a^{2} & b c \\ \frac{1}{b} & b^{2} & a c \\ \frac{1}{c} & c^{2} & a b \end{array}\right|$
\begin{aligned} &\text { Applying } \mathrm{C}_{3} \rightarrow C_{3} \div(a b c) \\ &=a b c\left|\begin{array}{ccc} \frac{1}{a} & a^{2} & \frac{b c}{a b c} \\ \frac{1}{b} & b^{2} & \frac{a c}{a b c} \\ \frac{1}{c} & c^{2} & \frac{a b}{a b c} \end{array}\right| \\ &=a b c\left|\begin{array}{ccc} \frac{1}{a} & a^{2} & \frac{1}{a} \\ \frac{1}{b} & b^{2} & \frac{1}{b} \\ \frac{1}{c} & c^{2} & \frac{1}{c} \end{array}\right| \end{aligned}
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
\begin{aligned} &=0 \times a b c \quad\left(\because C_{1}=C_{3}\right) \\ &=0 \end{aligned}
Hence $\left|\begin{array}{lll} \frac{1}{a} & a^{2} & b c \\ \frac{1}{b} & b^{2} & a c \\ \frac{1}{c} & c^{2} & a b \end{array}\right|=0$

Determinants Exercise MCQs Question 2(xi)

Answer:$\left|\begin{array}{ccc} a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z \end{array}\right|=0$
Hint: We will try to do any two column or row equal
Given:$\left|\begin{array}{ccc} a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z \end{array}\right|$
Solution:$\left|\begin{array}{ccc} a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z \end{array}\right|$
\begin{aligned} &\text { Applying } R_{3} \rightarrow 2 \times R_{3}+R_{1} \\ &\left|\begin{array}{ccc} a & b & c \\ a+2 x & b+2 y & c+2 z \\ 2 x+a & 2 y+b & 2 z+c \end{array}\right| \end{aligned}
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$=0 \quad\left(\because R_{2}=R_{3}\right)$
Hence $\left|\begin{array}{ccc} a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z \end{array}\right|=0$

Deteminants Exercise MCQs Question 2(xiv).

Answer:$\left|\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ \cos 180^{\circ} & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right|=0$
Hint: We will try to do any two column or row equal
Given:$\left|\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ \cos 180^{\circ} & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right|$
Solution:$\left|\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ \cos 180^{\circ} & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right|$
\begin{aligned} &=\left|\begin{array}{ccc} \sin ^{2} 23^{\circ} & \cos ^{2}\left(90^{\circ}-67^{\circ}\right) & -1 \\ -\sin ^{2} 67^{\circ} & -\cos ^{2}\left(90^{\circ}-23^{\circ}\right) & (-1)^{2} \\ \cos 180^{\circ} & \cos ^{2}\left(90^{\circ}-23^{\circ}\right) & \sin ^{2} 67^{\circ} \end{array}\right| \\ &\because \cos (90-\theta)=\sin \theta \\ &\cos 120^{\circ}=-1 \\ &=\left|\begin{array}{ccc} \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} & -1 \\ -\sin ^{2} 67^{\circ} & -\cos ^{2} 67^{0} & 1 \\ -1 & \cos ^{2} 67^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right| \end{aligned}
\begin{aligned} &C_{1} \rightarrow C_{1}+C_{2} \\ &=\left|\begin{array}{ccc} \sin ^{2} 23^{\circ}+\cos ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} & -1 \\ -\sin ^{2} 67^{\circ}+\left(-\cos ^{2} 67^{\circ}\right) & -\cos ^{2} 67^{\circ} & 1 \\ -1+\cos ^{2} 67^{\circ} & \cos ^{2} 67^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right| \\ &=\left|\begin{array}{ccc} 1 & \cos ^{2} 23^{\circ} & -1 \\ -\left(\sin ^{2} 67^{\circ}+\cos ^{2} 67^{\circ}\right) & -\cos ^{2} 67^{\circ} & 1 \\ -1+\cos ^{2} 67^{0} & \cos ^{2} 67^{0} & \sin ^{2} 67^{\circ} \end{array}\right| \\ &\because \sin ^{2} \theta+\cos ^{2} \theta=1 \end{aligned}
\begin{aligned} &=\left|\begin{array}{ccc} 1 & \cos ^{2} 23^{0} & -1 \\ -1 & -\cos ^{2} 67^{\circ} & 1 \\ -\sin ^{2} 67^{\circ} & \cos ^{2} 67^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right|\\ &\text { Taking common }(-1) \text { from } \mathrm{C}_{1}\\ &=(-1)\left|\begin{array}{ccc} -1 & \cos ^{2} 23^{0} & -1 \\ 1 & -\cos ^{2} 67^{\circ} & 1 \\ \sin ^{2} 67^{\circ} & \cos ^{2} 67^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right| \end{aligned}
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
\begin{aligned} &=(-1) \times 0 \\ &=0 \end{aligned}
Hence $\left|\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ \cos 180^{\circ} & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right|=0$

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