Category: Class 12 Mathematics Objective Questions

Mathematics Multiple Choice Questions on “Invertible Matrices”.

1. Which among the following is inverse of the matrix A=(begin{bmatrix}2&3\5&1end{bmatrix}) ?
a) (begin{bmatrix}frac{1}{13}&frac{3}{13}\ frac{5}{13}&frac{-2}{13}end{bmatrix})
b) (begin{bmatrix}frac{-1}{13}&frac{3}{13}\ frac{5}{13}&frac{-2}{13}end{bmatrix})
c) (begin{bmatrix}frac{-1}{13}&frac{3}{13}\1&frac{-2}{13}end{bmatrix})
d) (begin{bmatrix}frac{-1}{13}&frac{3}{13}\ frac{5}{13}&-2end{bmatrix})
Answer: b
Clarification: Consider the matrix A=(begin{bmatrix}2&3\5&1end{bmatrix})
Using elementary row operation, we write A=IA.
(begin{bmatrix}2&3\5&1end{bmatrix})=(begin{bmatrix}1&0\0&1end{bmatrix})A
(begin{bmatrix}-13&0\5&1end{bmatrix})=(begin{bmatrix}1&-3\0&1end{bmatrix})A   (Applying R₁→R₁-3R₂)
(begin{bmatrix}1&0\5&1end{bmatrix})=(begin{bmatrix}frac{-1}{13}&frac{3}{13}\0&1end{bmatrix})A   (Applying (R_1 rightarrow -frac{R_1}{13}))
(begin{bmatrix}1&0\0&1end{bmatrix})=(begin{bmatrix}frac{-1}{13}&frac{3}{13}\ frac{5}{13}&frac{-2}{13}end{bmatrix})A   (Applying R₂→R₂-5R₁)
(A^{-1}=begin{bmatrix}frac{-1}{13}&frac{3}{13}\ frac{5}{13}&frac{-2}{13}end{bmatrix}).

2. Which of the following matrices will not have an inverse?
a) (begin{bmatrix}2&4\-1&1end{bmatrix})
b) (begin{bmatrix}1&5&2\6&4&2\1&3&2end{bmatrix})
c) (begin{bmatrix}1&2\1&1end{bmatrix})
d) (begin{bmatrix}1&2&5\3&6&4end{bmatrix})
Answer: d
Clarification: The matrix A=(begin{bmatrix}1&2&5\3&6&4end{bmatrix}) will not have an inverse as it is a rectangular matrix. Rectangular matrix does not possess an inverse matrix.

3. If A and B are invertible matrices of the same order, then (AB)^-1=B^-1 A^-1.
a) True
b) False
Answer: a
Clarification: The given statement is true.
(AB) (AB)^-1=I (Using the formula AA^-1=I)
Multiplying both sides by A^-1, we get
A^-1 (AB) (AB)^-1=A^-1 I
(A^-1 A)B(AB)^-1=A^-1
IB(AB^-1)=A^-1
B(AB^-1)=A^-1
⇒B^-1 B(AB^-1)=B^-1 A^-1
(AB^-1)=B^-1 A^-1

4. A matrix A is invertible if it has all zeroes in one or more rows on L.H.S.
a) True
b) False
Answer: b
Clarification: The given statement is false. A matrix is non-invertible if it has all zeroes in one or more rows on L.H.S. This is because after applying all the elementary operations on the matrix, we should get an identity matrix on the L.H.S. to obtain an inverse of the given matrix, which is not possible if we obtain all zeroes in one or more rows.

5. The inverse of the matrix A=(begin{bmatrix}1&2&4\5&2&4\3&6&2end{bmatrix}) is
a) (begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\ frac{1}{40}&frac{-1}{8}&frac{1}{5}\ frac{3}{40}&1&frac{-1}{10}end{bmatrix})
b) (begin{bmatrix}frac{-1}{4}&frac{1}{4}&1\ frac{1}{40}&frac{-1}{8}&frac{1}{5}\ frac{3}{40}&0&frac{-1}{10}end{bmatrix})
c) (begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\ frac{1}{40}&frac{-1}{8}&frac{1}{5}\ frac{3}{40}&0&frac{-1}{10}end{bmatrix})
d) (begin{bmatrix}frac{-1}{4}&-frac{1}{4}&0\ frac{1}{40}&frac{1}{8}&frac{-1}{5}\ frac{3}{40}&0&frac{-1}{10}end{bmatrix})
Answer: c
Clarification: Consider the matrix A=(begin{bmatrix}1&2&4\5&2&4\3&6&2end{bmatrix})
Using the elementary row operation, we write A=IA
(begin{bmatrix}1&2&4\5&2&4\3&6&2end{bmatrix})=(begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})A
Applying R₁→R₁-R₂
(R_1 rightarrow frac{R_1}{-4})
(begin{bmatrix}1&0&0\5&2&4\3&6&2end{bmatrix})=(begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\0&1&0\0&0&1end{bmatrix})A
Applying R₂→R₂-5R₁ and R₃→R₃-3R₁
(begin{bmatrix}1&0&0\0&2&4\0&6&2end{bmatrix})=(begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\frac{5}{4}&frac{-1}{4}&0\frac{3}{4}&frac{-3}{4}&1end{bmatrix})A
Applying R₂→R₂-2R₃ and (R_2 rightarrow frac{R_2}{-10})
(begin{bmatrix}1&0&0\0&1&0\0&6&2end{bmatrix})=(begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\ frac{1}{40}&frac{-5}{40}&frac{1}{5}\frac{3}{4}&frac{-3}{4}&1end{bmatrix})A
Applying R₃→R₃-6R₂ and (R_2 rightarrow frac{R_2}{2})
(begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})=(begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\ frac{1}{40}&frac{-1}{8}&frac{1}{5}\ frac{3}{40}&0&frac{-1}{10}end{bmatrix})A
A^-1=(begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\ frac{1}{40}&frac{-1}{8}&frac{1}{5}\ frac{3}{40}&0&frac{-1}{10}end{bmatrix}).

6. Which of the following is the inverse of the matrix A=(begin{bmatrix}8&1\1&2end{bmatrix})?
a) (begin{bmatrix}frac{2}{15}&-frac{1}{15}\frac{1}{15}&frac{8}{15}end{bmatrix})
b) (begin{bmatrix}frac{1}{15}&-frac{1}{15}\-frac{1}{15}&frac{1}{15}end{bmatrix})
c) (begin{bmatrix}frac{2}{15}&-frac{1}{15}\-frac{1}{15}&frac{8}{15}end{bmatrix})
d) (begin{bmatrix}frac{2}{15}&frac{1}{15}\frac{1}{15}&frac{4}{15}end{bmatrix})
Answer: c
Clarification: Consider the matrix A=(begin{bmatrix}8&1\1&2end{bmatrix})
Using the elementary row operation, we write A=IA
Applying R₂→8R₂-R₁ and R₂→R₂/15, we get
(begin{bmatrix}8&1\0&1end{bmatrix})=(begin{bmatrix}1&0\-frac{1}{15}&frac{8}{15}end{bmatrix})A
Applying R₁→R₁-R₂ and R₁→R₁/8, we get
(begin{bmatrix}1&0\0&1end{bmatrix})=(begin{bmatrix}frac{2}{15}&-frac{1}{15}\-frac{1}{15}&frac{8}{15}end{bmatrix})A
A^-1=(begin{bmatrix}frac{2}{15}&-frac{1}{15}\-frac{1}{15}&frac{8}{15}end{bmatrix}).

7. Which among the below matrices has the inverse A^-1=(begin{bmatrix}1&-frac{5}{8}\0&frac{1}{8}end{bmatrix})
a) (begin{bmatrix}1&5\0&8end{bmatrix})
b) (begin{bmatrix}1&5\-1&8end{bmatrix})
c) (begin{bmatrix}1&5\0&16end{bmatrix})
d) (begin{bmatrix}1&8\0&8end{bmatrix})
Answer: a
Clarification: Consider the matrix A=(begin{bmatrix}1&5\0&8end{bmatrix})
Using the elementary column operations, we write A=AI
(begin{bmatrix}1&5\0&8end{bmatrix})=A(begin{bmatrix}1&0\0&1end{bmatrix})
Applying C₂→C₂-5C₁
(begin{bmatrix}1&0\0&8end{bmatrix})=A(begin{bmatrix}1&-5\0&1end{bmatrix})
Applying C₂→C₂/8
(begin{bmatrix}1&0\0&1end{bmatrix})=A(begin{bmatrix}1&-frac{5}{8}\0&frac{1}{8}end{bmatrix})
A^-1=(begin{bmatrix}1&-frac{5}{8}\0&frac{1}{8}end{bmatrix}).

8. Find the inverse of matrix A=(begin{bmatrix}5&1&3\4&2&6\5&4&2end{bmatrix})
a) (begin{bmatrix}0&-frac{1}{6}&0\-frac{11}{30}&frac{1}{12}&frac{3}{10}\-frac{1}{10}&frac{1}{4}&-frac{1}{10}end{bmatrix})
b) (begin{bmatrix}frac{1}{3}&-frac{1}{6}&0\-frac{11}{30}&frac{1}{12}&frac{3}{10}\-frac{1}{10}&frac{1}{4}&-frac{1}{10}end{bmatrix})
c) (begin{bmatrix}frac{1}{3}&-frac{1}{6}&0\-frac{11}{30}&1&frac{3}{10}\-frac{1}{10}&frac{1}{4}&-frac{1}{10}end{bmatrix})
d) (begin{bmatrix}frac{1}{3}&frac{1}{6}&0\frac{11}{30}&frac{1}{12}&frac{3}{10}\-frac{1}{10}&frac{1}{4}&frac{1}{10}end{bmatrix})
Answer: b
Clarification: Consider the matrix A=(begin{bmatrix}5&1&3\4&2&6\5&4&2end{bmatrix})
Using the elementary row operations, we write A=IA
(begin{bmatrix}5&1&3\4&2&6\5&4&2end{bmatrix})=(begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})A
Applying R₂→5R₂-4R₁ and R₃→R₃-R₁
(begin{bmatrix}5&1&3\0&6&18\0&3&-1end{bmatrix})=(begin{bmatrix}1&0&0\-4&5&0\-1&0&1end{bmatrix})A
Applying R₁→R₂-6R₁ and R₃→R₂-2R₃
(begin{bmatrix}-30&0&0\0&6&18\0&0&20end{bmatrix})=(begin{bmatrix}-10&5&0\-4&5&0\-2&5&-2end{bmatrix})A
Applying R₂→20R₂-18R₃
(begin{bmatrix}-30&0&0\0&120&0\0&0&20end{bmatrix})=(begin{bmatrix}-10&5&0\-44&10&36\-2&5&-2end{bmatrix})A
Applying R₁→R₁/(-30), R₂→R₂/120, R₃→R₃/20
(begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})=(begin{bmatrix}frac{1}{3}&-frac{1}{6}&0\-frac{11}{30}&frac{1}{12}&frac{3}{10}\-frac{1}{10}&frac{1}{4}&-frac{1}{10}end{bmatrix})A
A^-1=(begin{bmatrix}frac{1}{3}&-frac{1}{6}&0\-frac{11}{30}&frac{1}{12}&frac{3}{10}\-frac{1}{10}&frac{1}{4}&-frac{1}{10}end{bmatrix}).

9. Which of the following is not a property of invertible matrices if A and B are matrices of the same order?
a) (AB)^-1=A^-1 B^-1
b) (AA^-1)=(A^-1 A)=I
c) (AB)^-1=B^-1 A^-1
d) AB=BA=I
Answer: a
Clarification: (AB)^-1=A^-1 B^-1 is incorrect. The correct formula is (AB)^-1=B^-1 A^-1. B^-1 A^-1 ≠ A^-1 B^-1 as matrix multiplication is not commutative.

10. Find the inverse of A=(begin{bmatrix}5&3\4&1end{bmatrix}).
a) (begin{bmatrix}-frac{1}{7}&frac{3}{7}\frac{4}{7}&-frac{5}{7}end{bmatrix})
b) (begin{bmatrix}-frac{1}{7}&frac{3}{7}\frac{4}{7}&frac{5}{7}end{bmatrix})
c) (begin{bmatrix}-frac{1}{7}&-frac{3}{7}\frac{4}{7}&-frac{5}{7}end{bmatrix})
d) (begin{bmatrix}0&frac{3}{7}\frac{4}{7}&frac{5}{7}end{bmatrix})
Answer: a
Clarification: Consider the matrix A=(begin{bmatrix}5&3\4&1end{bmatrix})
By using the elementary row operations, we write A=IA
(begin{bmatrix}5&3\4&1end{bmatrix})=(begin{bmatrix}1&0\0&1end{bmatrix})A
Applying R₁→R₁-3R₂ and R₁→R₁/(-7), we get
(begin{bmatrix}1&0\4&1end{bmatrix})=(begin{bmatrix}-frac{1}{7}&frac{3}{7}\0&1end{bmatrix})
Applying R₂→R₂-4R₁, we get
(begin{bmatrix}1&0\0&1end{bmatrix})=(begin{bmatrix}-frac{1}{7}&frac{3}{7}\frac{4}{7}&-frac{5}{7}end{bmatrix})A
⇒A^-1=(begin{bmatrix}-frac{1}{7}&frac{3}{7}\frac{4}{7}&-frac{5}{7}end{bmatrix}).

Mathematics Multiple Choice Questions on “Second Order Derivatives”.

1. Find the second order derivative of y=9 log⁡ t³.
a) (frac{27}{t^2})
b) –(frac{27}{t^2})
c) –(frac{1}{t^2})
d) –(frac{27}{2t^2})
Answer: b
Clarification: Given that, y=9 log⁡t³
(frac{dy}{dx}=9.frac{1}{t^3}.3t^2=frac{27}{t})
(frac{d^2 y}{dx^2}=27(-frac{1}{t^2})=-frac{27}{t^2}).

2. Find (frac{d^2y}{dx^2}), if y=tan²⁡x+3 tan⁡x.
a) sec²⁡⁡x tan⁡x (2 tan⁡x+sec⁡x+3)
b) 2 sec²⁡⁡x tan⁡x (2 tan⁡x-sec⁡x+3)
c) 2 sec²⁡⁡x tan⁡x (2 tan⁡x+sec⁡x+3)
d) 2 sec²⁡⁡x tan⁡x (2 tan⁡x+sec⁡x-3)
Answer: c
Clarification: Given that, y=tan²⁡⁡x+3 tan⁡x
(frac{dy}{dx})=2 tan⁡x sec²⁡⁡x+3 sec²⁡x=sec²⁡⁡x (2 tan⁡x+3)
By using the u.v rule, we get
(frac{d^2 y}{dx^2}=frac{d}{dx}) (sec²⁡⁡x).(2 tan⁡x+3)+(frac{d}{dx}) (2 tan⁡x+3).sec²⁡⁡x
(frac{d^2 y}{dx^2})=2 sec²⁡⁡x tan⁡x (2 tan⁡x+3)+sec²⁡⁡x (2 sec⁡x tanx)
=2 sec²⁡x tan⁡x (2 tan⁡x+sec⁡x+3).

3. If y=6x²+3, then (left (frac{dy}{dx}right )^2=frac{d^2 y}{dx^2}).
a) True
b) False
Answer: b
Clarification: The given statement is false. Given that, y=6x²+3
(frac{dy}{dx})=12x
⇒(left (frac{dy}{dx}right )^2=(12x)^2=144x^2)
(frac{d^2 y}{dx^2}=frac{d}{dx}) (12x)=12
∴(left (frac{dy}{dx}right )^2≠frac{d^2 y}{dx^2})

4. Find the second order derivative of y=2e^2x-3 log⁡(2x-3).
a) 8e^2x+(frac{1}{(2x-3)^2})
b) 8e^2x–(frac{12}{(2x-3)^2})
c) e^2x+(frac{12}{(2x-3)^2})
d) 8e^2x+(frac{12}{(2x-3)^2})
Answer: d
Clarification: Given that, y=2e^2x-3 log⁡(2x-3)
(frac{dy}{dx})=4e^2x-3.(frac{1}{(2x-3)}).2=4e^2x–(frac{6}{(2x-3)})
(frac{d^2 y}{dx^2}=frac{d}{dx} (frac{dy}{dx}))=8e^2x+(frac{12}{(2x-3)^2})

5. Find (frac{d^2 y}{dx^2}), if y=2 sin^-1⁡(cos⁡x).
a) 0
b) sin^-1((frac{1}{cos⁡x}))
c) 1
d) -1
Answer: a
Clarification: Given that, y=2 sin^-1⁡(cos⁡x)
(frac{dy}{dx}=2.frac{1}{sqrt{1-cos^2⁡x}}).-sin⁡x=-2 (∵(sqrt{1-cos^2⁡x})=sin⁡x)
(frac{d^2 y}{dx^2})=(frac{d}{dx} (frac{dy}{dx})=frac{d}{dx}) (-2)=0

6. If y=log⁡(2x³), find (frac{d^2 y}{dx^2}).
a) –(frac{2}{x^2})
b) (frac{3}{x^2})
c) (frac{2}{x^2})
d) –(frac{3}{x^2})
Answer: d
Clarification: Given that, y=log⁡(2x³)
(frac{dy}{dx}=frac{1}{(2x^3)}.6x^2=frac{3}{x})
(frac{d^2 y}{dx^2}=-frac{3}{x^2})

7. Find (frac{d^2 y}{dx^2})-6 (frac{dy}{dx}) if y=4x⁴+2x.
a) ((4x^2+8x-1))
b) (12(4x^2+8x-1))
c) –(12(4x^2+8x-1))
d) (12(4x^2-8x-1))
Answer: d
Clarification: Given that, (y=4x^4+2x)
(frac{dy}{dx})=16x³+2
(frac{d^2 y}{dx^2})=48x²
(frac{d^2 y}{dx^2})-6 (frac{dy}{dx}=48x^2-96x^3-12)
=12(4x²-8x-1)

8. Find the second order derivative y=e^2x+sin^-1⁡e^x .
a) e^2x+(frac{e^x}{(1-e^2x)^{3/2}})
b) 4e^2x+(frac{1}{(1-e^2x)^{3/2}})
c) 4e^2x–(frac{e^x}{(1-e^2x)^{3/2}})
d) 4e^2x+(frac{e^x}{(1-e^2x)^{3/2}})
Answer: d
Clarification: Given that, y=e^2x+sin^-1⁡e^x
(frac{dy}{dx})=2e^2x+(frac{1}{sqrt{1-e^{2x}}} e^x)
(frac{d^2 y}{dx^2} = 4e^2x+bigg(frac{frac{d}{dx} (e^x) sqrt{1-e^{2x}} – frac{d}{dx} (sqrt{1-e^{2x}}).e^x}{(sqrt{1-e^{2x}})^2}bigg))
(=4e^{2x}+frac{(e^x sqrt{1-e^{2x}})-e^x left(frac{1}{2sqrt{1-e^{2x}}}.-2e^{2x}right)}{1-e^{2x}})
(=4e^{2x}+frac{(e^x (1-e^{2x})+e^{3x})}{(1-e^{2x})^{frac{3}{2}}})
(=4e^{2x}+frac{e^x (1-e^{2x}+e^{2x})}{(1-e^{2x})^{frac{3}{2}}})
4e^2x+(frac{e^x}{(1-e^2x)^{3/2}}).

9. Find the second order derivative of y=3x² 1 + log⁡(4x)
a) 3+(frac{1}{x^2})
b) 3-(frac{1}{x^2})
c) 6-(frac{1}{x^2})
d) 6+(frac{1}{x^2})
Answer: c
Clarification: Given that, y=3x²+log⁡(4x)
(frac{dy}{dx}=6x+frac{1}{4x}.4=6x+frac{1}{x}=frac{6x^2+1}{x})
(frac{d^2 y}{dx^2}=frac{frac{d}{dx} (6x^2+1).(x)-frac{d}{dx} (x).(6x^2+1)}{x^2} Big(using, frac{d}{dx} (frac{u}{v})=frac{(frac{d}{dx} (u).v-frac{d}{dx} (v).u)}{v^2}Big))
(frac{d^2 y}{dx^2}=frac{(12x.x-6x^2-1)}{x^2} )
(frac{d^2 y}{dx^2}=frac{6x^2-1}{x^2} = 6-frac{1}{x^2}).

10. Find the second order derivative if y=e^2x2.
a) 4e^2x2 (4x²+3)
b) 4e^2x2 (4x²-1)
c) 4e^2x2 (4x²+1)
d) e^2x2 (4x²+1)
Answer: c
Clarification: Given that, y=e^2x2
(frac{dy}{dx})=e^2x2.4x
By using u.v rule, we get
(frac{d^2 y}{dx^2}=frac{d}{dx} (e^{{2x}^2}).4x+frac{d}{dx} (4x).e^{{2x}^2})
16x² e^2x2+4e^2x2=4e^2x2 (4x²+1)

Mathematics Multiple Choice Questions on “Fundamental Theorem of Calculus-1”.

1. Find (int_0^8x ,dx).
a) 32
b) 34
c) 21
d) 24
Answer: a
Clarification: Let I=(int_0^8x ,dx)
F(x)=(int x ,dx=frac{x^2}{2})
Using the second fundamental theorem of calculus, we get
I=F(8)-f(0)
∴(int_0^8x ,dx=frac{8^2}{2}-0=32)

2. Find (int_0^{frac{π}{2}} ,5 ,sin⁡x ,dx).
a) -5
b) 9
c) 5
d) -9
Answer: c
Clarification: Let (I=int_0^{frac{π}{2}} ,5 ,sin⁡x ,dx)
F(x)=(int5 ,sin⁡x ,dx=-5 ,cos⁡x)
Applying the limits by using the fundamental theorem of calculus, we get
I=F((frac{π}{2}))-F(0)
∴(int_0^{frac{π}{2}} ,5 ,sin⁡x ,dx=-5[cos⁡frac{π}{2}-cos⁡0])
=-5[0-1]=5

3. Find the value of (int_4^5 ,log⁡x ,dx).
a) 5 log⁡5-log⁡4+1
b) 5 log⁡5-4 log⁡4-1
c) 4 log⁡5-4 log⁡4-1
d) 5-4 log⁡4-log⁡5
Answer: b
Clarification: Let I=(int_4^5 ,log⁡x ,dx).
F(x)=∫ log⁡x dx
By using the formula (int ,u.v dx=u int v ,dx-int u'(int ,v ,dx)), we get
(int log ⁡x ,dx=log⁡x int ,dx-int(log⁡x)’int ,dx)
F(x)=x log⁡x-∫ dx=x(log⁡x-1).
Applying the limits using the fundamental theorem of calculus, we get
I=F(5)-F(4)=(5 log⁡5-5)-(4 log⁡4-4)
=5 log⁡5-4 log⁡4-1.

4. Find (int_0^{frac{π}{4}} ,9 ,cos^2⁡x ,dx).
a) (frac{9}{2}left (frac{π}{6}-1right))
b) (frac{9}{4}left (frac{π}{2}+1right))
c) (frac{9}{4}left (frac{π}{2}-1right))
d) (left (frac{π}{2}-1right))
Answer: c
Clarification: Let I=(int_0^{frac{π}{4}} ,9 ,cos^2⁡x ,dx).
F(x)=(int ,9 ,cos^2⁡x ,dx)
=9(int(frac{1+cos⁡2x}{2})dx)
=(frac{9}{2} (x-frac{sin⁡2x}{2}))
Applying the limits, we get
I=(F(frac{π}{4})-F(0)=frac{9}{2} left (frac{π}{4}-frac{sin⁡2(frac{π}{4})}{2}right)-frac{9}{2} (0-frac{sin⁡0}{2}))
=(frac{9}{2}left (frac{π}{4}-frac{sin⁡π/2}{2}right )=frac{9}{4} (π/2-1))

5. Find (int_0^2 ,e^{2x} ,dx).
a) (frac{e^4-1}{6})
b) (frac{e^4+1}{2})
c) (frac{e-1}{2})
d) (frac{e^4-1}{2})
Answer: d
Clarification: Let (I=int_0^2 ,e^2x ,dx)
F(x)=(int e^{2x} dx)
=(frac{e^{2x}}{2})
Applying the limits, we get
I=F(2)-F(0)
=(frac{e^2(2)}{2}-frac{e^2(0)}{2}=frac{(e^4-1)}{2}).

6. Find (int_{π/4}^{π/2} ,2sinx ,sin⁡(cos⁡x) ,dx).
a) 2(1-cos⁡(frac{1}{sqrt{2}}))
b) (cos⁡(frac{1}{sqrt{2}})-cos⁡1)
c) 2(cos⁡(frac{1}{sqrt{2}})+1)
d) (cos⁡(frac{1}{sqrt{2}})+cos⁡1)
Answer: a
Clarification: Let (I=int_{π/4}^{π/2} ,2sinx ,sin⁡(cos⁡x) ,dx)
F(x)=(int 2 ,sin⁡x ,sin⁡(cos⁡x)dx)
Let cos⁡x=t
Differentiating w.r.t x, we get
sin⁡x dx=dt
∴(int 2 ,sin⁡x ,sin⁡(cos⁡x)dx=int 2 ,sin⁡t ,dt=-2 ,cos⁡t)
Replacing t with cos⁡x, we get
∴∫ 2 sin⁡x sin⁡(cos⁡x)dx=-2 cos⁡(cos⁡x)
By applying the limits, we get
(I=F(frac{π}{4})-F(frac{π}{2})=-2 cos⁡(frac{cos⁡π}{4})+2 cos⁡(frac{cos⁡π}{2}))
=2(1-cos⁡(frac{1}{sqrt{2}}))

7. Find (int_{-2}^1 ,5x^4 ,dx).
a) 54
b) 75
c) 33
d) 36
Answer: c
Clarification: (I=int_{-2}^1 ,5x^4 ,dx)
F(x)=(int5x^4 ,dx=5(frac{x^5}{5})=x^5)
Applying the limits by using the fundamental theorem of calculus, we get
I=F(1)-F(-2)
=(1)⁵-(-2)⁵=1+32=33.

8. Find (int_0^3 ,e^x ,dx).
a) e³+1
b) -e³-1
c) e³-1
d) 3e³-2
Answer: c
Clarification: Let I=(int_0^3 ,e^x ,dx)
F(x)=(int ,e^x ,dx=e^x)
Applying the limits, we get
I=F(3)-F(0)
=e³-e⁰=e³-1.

9. Find (int_0^{π/4} ,2 ,tan⁡x ,dx).
a) log⁡2
b) log⁡(sqrt{2})
c) 2 log⁡2
d) 0
Answer: a
Clarification: (I=int_0^{π/4} ,2 ,tan⁡x ,dx)
F(x)=∫ 2 tan⁡x dx
=2∫ tan⁡x dx
=2 log⁡|sec⁡x|
Therefore, by using the fundamental theorem of calculus, we get
I=F(π/4)-F(0)
(=2left(log⁡|sec frac{⁡π}{4}|-log⁡|sec⁡0|right)=2 log⁡sqrt{2}-log⁡1)
(=2 log⁡sqrt{2}=log⁡(sqrt{2})^2=log⁡2)
I=(frac{8}{3} log⁡2-frac{8}{3}-0+frac{1}{3}=frac{8}{3} log⁡2-frac{7}{3}).

10. Find (int_{-1}^1 ,2xe^x ,dx).
a) (frac{4}{e})
b) 4e
c) –(frac{4}{e})
d) -4e
Answer: a
Clarification: (I=int_{-1}^1 ,2xe^x ,dx)
F(x)=(int 2xe^x dx)
By using the formula, (int u.v ,dx=u int v ,dx-int u'(int v ,dx))
F(x)=2x(int e^x dx-int(2x)’int e^x ,dx)
=(2xe^x-int 2e^x dx)
=(2e^x (x-1))
Therefore, by using the fundamental theorem of calculus, we get
I=F(1)-F(-1)
I=2e¹ (1-1)-2e^-1 (-1-1)
I=(frac{4}{e}).

Mathematics Question Papers for Class 12 on “Product of Two Vectors-2”.

1. Evaluate the product ((2vec{a}+5vec{b}).(4vec{a}-5vec{b})).
a) (|vec{a}|^2+2vec{a}.vec{b}-15|vec{b}|^2)
b) (8|vec{a}|^2+2vec{a}.vec{b}-15|vec{b}|^2)
c) (8|vec{a}|^2-4vec{a}.vec{b}-15|vec{b}|^2)
d) (|vec{a}|^2+vec{a}.vec{b}-5|vec{b}|^2)
Answer: b
Clarification: To evaluate: ((2vec{a}+5vec{b}).(4vec{a}-5vec{b}))
=(2vec{a}.4vec{a}-2vec{a}.5vec{b}+3vec{b}.4vec{a}-3vec{b}.5vec{b})
=(8|vec{a}|^2+2vec{a}.vec{b}-15|vec{b}|^2)

2. Find the magnitude of (vec{a}) and (vec{b}) which are having the same magnitude and such that the angle between them is 60° and their scalar product is (frac{1}{4}).
a) (|vec{a}|=|vec{b}|=frac{1}{2√2})
b) (|vec{a}|=|vec{b}|=frac{1}{√2})
c) (|vec{a}|=|vec{b}|=frac{1}{2√3})
d) (|vec{a}|=|vec{b}|=frac{2}{√3})
Answer: a
Clarification: Given that: a) (|vec{a}|=|vec{b}|)
b) θ=60°
c) (vec{a}.vec{b}=frac{1}{4})
∴(|vec{a}||vec{b}| cos⁡θ=frac{1}{4})
=(|vec{a}|^2 cos⁡60°=frac{1}{4})
⇒(|vec{a}|^2=frac{1}{4}.frac{1}{2})
∴(|vec{a}|=|vec{b}|=frac{1}{2√2}).

3. If (vec{a}=hat{i}-hat{j}+3hat{k}, ,vec{b}=5hat{i}-2hat{j}+hat{k} ,and ,vec{c}=hat{i}-hat{j}) are such that (vec{a}+μvec{b}) is perpendicular to (vec{c}), then the value of μ.
a) (frac{7}{2})
b) –(frac{7}{2})
c) –(frac{3}{2})
d) (frac{7}{9})
Answer: b
Clarification: Given that: (vec{a}=hat{i}-hat{j}+3hat{k}, ,vec{b}=5hat{i}-2hat{j}+hat{k} ,and ,vec{c}=hat{i}-hat{j})
Also given, (vec{a}+μvec{b}) is perpendicular to (vec{c})
Therefore, ((vec{a}+μvec{b}).vec{c}=0)
i.e. ((hat{i}-hat{j}+3hat{k}+μ(5hat{i}-2hat{j}+hat{k})).(hat{i}-hat{j}))=0
(((1+5μ) ,hat{i}-(1+2μ) ,hat{j}+(μ+3) ,hat{k}).(hat{i}-hat{j}))=0
1+5μ+1+2μ=0
μ=-(frac{7}{2}).

4. Find the angle between (vec{a} ,and ,vec{b}) if (|vec{a}|=2,|vec{b}|=frac{1}{2√3}) and (vec{a}×vec{b}=frac{1}{2}).
a) (frac{2π}{3})
b) (frac{4π}{5})
c) (frac{π}{3})
d) (frac{π}{2})
Answer: c
Clarification: Given that, (|vec{a}|=2, ,|vec{b}|=frac{1}{2√3}) and (vec{a}×vec{b}=frac{1}{2})
We know that, (vec{a}×vec{b}=vec{a}.vec{b} ,sin⁡θ)
∴ (sin⁡θ=frac{(vec{a}×vec{b})}{|vec{a}|.|vec{b}|})
sin⁡θ=(frac{frac{1}{2}}{2×frac{1}{2√3}}=frac{sqrt{3}}{2})
θ=(sin^{-1}⁡frac{sqrt{3}}{2}=frac{π}{3})

5. Find the vector product of the vectors (vec{a}=2hat{i}+4hat{j}) and (vec{b}=3hat{i}-hat{j}+2hat{k}).
a) (hat{i}-19hat{j}-4hat{k})
b) (3hat{i}+19hat{j}-14hat{k})
c) (3hat{i}-19hat{j}-14hat{k})
d) (3hat{i}+5hat{j}+4hat{k})
Answer: c
Clarification: Given that, (vec{a}=2hat{i}+4hat{j}) and (vec{b}=3hat{i}-hat{j}+2hat{k})
Calculating the vector product, we get
(vec{a}×vec{b}=begin{vmatrix}hat{i}&hat{j}&hat{k}\2&4&-5\3&-1&2end{vmatrix})
=(hat{i}(8-5)-hat{j}(4-(-15))+hat{k}(-2-12))
=(3hat{i}-19hat{j}-14hat{k})

6. If (vec{a} ,and ,vec{b}) are two non-zero vectors then ((vec{a}-vec{b})×(vec{a}+vec{b}))=_________
a) (2(vec{a}×vec{b}))
b) ((vec{a}×vec{b}))
c) –(4(vec{a}×vec{b}))
d) (3(vec{a}×vec{b}))
Answer: a
Clarification: Consider ((vec{a}-vec{b})×(vec{a}+vec{b}))
=((vec{a}-vec{b})×vec{a}+(vec{a}+vec{b})×vec{b})
=(vec{a}×vec{a}-vec{b}×vec{a}+vec{a}×vec{b}-vec{b}×vec{b})
We know that, (vec{a}×vec{a}=0,vec{b}×vec{b}=0 ,and ,vec{a}×vec{b}=-vec{b}×vec{a})
∴ (vec{a}×vec{a}-vec{b}×vec{a}+vec{a}×vec{b}-vec{b}×vec{b}=0+2(vec{a}×vec{b})+0)
Hence, ((vec{a}-vec{b})×(vec{a}+vec{b})=2(vec{a}×vec{b}))

7. Find the product ((vec{a}+vec{b}).(7vec{a}-6vec{b})).
a) (2|vec{a}|^2+6vec{a}.vec{b}-3|vec{b}|^2)
b) (8|vec{a}|^2+5vec{a}.vec{b}-5|vec{b}|^2)
c) (2|vec{a}|^2+6vec{a}.vec{b}-8|vec{b}|^2)
d) (7|vec{a}|^2+vec{a}.vec{b}-6|vec{b}|^2)
Answer: d
Clarification: To evaluate: ((vec{a}+vec{b}).(7vec{a}-6vec{b}))
=(vec{a}.7vec{a}-vec{a}.6vec{b}+vec{b}.7vec{a}-6vec{b}.vec{b})
=(7|vec{a}|^2+vec{a}.vec{b}-6|vec{b}|^2)

8. Find the vector product of the vectors (vec{a}=-hat{j}+hat{k}) and (vec{b}=-hat{i}-hat{j}-hat{k}).
a) (2hat{i}-hat{j}+hat{k})
b) (2hat{i}-hat{j}-4hat{k})
c) (hat{i}+hat{j}-hat{k})
d) (2hat{i}-hat{j}-hat{k})
Answer: d
Clarification: Given that, (vec{a}=-hat{j}+hat{k}) and (vec{b}=-hat{i}-hat{j}-hat{k})
Calculating the vector product, we get
(vec{a}×vec{b}=begin{vmatrix}hat{i}&hat{j}&hat{k}\0&-1&1\-1&-1&-1end{vmatrix})
=(hat{i}(1-(-1))-hat{j}(0-(-1))+hat{k}(0-1))
=(2hat{i}-hat{j}-hat{k})

9. If (vec{a}=2hat{i}+3hat{j}+4hat{k}) and (vec{b}=4hat{i}-2hat{j}+3hat{k}). Find (|vec{a}×vec{b}|).
a) (sqrt{685})
b) (sqrt{645})
c) (sqrt{679})
d) (sqrt{689})
Answer: b
Clarification: Given that, (vec{a}=2hat{i}+3hat{j}+4hat{k}) and (vec{b}=4hat{i}-2hat{j}+3hat{k})
∴ (vec{a}×vec{b}=begin{vmatrix}hat{i}&hat{j}&hat{k}\2&3&4\4&-2&3end{vmatrix})
=(hat{i}(9—8)-hat{j}(6-16)+hat{k}(-4-12))
=(17hat{i}+10hat{j}-16hat{k})
∴(|vec{a}×vec{b}|=sqrt{17^2+10^2+(-16)^2})
=(sqrt{289+100+256})
=(sqrt{645})

10. Find the angle between the vectors if (|vec{a}|=|vec{b}|=3sqrt{2}) and (vec{a}.vec{b}=9sqrt{3}).
a) (frac{π}{6})
b) (frac{π}{5})
c) (frac{π}{3})
d) (frac{π}{2})
Answer: a
Clarification: We know that, (vec{a}.vec{b}=|vec{a}|.|vec{b}| ,cos⁡θ)
Given that, (|vec{a}|=|vec{b}|=3sqrt{2} ,and ,vec{a}.vec{b}=9sqrt{3})
(cos⁡θ=frac{vec{a}.vec{b}}{|vec{a}|.|vec{b}|}=frac{9sqrt{3}}{(3sqrt{2})^2}=frac{sqrt{3}}{2})
(θ=cos^{-1}frac{sqrt{3}}{2}=frac{π}{6}).

Mathematics Question Papers for Class 12,

Mathematics Multiple Choice Questions on “Calculus Application – Tangents and Normals – 1”.

1. What is the equation of the tangent at a specific point of y² = 4ax at (0, 0)?
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: a
Clarification: Equation of the given parabola is y² = 4ax ……….(1)
Differentiating both side of (1) with respect to x we get,
2y(dy/dx) = 4a
Or dy/dx = 2a/y
Clearly dy/dx does not exist at (0, 0). Hence, the tangent to the parabola (1) at (0, 0) is parallel to y axis.
Again, the tangent passes through (0, 0). Therefore, the required tangent to the parabola (1) at (0, 0) is the y-axis and hence the required equation of the tangent is x = 0.

2. If X and Y are given as current co-ordinates, what is the equation of the tangent at a specific point of x³ – 3axy + y³ = 0 at (x, y)?
a) (x² – ay)X + (y² – ax)Y = -2axy
b) (x² – ay)X + (y² – ax)Y = 2axy
c) (x² – ay)X + (y² – ax)Y = axy
d) (x² – ay)X + (y² – ax)Y = -axy
Answer: c
Clarification: Equation of the given curve is, x³ – 3axy + y³ = 0 ……….(1)
Differentiating both sides with respect to x we get,
3x² – 3a(x(dy/dx) + y) + 3y²(dy/dx) = 0
Or dy/dx = (ay – x²)/(y² – ax)
So, it is clear that this can be written as,
Y – y = (dy/dx)(X – x)
Or Y – y = [(ay – x²)/(y² – ax)](X – x)
Simplifying the above equation by cross multiplication, we get,
(x² – ay)X + (y² – ax)Y = x³ – 3axy + y³ + axy
Using (1),
(x² – ay)X + (y² – ax)Y = axy

3. What will be the equation of the normal to the hyperbola xy = 4 at the point (2, 2)?
a) x + y = 0
b) x – y = 0
c) 2x – y = 0
d) x + 2y = 0
Answer: b
Clarification: Equation of the given hyperbola is, xy = 4 ……….(1)
Differentiating both side of (1) with respect to y, we get,
y*(dx/dy) + x(1) = 0
Or dx/dy = -(x/y)
Thus, the required equation of the normal to the hyperbola at (2, 2) is,
y – 2 = -[dx/dy]_{(2, 2)} (x – 2) = -(-2/2)(x – 2)
So, from here,
y – 2 = x – 2
Or x – y = 0

4. At which point does the normal to the hyperbola xy = 4 at (2, 2) intersects the hyperbola again?
a) (-2, -2)
b) (-2, 2)
c) (2, -2)
d) (0, 2)
Answer: a
Clarification: Equation of the given hyperbola is, xy = 4 ……….(1)
Differentiating both side of (1) with respect to y, we get,
y*(dx/dy) + x(1) = 0
Or dx/dy = -(x/y)
Thus, the required equation of the normal to the hyperbola at (2, 2) is,
y – 2 = -[dx/dy]_{(2, 2)} (x – 2) = -(-2/2)(x – 2)
So, from here,
y – 2 = x – 2
Or x – y = 0 ……….(2)
Solving the equation (1) and (2) we get,
x = 2 and y = 2 or x = -2 and y = -2
Thus, the line (2) intersects the hyperbola (1) at (2, 2) and (-2, -2).
Hence, the evident is that the normal at (2, 2) to the hyperbola (1) again intersects it at (-2, -2).

5. What will be the equation of the tangent to the circle x² + y² – 6x + 4y – 7 = 0, which are perpendicular to the straight line 2x – y + 3 = 0?
a) x + 2y – 9 = 0
b) x + 2y + 9 = 0
c) x + 2y – 10 = 0
d) x + 2y + 10 = 0
Answer: a
Clarification: The equation of any straight line perpendicular to the line 2x – y + 3 = 0 is,
x + 2y + k = 0 ……….(1)
Now, the co-ordinate of the center of the circle (3, -2) and its radius is,
√(9 + 4 – (-7) = 2√5
If straight line (1) be tangent to the given circle then, the perpendicular distance of the point (3, -2) from the line (1) = radius of the circle
Thus, ±(3 + 2(-2) + k)/√(1 + 4)
Or k – 1 = 2√5 * √5
So, k = 1 ± 10
= 11 or -9
Putting the value of k in (1) we get,
x + 2y + 11 = 0 and x + 2y – 9 = 0

6. What is the equation of the tangent to the parabola y² = 8x, which is inclined at an angle of 45° with the x axis?
a) x + y – 2 = 0
b) x + y + 2 = 0
c) x – y + 2 = 0
d) x – y – 2 = 0
Answer: c
Clarification: Equation of the given parabola is, y² = 8x ……….(1)
Differentiating both sides with respect to x,
2y(dy/dx) = 8
Or dy/dx = 4/y
Thus, equation of the tangent to the parabola (1) at (x₁, y₁) = (2t², 4t) is,
y – y₁ = [dy/dx]_{(x1, y1)} (x – 2t²)
y – 4t = [dy/dx]_{(2(t²), 4t)} (x – 2t²)
Putting the value of y = 4t in the equation dy/dx = 4/y, we get,
y – 4t = 4/4t(x – 2t²) ……….(2)
If the tangent to the parabola y² = 8x, which is inclined at an angle of 45° with the x axis,
Then, slope of tangent (2) = tan 45° = 1
Thus, 4/4t = 1
Or t = 1
Thus, required equation of the tangent is,
y– 4 = 1(x – 2)
Putting, t = 1 in (2),
So, x – y + 2 = 0

7. What will be the equation of the tangent to the circle x² + y² – 6x + 4y – 7 = 0, which are perpendicular to the straight line 2x – y + 3 = 0?
a) x – 2y + 11 = 0
b) x – 2y – 11 = 0
c) x + 2y + 11 = 0
d) x + 2y – 11 = 0
Answer: c
Clarification: The equation of any straight line perpendicular to the line 2x – y + 3 = 0 is,
x + 2y + k = 0 ……….(1)
Now, the co-ordinate of the center of the circle (3, -2) and its radius is,
√(9 + 4 – (-7) = 2√5
If straight line (1) be tangent to the given circle then, the perpendicular distance of the point (3, -2) from the line (1) = radius of the circle
Thus, ±(3 + 2(-2) + k)/√(1 + 4)
Or k – 1 = 2√5 * √5
So, k = 1 ± 10
= 11 or -9
Putting the value of k in (1) we get,
x + 2y + 11 = 0 and x + 2y – 9 = 0

8. What will be the equation of normal to the hyperbola 3x² – 4y² = 12 at the point (x₁, y₁)?
a) 3x₁y + 4y₁x + 7x₁y₁ = 0
b) 3x₁y + 4y₁x – 7x₁y₁ = 0
c) 3x₁y – 4y₁x – 7x₁y₁ = 0
d) 3x₁y – 4y₁x + 7x₁y₁ = 0
Answer: b
Clarification: Equation of the given hyperbola is, 3x² – 4y² = 12 ……….(1)
Differentiating both sides of (1) with respect to y we get,
3*2x(dy/dx) – 4*(2y) = 0
Or dx/dy = 4y/3x
Therefore, the equation of the normal to the hyperbola (1) at the point (x₁, y₁) on it is,
y – y₁ = -[dx/dy]_{(x1, y1)} (x – x₁) = -4y₁/3x₁(x – x₁)
Or 3x₁y + 4y₁x – 7x₁y₁ = 0

9. What is the nature of the straight line x + y + 7 = 0 to the hyperbola 3x² – 4y² = 12 whose normal is at the point (x₁, y₁)?
a) Chord to hyperbola
b) Tangent to hyperbola
c) Normal to hyperbola
d) Segment to hyperbola
Answer: c
Clarification: Equation of the given hyperbola is, 3x² – 4y² = 12 ……….(1)
Differentiating both sides of (1) with respect to y we get,
3*2x(dy/dx) – 4*(2y) = 0
Or dx/dy = 4y/3x
Therefore, the equation of the normal to the hyperbola (1) at the point (x₁, y₁) on it is,
y – y₁ = -[dx/dy]_{(x1, y1)} (x – x₁) = -4y₁/3x₁(x – x₁)
Or 3x₁y + 4y₁x – 7x₁y₁ = 0
Now, if possible, let us assume that the straight line
x + y + 7 = 0 ………..(2)
This line is normal to the hyperbola (1) at the point (x₁, y₁). Then, the equation (2) and (3) must be identical. Hence, we have,
3x₁/1 = 4y₁/1 = -7x₁y₁/7
So, x₁ = -4 and y₁ = -3
Now, 3x₁² – 4y₁² = 3(-4)² – 4(-3)² = 12
This shows the point (-4, -3) lies on the hyperbola (1).
Thus, it is evident that the straight line (3) is normal to the hyperbola (1).

10. What is the foot of the normal if the straight line x + y + 7 = 0 is normal to the hyperbola 3x² – 4y² = 12 whose normal is at the point (x₁, y₁)?
a) (4, 3)
b) (-4, 3)
c) (4, -3)
d) (-4, -3)
Answer: d
Clarification: Equation of the given hyperbola is, 3x² – 4y² = 12 ……….(1)
Differentiating both sides of (1) with respect to y we get,
3*2x(dy/dx) – 4*(2y) = 0
Or dx/dy = 4y/3x
Therefore, the equation of the normal to the hyperbola (1) at the point (x₁, y₁) on it is,
y – y₁ = -[dx/dy]_{(x1, y1)} (x – x₁) = -4y₁/3x₁(x – x₁)
Or 3x₁y + 4y₁x – 7x₁y₁ = 0
Now, if possible, let us assume that the straight line
x + y + 7 = 0 ………..(2)
This line is normal to the hyperbola (1) at the point (x₁, y₁). Then, the equation (2) and (3) must be identical. Hence, we have,
3x₁/1 = 4y₁/1 = -7x₁y₁/7
So, x₁ = -4 and y₁ = -3
Now, 3x₁² – 4y₁² = 3(-4)² – 4(-3)² = 12
This shows the point (-4, -3) lies on the hyperbola (1).
So, it’s the normal to the hyperbola.
Thus, it is evident that the straight line (3) is normal to the hyperbola (1); the co-ordinate foot is (-4, -3).

Mathematics Multiple Choice Questions on “Determinant – 1”.

1. Evaluate (begin{vmatrix}2&5\-1&-1end{vmatrix}).
a) 3
b) -7
c) 5
d) -2
Answer: a
Clarification: Expanding along R₁, we get
∆=2(-1)-5(-1)=-2+5
=3.

2. Evaluate (begin{vmatrix}5&-4\1&sqrt{3}end{vmatrix}).
a) 4(sqrt{3})+4
b) 4(sqrt{3})+5
c) 5(sqrt{3})+4
d) 5(sqrt{3})-4
Answer: c
Clarification: Evaluating along R₁, we get
∆=5((sqrt{3}))-(-4)1=5(sqrt{3})+4.

3. Evaluate (begin{vmatrix}-sinθ&-1\1&sin⁡θend{vmatrix}).
a) cos²⁡θ
b) -cos²⁡θ
c) cos⁡2θ
d) cos⁡θ
Answer: a
Clarification: Expanding along R₁, we get
∆=-sinθ(sinθ)-(-1)1=-sin²⁡θ+1=cos²⁡θ.

4. Evaluate (begin{vmatrix}i&-1\-1&-iend{vmatrix}).
a) 4
b) 3
c) 2
d) 0
Answer: d
Clarification: Expanding along R₁, we get
∆=-i(i)-(-1)(-1)=-i²-1=-(-1)-1=0.

5. Evaluate (begin{vmatrix}1&1&-2\3&4&5\-1&2&1end{vmatrix}).
a) -6
b) -34
c) 34
d) 22
Answer: b
Clarification: ∆=(begin{vmatrix}1&1&-2\3&4&5\-1&2&1end{vmatrix})
Expanding along the first row, we get
∆=1(begin{vmatrix}4&5\2&1end{vmatrix})-1(begin{vmatrix}3&5\-1&1end{vmatrix})-2(begin{vmatrix}3&4\-1&2end{vmatrix})
=1(4-5(2))-1(3-5(-1))-2(6-4(-1))
=(4-10)-(3+5)-2(6+4)
=-6-8-20=-34.

6. Evaluate (begin{vmatrix}5&4&3\3&4&1\5&6&1end{vmatrix}).
a) 4
b) -24
c) -8
d) 8
Answer: c
Clarification: Expanding along the first row, we get
∆=5(begin{vmatrix}4&1\6&1end{vmatrix})-4(begin{vmatrix}3&1\5&1end{vmatrix})+3(begin{vmatrix}3&4\5&6end{vmatrix})
=5(4-6)-4(3-5)+3(18-20)
=5(-2)-4(-2)+3(-2)=-10+8-6=-8.

7. Evaluate (begin{vmatrix}8x+1&2x-2\x^2-1&3x+5end{vmatrix}).
a) -2x³-26x²+45x+3
b) -2x³+26x²+45x+3
c) -2x³+26x²+45x-3
d) -2x³-26x²-45x+3
Answer: b
Clarification: Expanding along the first row, we get
∆=8x+1(3x+5)-(2x-2)(x²-1)
=(24x²+43x+5)-(2x³-2x²-2x+2)
=-2x³+26x²+45x+3.

8. If A=(begin{bmatrix}2&5&9\6&1&3\4&8&2end{bmatrix}), find |A|.
a) 352
b) 356
c) 325
d) 532
Answer: a
Clarification: Given that, A=(begin{bmatrix}2&5&9\6&1&3\4&8&2end{bmatrix})
⇒|A|=(begin{vmatrix}2&5&9\6&1&3\4&8&2end{vmatrix})
Evaluating along the first row, we get
∆=2(begin{vmatrix}1&3\8&2end{vmatrix})-5(begin{vmatrix}6&3\4&2end{vmatrix})+9(begin{vmatrix}6&1\4&8end{vmatrix})
∆=2(2-24)-5(12-12)+9(48-4)
∆=2(-22)-0+9(44)
∆=-44+9(44)=44(-1+9)=352

9. Evaluate (begin{vmatrix}sqrt{3}&sqrt{2}\-1&2sqrt{3}end{vmatrix}).
a) 6-3(sqrt{2})
b) 6-(sqrt{2})
c) 6+3(sqrt{2})
d) 6+(sqrt{2})
Answer: d
Clarification: ∆=(begin{vmatrix}sqrt{3}&sqrt{2}\-1&2sqrt{3}end{vmatrix})
∆=((sqrt{3})×2(sqrt{3}))+(sqrt{2})
∆=6+(sqrt{2}).

10. Find the value of x if (begin{vmatrix}3&x\2&x^2 end{vmatrix})=(begin{vmatrix}5&3\3&2end{vmatrix}).
a) x=1, –(frac{1}{3})
b) x=-1, –(frac{1}{3})
c) x=1, (frac{1}{3})
d) x=-1, (frac{1}{3})
Answer: a
Clarification: Given that (begin{vmatrix}3&x\2&x^2 end{vmatrix})=(begin{vmatrix}5&3\3&2end{vmatrix})
⇒3x²-2x=5(2)-3(3)
⇒3x²-2x=1
Solving for x, we get
x=1, –(frac{1}{3}).