Question Set A

cbse 12th Standard Maths Matrices Important Questions And Answers Study Material - QB365

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Questions:

Generic 2

4 x 2 = 8

If \(A^{\prime}=\left[\begin{array}{rr}-2 & 3 \\ 1 & 2\end{array}\right]\) and \(B=\left[\begin{array}{rr}-1 & 0 \\ 1 & 2\end{array}\right]\), then find \((A+2 B)^{\prime} \)

Find X and Y, if \(2 x+3y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right] and \ 3 x+2 y=\left[\begin{array}{rr}2 & -2 \\ -1 & 5\end{array}\right]\)

Evaluate the integral: \(\int {e^{tan^{-1}x}\over1 + x^2}dx\)

Evaluate the integral: \(\int { {x^2\over1+x^3}dx. } \)

Generic 3

6 x 3 = 18

Integrate the rational functions in \(\frac { 2x }{ \left( { x }^{ 2 }+1 \right) ({ x }^{ 2 }+3) } \)

Integrate the functions in \(\frac { 1 }{ \sqrt { 9-{ 25x }^{ 2 } } } \)

Integrate \(\frac { 1 }{ 1-\tan { x } } \)

Integrate \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\)

If \(A=\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{matrix} \right] \)and \(B=\left[ \begin{matrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{matrix} \right] \), then find 2A-B.

10.

Given:\(3\begin{bmatrix} x & y \\ z & w \end{bmatrix}=\begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix}+\begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix},\) find the values of x, y, z and w.

Generic 5

4 x 5 = 20

11.

\(\text { If } A=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2 \end{array}\right], B=\left[\begin{array}{rr} 1 & 3 \\ 0 & 2 \\ -1 & 4 \end{array}\right] \text { and } C=\left[\begin{array}{cccc} 1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1 \end{array}\right]\) find A(BC), (AB)C and show that (AB)C = A(BC).

12.

Find X and Y, if \(X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right] and\ X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right] \)

13.

Evaluate : \(\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { dx }{ 1+\sqrt { \cot { x } } } } \)

14.

Evaluate: \(\int _{ 1 }^{ 3 }{ ({ 2x }^{ 2 }+5x) } dx\) as a limit of a sums.

Multiple Choice Question

10 x 1 = 10

15.

For the matrix A = \(\left[\begin{array}{ccc} 2 & -1 & 1 \\ \lambda & 2 & 0 \\ 1 & -2 & 3 \end{array}\right]\) to be invertible, the value of \(\lambda\) is

(a)

(b)

(c)

R - {10}

(d)

R - {-10}

16.

lf A is a square matrix of order 2 and |A| = -2, then value of |5A'| is

(a)

-50

(b)

-10

(c)

(d)

17.

A matrix has 18 elements, then possible number of orders of a matrix are

(a)

(b)

(c)

(d)

18.

The matrix A satisfies the equation \(\left[\begin{array}{rr} 0 & 2 \\ -1 & 1 \end{array}\right] A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) then matrix A is

(a)

\(\left[\begin{array}{rr} 2 & 0 \\ 1 & -1 \end{array}\right]\)

(b)

\(\left[\begin{array}{rr} 1 & -2 \\ 1 & 0 \end{array}\right]\)

(c)

\(\left[\begin{array}{cc} \frac{1}{2} & -1 \\ \frac{1}{2} & 0 \end{array}\right]\)

(d)

\(\left[\begin{array}{rr} 1 & 2 \\ -1 & 0 \end{array}\right]\)

19.

\(\int_{-\pi / 4}^{\pi / 4} \frac{d x}{1+\cos 2 x}\) is equal to

(a)

(b)

(c)

(d)

20.

The matrix \(\left[\begin{array}{ccc}0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0\end{array}\right]\) is a

(a)

diagonal matrix

(b)

symmetric matrix

(c)

skew-symmetric matrix

(d)

scalar matrix

21.

The value of \(\int _{ 0 }^{ 1 }{ \left( \frac { 2x-1 }{ 1+x-{ x }^{ 2 } } \right) } dx\) is

(a)

(b)

(c)

-1

(d)

\(\frac{\pi}{4}\)

22.

\(\int { \frac { dx }{ { e }^{ x }+{ e }^{ -x } } } \) is equal to

(a)

tan^–1 (e^x) + C

(b)

tan^–1 (e^–x) + C

(c)

log (e^x – e^–x) + C

(d)

log (e^x + e^–x) + C

23.

If \(\frac { d }{ dx } f(x)=4{ x }^{ 3 }-\frac { 3 }{ { x }^{ 4 } } \) such that f (2) = 0. Then f (x) is

(a)

\({ x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } -\frac { 129 }{ 8 } \)

(b)

\({ x }^{ 3 }+\frac { 1 }{ { x }^{ 4 } } +\frac { 129 }{ 8 } \)

(c)

\({ x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } +\frac { 129 }{ 8 } \)

(d)

\({ x }^{ 3 }+\frac { 1 }{ { x }^{ 4 } } -\frac { 129 }{ 8 } \)

24.

If ∫sec²(7 – 4x)dx = a tan (7 – 4x) + C, then value of a is

(a)

(b)

-4

(c)

(d)

\(-\frac { 1 }{ 4 } \)

Case Study Questions

1 x 4 = 4

25.

Three shopkeepers A, Band C go to a store to buy stationary. A purchase 12 dozen notebooks, 5 dozen pens and 6 dozen pencils. B purchases 10 dozen notebooks, 6 dozen pens and 7 dozen pencils. C purchases 11 dozen notebooks, 13 dozen pens and 8 dozen pencils. A notebook costs Rs.40, a pen costs Rs.12 and a pencil costs Rs.3.Based on the above information, answer the following questions
(i) The number of items purchased by shopkeepers A, Band C represented in matrix form as

(ii) If Y represents the matrix formed by the cost of each item, then XY equal

(iii) Bill of A is equal to

(a) Rs.6740

(b) Rs.8140

(c) Rs. 5740

(d) Rs.6696

(iv) If A² = A, then (A + 1)3 - 7A =

(a) A

(b) -I

(c) I

(d) A+ I

(v) If A and B are 3 x 3 matrices such that A² - B² = (A - B) (A + B), then

(a) either A or B is zero matrix

(b) either A or B is unit matrix

(c) Rs. A= B

(d) AB = BA

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Answers Key:

Generic 2

4 x 2 = 8

\(\text { We have, } A^{\prime}=\left[\begin{array}{cc} -2 & 3 \\ 1 & 2 \end{array}\right] \text { and } B=\left[\begin{array}{cc} -1 & 0 \\ 1 & 2 \end{array}\right] \)
\(\therefore A=\left(A^{\prime}\right)^{\prime}=\left[\begin{array}{rr} -2 & 3 \\ 1 & 2 \end{array}\right]^{\prime}=\left[\begin{array}{rr} -2 & 1 \\ 3 & 2 \end{array}\right]\)
[interchanging the elements of rows and columns]
\(\text { Now, } \quad A+2 B=\left[\begin{array}{rr} -2 & 1 \\ 3 & 2 \end{array}\right]+2\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right] \)
\(=\left[\begin{array}{rr} -2 & 1 \\ 3 & 2 \end{array}\right]+\left[\begin{array}{rr} -2 & 0 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{rr} -2-2 & 1+0 \\ 3+2 & 2+4 \end{array}\right]=\left[\begin{array}{rr} -4 & 1 \\ 5 & 6 \end{array}\right] \)
\(\Rightarrow(A+2 B)^{\prime}=\left[\begin{array}{rr} -4 & 1 \\ 5 & 6 \end{array}\right]^{\prime}=\left[\begin{array}{rr} -4 & 5 \\ 1 & 6 \end{array}\right] \)
[interchanging the elements ofrows and columns]

We have, \(2 X+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]\) and
\(3 X+2 Y=\left[\begin{array}{rr} 2 & -2 \\ -1 & 5 \end{array}\right]\)
On multiplying Eq. (i) by 2 and Eq. (ii) by 3 , we get
\(4 X+6 Y=\left[\begin{array}{ll} 4 & 6 \\ 8 & 0 \end{array}\right]\) and \( 9 X+6 Y=\left[\begin{array}{rr}6 & -6 \\ -3 & 15\end{array}\right]\)
On subtracting Eq. (iii) from Eq. (iv), we get
\( (9 X+6 Y)-(4 X+6 Y)=\left[\begin{array}{rr} 6 & -6 \\ -3 & 15 \end{array}\right]-\left[\begin{array}{ll} 4 & 6 \\ 8 & 0 \end{array}\right] \)
\(\Rightarrow 9 X+6 Y-4 X-6 Y=\left[\begin{array}{rr} 6-4 & -6-6 \\ -3-8 & 15-0 \end{array}\right]=\left[\begin{array}{rr} 2 & -12 \\ -11 & 15 \end{array}\right] \)
\(\Rightarrow X=\frac{1}{5}\left[\begin{array}{rr}2 & -12 \\ -11 & 15\end{array}\right]=\left[\begin{array}{cc}\frac{2}{5} & \frac{-12}{5} \\ \frac{-11}{5} & 3\end{array}\right]\)
On substituting the value of X in Eq. (i), we get
\( \Rightarrow\left[\begin{array}{cc} \frac{2}{5} & \frac{-12}{5} \\ \frac{-11}{5} & 3 \end{array}\right]+3 Y=\left[\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}\right] \)
\(\Rightarrow \left[\begin{array}{cc} \frac{4}{5} & \frac{-24}{5} \\ \frac{-22}{5} & 6 \end{array}\right]+3 Y=\left[\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}\right] \)
\(\Rightarrow Y=\frac{1}{3}\left[\begin{array}{cc} 2-\frac{4}{5} & 3+\frac{24}{5} \\ 4+\frac{22}{5} & 0-6 \end{array}\right]\)
\(\Rightarrow Y=\frac{1}{3}\left[\begin{array}{cc} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \end{array}\right]=\left[\begin{array}{cc} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{array}\right] \)

\(\int \frac{e^{\tan ^{4} x}}{1+x^{2}} d x=\int e^{t} d t=e^{t}+C=e^{\tan ^{-1} x}+C \)

\(\int \frac{x^{2}}{1+x^{3}} d x =\frac{1}{3} \int \frac{1}{t} d t=\frac{1}{3} \log |t|+C \)
\(=\frac{1}{3} \log \left|1+x^{3}\right|+C\)

Generic 3

6 x 3 = 18

Put \({ x }^{ 2 }=t\) So that \(2xdx=dt.\)
\(\therefore \int { \frac { 2x }{ \left( { x }^{ 2 }+1 \right) ({ x }^{ 2 }+3) } } dx\)
\(=\int { \frac { dt }{ (t+1)(t+3) } = } \frac { 1 }{ 2 } \int { \left( \frac { 1 }{ t+1 } -\frac { 1 }{ t+3 } \right) } dt\)
[Partial Fractions] [Do it]
\(=\frac { 1 }{ 2 } \log { \left| t+1 \right| } -\log { \left| t+3 \right| } +C\)
\(=\frac { 1 }{ 2 } \log { \left| \frac { t+1 }{ t+3 } \right| } +C\)
\(=\frac { 1 }{ 2 } \log { \left| \frac { { x }^{ 2 }+1 }{ { x }^{ 2 }+3 } \right| } +C\)
\(=\frac { 1 }{ 2 } \log { \left| \frac { { x }^{ 2 }+1 }{ { x }^{ 2 }+3 } \right| } +C\)
\(\left[ \therefore { x }^{ 2 }\ge 0\Rightarrow { x }^{ 2 }+1>0.Similarly\quad { x }^{ 2 }+3>0 \right] \)

\(\text { Let } 5 x=t \)
\(\therefore 5 d x=d t \)
\(\Rightarrow \int \frac{1}{\sqrt{9-25 x^{2}}} d x =\frac{1}{5} \int \frac{1}{9-t^{2}} d t \)
\(=\frac{1}{5} \int \frac{1}{\sqrt{3^{2}-t^{2}}} d t \)
\(=\frac{1}{5} \sin ^{-1}\left(\frac{t}{3}\right)+\mathrm{C} \)
\(=\frac{1}{5} \sin ^{-1}\left(\frac{5 x}{3}\right)+\mathrm{C}\)

\(I= \int { \frac { 1 }{ 1-\tan { x } } } dx.\)
\(=\int { \frac { 1 }{ 1-\frac { \sin { x } }{ \cos { x } } } } dx\)
\(=\int { \frac { \cos { x } }{ \cos { x } -\sin { x } } } dx\)
\(=\frac { 1 }{ 2 } \int { \frac { 2\cos { x } }{ \cos { x } -\sin { x } } } dx\)
\(=\frac { 1 }{ 2 } \frac { (\cos { x } -\sin { x } )-(-\sin { x } -\cos { x } ) }{ \cos { x } -\sin { x } } dx\)
\(=\frac { 1 }{ 2 } \int { 1.dx-\frac { 1 }{ 2 } } \int { \frac { -\sin { x } -\cos { x } }{ \cos { x } -\sin { x } } } dx\)
\(=\frac { 1 }{ 2 } x-\frac { 1 }{ 2 } \int { \frac { -\sin { x } -\cos { x } }{ \cos { x } -\sin { x } } } dx\)
Let \({ I }^{ \prime }=\frac { -\sin { x } -\cos { x } }{ \cos { x } -\sin { x } } dx\)
Put \(\cos { x } -\sin { x } =t\) so that \((\cos { x } -\sin { x } )dx=dt.\)
\(\therefore { I }^{ \prime }=\int { \frac { dt }{ t } } =\log { \left| t \right| }\)
\( = \log { \left| \cos { x } -\sin { x } \right| . } \)
From (1),
\(I=\frac { 1 }{ 2 } x-\frac { 1 }{ 2 } \log { \left| \cos { x } -\sin { x } \right| } +C\)

\(I= \int { \frac { 2\cos { x } -3\sin { x } }{ 2\sin { x } +3\cos { x } } } dx\)
Put \(2\sin { x } +3\ \cos { x } =t\) so that
\((2\cos { x } -3\sin { x } )dx=dt\)
\(\therefore I=\frac { 1 }{ 2 } \int { \frac { dt }{ t } } =\frac { 1 }{ 2 } \log { \left| t \right| } +C\)
\(=\frac { 1 }{ 2 } \log { \left| 2\sin { x } +3\cos { x } \right| } +C.\)

We have
\(2 A-B=2\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right]-\left[\begin{array}{ccc} 3 & -1 & 3 \\ -1 & 0 & 2 \end{array}\right]\)
\(=\left[\begin{array}{lll} 2 & 4 & 6 \\ 4 & 6 & 2 \end{array}\right]+\left[\begin{array}{ccc} -3 & 1 & -3 \\ 1 & 0 & -2 \end{array}\right]\)
\(=\left[\begin{array}{lll} 2-3 & 4+1 & 6-3 \\ 4+1 & 6+0 & 2-2 \end{array}\right]=\left[\begin{array}{ccc} -1 & 5 & 3 \\ 5 & 6 & 0 \end{array}\right]\)

10.

We have: \(3\begin{bmatrix} x & y \\ z & w \end{bmatrix}=\begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix}+\begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix},\)
\(\Rightarrow \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix}=\begin{bmatrix} x+4 & x+y+6 \\ z+w-1 & 2w+3 \end{bmatrix},\)
Equating corresponding elements:
3x = x + y \(\Rightarrow \) 2x = 4 \(\Rightarrow \) x = 2
3y = x + y + 6 \(\Rightarrow \) 2y = 2 + 6 \(\Rightarrow \) 2y = 8 \(\Rightarrow \) y = 4
3w = 2w + 3 \(\Rightarrow \) w = 3 and
3z = z + w - 1\(\Rightarrow \) 2z = 3 - 1 = 2 \(\Rightarrow \) z = 1.
Hence, x = 2, y = 4, z = 1 and w = 3.

Generic 5

4 x 5 = 20

11.

\(\text {We have } A B=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2 \end{array}\right]\left[\begin{array}{rr} 1 & 3 \\ 0 & 2 \\ -1 & 4 \end{array}\right]=\left[\begin{array}{cc} 1+0+1 & 3+2-4 \\ 2+0-3 & 6+0+12 \\ 3+0-2 & 9-2+8 \end{array}\right]=\left[\begin{array}{cc} 2 & 1 \\ -1 & 18 \\ 1 & 15 \end{array}\right]\)
\(\text { (AB) }(C)=\left[\begin{array}{cc} 2 & 1 \\ -1 & 18 \\ 1 & 15 \end{array}\right]\left[\begin{array}{cccc} 1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1 \end{array}\right]=\left[\begin{array}{cccc} 2+2 & 4+0 & 6-2 & -8+1 \\ -1+36 & -2+0 & -3-36 & 4+18 \\ 1+30 & 2+0 & 3-30 & -4+15 \end{array}\right]\)
\(=\left[\begin{array}{cccc} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{array}\right]\)
\(\text { Now } \quad \mathrm{BC}=\left[\begin{array}{rr} 1 & 3 \\ 0 & 2 \\ -1 & 4 \end{array}\right]\left[\begin{array}{llrr} 1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1 \end{array}\right]=\left[\begin{array}{rrrr} 1+6 & 2+0 & 3-6 & -4+3 \\ 0+4 & 0+0 & 0-4 & 0+2 \\ -1+8 & -2+0 & -3-8 & 4+4 \end{array}\right]\)
\(=\left[\begin{array}{cccc} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8 \end{array}\right]\)
\(\text { Therefore } \quad A(B C)=\left[\begin{array}{rrr} 1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2 \end{array}\right]\left[\begin{array}{cccc} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8 \end{array}\right]\)
\(=\left[\begin{array}{cccc} 7+4-7 & 2+0+2 & -3-4+11 & -1+2-8 \\ 14+0+21 & 4+0-6 & -6+0-33 & -2+0+24 \\ 21-4+14 & 6+0-4 & -9+4-22 & -3-2+16 \end{array}\right]\)
\(=\left[\begin{array}{cccc} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{array}\right] . \text { Clearly, (AB) } C=A\)

12.

\(X+Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right] \ldots .(1) \)
\(X-Y=\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right] . .(2)\)
Add and subtract the given equations
\(2 X=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right]+\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]=\left[\begin{array}{ll} 7+3 & 0+0 \\ 2+0 & 5+3 \end{array}\right]=\left[\begin{array}{cc} 10 & 0 \\ 2 & 8 \end{array}\right] \)
\(\therefore X=\frac{1}{2}\left[\begin{array}{cc} 10 & 0 \\ 2 & 8 \end{array}\right]=\left[\begin{array}{ll} 5 & 0 \\ 1 & 4 \end{array}\right] \)
\(\text { Now, } X+Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right] \)
\(\Rightarrow\left[\begin{array}{ll} 5 & 0 \\ 1 & 4 \end{array}\right]+Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right] \)
\(\Rightarrow Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right]-\left[\begin{array}{ll} 5 & 0 \\ 1 & 4 \end{array}\right] \)
\(\Rightarrow Y=\left[\begin{array}{ll} 7-5 & 0-0 \\ 2-1 & 5-4 \end{array}\right]\)
\(\therefore Y=\left[\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right]\)

13.

\(I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { dx }{ 1+\sqrt { \cot { x } } } } \)
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \sin { x } } }{ \sqrt { \sin { x } } +\sqrt { \cos { x } } } } dx\)
Apply property \(\int _{ a }^{ b }{ f\left( x \right) } =\int _{ a }^{ b }{ f\left( a+b-x \right) } \)
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \sin { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } }{ \sqrt { \sin { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } +\sqrt { \cos { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } } } dx\)
\(\therefore I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \cos { x } } }{ \sqrt { \cos { x } } +\sqrt { \sin { x } } } } dx\)
Adding, we get \(2I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ dx } =\left[ x \right] _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }\)

14.

\(I=\int _{ 1 }^{ 3 }{ ({ 2x }^{ 2 }+5x) } dx\)
\(\lim _{ h\rightarrow 0 }{ h } [f(1)+f(1+h)+f(1+2h)+.........+f(1+(n-1)h)]\)
\(where\quad f(x)={ { 2x }^{ 2 }+5x }\quad and\quad h=\frac { 2 }{ n } or\quad nh=2\)
\(f(1)=7\)
\(f(1+h)=2(1+h{ ) }^{ 2 }+5(1+h)\)
\(=7+9h+2{ h }^{ 2 }\)
\(f(1+2h)=2(1+2h{ ) }^{ 2 }+5(1+2h)\)
\(=7+18h+2.{ { 2 }^{ 2 }h }^{ 2 }\)
\(f(1+3h)=2(1+3h{ ) }^{ 2 }+5(1+3h)\)
\(=7+27h+2.{ { 3 }^{ 2 }h }^{ 2 }\)
\(=7+27h+2.{ { 3 }^{ 2 }h }^{ 2 }\)
\(I=\lim _{ h\rightarrow 0 }{ h } [7n+9h\frac { n(n-1) }{ 2 } +2{ h }^{ 2 }.\frac { n(n-1)(2n-1) }{ 6 } \)
\(=\lim _{ h\rightarrow 0 }{ h } \left[ 7nh+\frac { 9 }{ 2 } nh(nh-h)+\frac { 1 }{ 3 } nh(nh-h)(2nh-h) \right] \)
\(=14+18+\frac { 16 }{ 3 } =\frac { 112 }{ 3 } \)

Multiple Choice Question

10 x 1 = 10

15.

(d)
R - {-10}

16.

(a)
-50

17.

18.

19.

(a)
1

20.

\(\text { Let } A=\left[\begin{array}{ccc} 0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0 \end{array}\right] \text { . Then, } A^{\prime}=-A\)

21.

(b)
0

22.

(a)
tan^–1 (e^x) + C

23.

24.

∫sec²(7 – 4x)dx =\(\frac { tan(7-4x) }{ -4 } +C=-\frac { 1 }{ 4 } \)tan(7 - 4x) + C

Case Study Questions

1 x 4 = 4

25.

(i) (a) : Number of items purchased by shopkeepers A, B and C can be written in matrix form as

(ii) (b) : Since \(Y=\left[\begin{array}{c} 40 \\ 12 \\ 3 \end{array}\right]\)\(Note book\\ Pen\\ Penc\)
\(\therefore \quad X Y=\left[\begin{array}{lll} 144 & 60 & 72 \\ 120 & 72 & 84 \\ 132 & 156 & 96 \end{array}\right]\left[\begin{array}{c} 40 \\ 12 \\ 3 \end{array}\right]\)
\(=\left[\begin{array}{c} 5760+720+216 \\ 4800+864+252 \\ 5280+1872+288 \end{array}\right]=\left[\begin{array}{l} 6696 \\ 5916 \\ 7440 \end{array}\right]\)
(iii) (d) : Bill of A is Rs.6696.
(iv) (c) : (A + I)² = A² + 2A + 1= 3A + I
\(\Rightarrow\) (A + I)³ = (3A + I) (A + I)
= 3A² + 4A + I = 7A + I
\(\therefore\) (A+I)³-7A = I
(v) (d) : A² - B² = (A - B) (A + B) = A² + AB - BA - B²
\(\therefore\) AB=BA.