Category: Class 12 Mathematics Objective Questions

Mathematics Multiple Choice Questions on “Properties of Determinants”.

1. Which of the following is not a property of determinant?
a) The value of determinant changes if all of its rows and columns are interchanged
b) The value of determinant changes if any two rows or columns are interchanged
c) The value of determinant is zero if any two rows and columns are identical
d) The value of determinant gets multiplied by k, if each element of row or column is multiplied by k
Answer: a
Clarification: The value of determinant remains unchanged if all of its rows and columns are interchanged i.e. |A|=|A’|, where A is a square matrix and A’ is the transpose of the matrix A.

2. Find the determinant of the matrix A=(begin{bmatrix}1&x&y\1&x&-y\1&-x^2&y^2end{bmatrix}).
a) (x+1)
b) -2xy(x+1)
c) xy(x+1)
d) 2xy(x+1)
Answer: b
Clarification: Given that, A=(begin{bmatrix}1&x&y\1&x&-y\1&-x^2&y^2end{bmatrix})
Δ=(begin{vmatrix}1&x&y\1&x&-y\1 &-x^2&y^2 end{vmatrix})
Taking x common C₂ and y common from C₃, we get
Δ=xy(begin{vmatrix}1&1&1\1&1&-1\1&-x&yend{vmatrix})
Expanding along R₁, we get
Δ=xy{1(y-x)-1(y+1)+1(-x-1)}
Δ=xy(y-x-y-1-x-1)
Δ=xy(-2x-2)=-2xy(x+1).

3. Evaluate (begin{vmatrix}x^2&x^3&x^4\x&y&z\x^2&x^3&x^4 end{vmatrix}).
a) 0
b) 1
c) xyz
d) x² yz³
Answer: a
Clarification: Δ=(begin{vmatrix}x^2&x^3&x^4\x&y&z\x^2&x^3&x^4 end{vmatrix})
If the elements of any two rows or columns are identical, then the value of determinant is zero. Here, the elements of row 1 and row 3 are identical. Hence, its determinant is 0.

4. Evaluate (begin{vmatrix}cos⁡θ&-cos⁡θ&1\sin^2⁡θ&cos^2⁡θ&1\sin⁡θ&-sin⁡θ&1end{vmatrix}).
a) sin⁡θ+cos²⁡θ
b) -sin⁡θ-cos²⁡⁡θ
c) -sin⁡θ+cos²⁡⁡θ
d) sin⁡θ-cos²⁡⁡θ
Answer: d
Clarification: Δ=(begin{vmatrix}cos⁡θ&-cos⁡θ&1\sin^2⁡θ&cos^2⁡θ&1\sin⁡θ&-sin⁡θ&1end{vmatrix})
Applying C₁→C₁+C₂
Δ=(begin{vmatrix}cos⁡θ-cos⁡θ&-cos⁡θ&1\sin^2⁡θ+cos^2⁡θ&cos^2⁡θ&1\sinθ-sin⁡θ&-sin⁡θ&1end{vmatrix})=(begin{vmatrix}0&-cos⁡θ&1\1&cos^2⁡θ&1\0&-sin⁡θ&1end{vmatrix})
Expanding along C₁, we get
0-1(cos²⁡⁡θ+sinθ)=sin⁡θ-cos²⁡⁡θ.

5. Evaluate (begin{vmatrix}b-c&b&c\a&c-a&c\a&b&a-bend{vmatrix}).
a) 2abc
b) 2a{(b-c)(c-a+b)}
c) 2b{(a-c)(a+b+c)}
d) 2c{(b-c)(a-c+b)}
Answer: b
Clarification: Δ=(begin{vmatrix}b-c&b&c\a&c-a&c\a&b&a-bend{vmatrix})
Applying C₂→C₂-C₃
Δ=(begin{vmatrix}b-c&b-c&c\a&-a&c\a&-a&a-bend{vmatrix})
Applying C₁→C₁-C₂
Δ=(begin{vmatrix}0&b-c&c\2a&-a&c\2a&-a&a-bend{vmatrix})
Applying R₂→R₂-R₃
Δ=(begin{vmatrix}0&b-c&c\0&0&c-a+b\2a&-a&a-bend{vmatrix})
Expanding along C₁, we get
Δ=2a{(b-c)(c-a+b)}

6. If A=(begin{bmatrix}1&3\2&1end{bmatrix}), then ________
a) |2A|=4|A|
b) |2A|=2|A|
c) |A|=2|A|
d) |A|=|4A|
Answer: a
Clarification: Given that, A=(begin{bmatrix}1&3\2&1end{bmatrix})
2A=2(begin{bmatrix}1&3\2&1end{bmatrix})=(begin{bmatrix}2&6\4&2end{bmatrix})
|2A|=(begin{vmatrix}2&6\4&2end{vmatrix})=(4-24)=-20
4|A|=4(begin{vmatrix}1&3\2&1end{vmatrix})=4(1-6)=4(-5)=-20
∴|2A|=4|A|.

7. Evaluate (begin{vmatrix}-a&b&c\-2a+4x&2b-4y&2c+4z\x&-y&zend{vmatrix}).
a) 0
b) abc
c) 2abc
d) -1
Answer: a
Clarification: Δ=(begin{vmatrix}-a&b&c\-2a+4x&2b-4y&2c+4z\x&-y&zend{vmatrix})
Using the properties of determinants, the given determinant can be expressed as a sum of two determinants.
Δ=(begin{vmatrix}-a&b&c\-2a&2b&2c\x&-y&zend{vmatrix})+(begin{vmatrix}-a&b&c\4x&-4y&4z\x&-y&zend{vmatrix})
Δ=2(begin{vmatrix}-a&b&c\-a&b&c\x&-y&zend{vmatrix})+4(begin{vmatrix}-a&b&c\x&-y&z\x&-y&zend{vmatrix})
Since two rows are similar in each of the determinants, the determinant is 0.

8. Find the determinant of A=(begin{bmatrix}c^2&cb&ca\ab&a^2&-ac\ab&bc&-b^2end{bmatrix})
a) abc(a³+b³+c³+abc)
b) abc(a³+b³+c³-abc)
c) abc(a³+b³+c³+abc)
d) (a³-b³+c³-abc)
Answer: b
Clarification: Given that, A=(begin{bmatrix}c^2&cb&ca\ab&a^2&-ac\ab&bc&-b^2end{bmatrix})
Taking c a, b common from R₁, R₂, R₃ respectively, we get
Δ=abc(begin{bmatrix}c&b&a\b&a&-c\a&c&-bend{bmatrix})
Δ=abc{(c(-ab+c²)-b(-b²+ac)+a(bc-a²)
Δ=abc(-abc+c³+b³-abc+abc-a³)
Δ=abc(a³+b³+c³-abc).

9. Evaluate (begin{vmatrix}1+m&n&q\m&1+n&q\n&m&1+qend{vmatrix}).
a) -1(1+m+n+q)
b) 1+m+n+q
c) 1+2q
d) 1+q
Answer: a
Clarification: Given that, Δ=(begin{vmatrix}1+m&n&q\m&1+n&q\n&m&1+qend{vmatrix})
Applying C₁→C₁+C₂+C₃
Δ=(begin{vmatrix}1+m+n+q&n&q\1+m+n+q&1+n&q\1+m+n+q&m&1+qend{vmatrix})=(1+m+n+q)(begin{vmatrix}1&n&q\1&1+n&q\1&m&1+qend{vmatrix})
Applying R₁→R₂-R₁
Δ=(1+m+n+q)(begin{vmatrix}0&1&0\1&1+n&q\1&m&1+qend{vmatrix})
Expanding along the first row, we get
Δ=(1+m+n+q)(0-1(1+q-q)+0)
Δ=-1(1+m+n+q).

10. Evaluate (begin{vmatrix}4&8&12\6&12&18\7&14&21end{vmatrix}).
a) 168
b) -1
c) -168
d) 0
Answer: d
Clarification: Δ=(begin{vmatrix}4&8&12\6&12&18\7&14&21end{vmatrix})
Taking 4, 6 and 7 from R₁, R₂, R₃ respectively
Δ=4×6×7(begin{vmatrix}1&2&3\1&2&3\1&2&3end{vmatrix})
Since the elements of all rows are identical, the determinant is zero.

Mathematics Interview Questions and Answers on “Derivatives Application – Rate of Change of Quantities”.

1. If the rate of change of radius of a circle is 6 cm/s then find the rate of change of area of the circle when r=2 cm.
a) 74.36 cm²/s
b) 75.36 cm²/s
c) 15.36 cm²/s
d) 65.36 cm²/s
Answer: b
Clarification: The rate of change of radius of the circle is (frac{dr}{dt})=6 cm/s
The area of a circle is A=πr²
Differentiating w.r.t t we get,
(frac{dA}{dt}=frac{d}{dt}) (πr²)=2πr (frac{dr}{dt})=2πr(6)=12πr.
(frac{dA}{dt})|_r=2=24π= 24×3.14=75.36 cm²/s

2. The edge of a cube is increasing at a rate of 7 cm/s. Find the rate of change of area of the cube when x=6 cm.
a) 578 cm²/s
b) 498 cm²/s
c) 504 cm²/s
d) 688 cm²/s
Answer: c
Clarification: Let the edge of the cube be x. The rate of change of edge of the cube is given by (frac{dx}{dt})=7cm/s.
The area of the cube is A=6x²
∴(frac{dA}{dt}=frac{d}{dt} )(6x²)=12x.(frac{dx}{dt})=12x×7=84x
(frac{dA}{dt})|__x=6=84×6=504 cm²/s.

3. The rate of change of area of a square is 40 cm²/s. What will be the rate of change of side if the side is 10 cm.
a) 2 cm/s
b) 4 cm/s
c) 8 cm/s
d) 6 cm/s
Answer: a
Clarification: Let the side of the square be x.
A=x², where A is the area of the square
Given that, (frac{dA}{dt})=2x (frac{dx}{dt})=40 cm²/s.
(frac{dx}{dt}=frac{20}{x} )cm/s
(frac{dx}{dt}=frac{20}{10})=2 cm/s.

4. The total cost P(x) in rupees associated with a product is given by P(x)=0.4x²+2x-10. Find the marginal cost if the no. of units produced is 5.
a) Rs.3
b) Rs.4
c) Rs.5
d) Rs.6
Answer: d
Clarification: The Marginal cost is the rate of change of revenue w.r.t the no. of units produced, we get
(frac{dP(x)}{dt})=0.8x+2
cost(MC)=(frac{dP(x)}{dt}|)_x=5=0.8x+2=0.8(5)+2=4+2=6.

5. At what rate will the lateral surface area of the cylinder increase if the radius is increasing at the rate of 2 cm/s when the radius is 5 cm and height is 10 cm?
a) 40 cm/s
b) 40π cm/s
c) 400π cm/s
d) 20π cm/s
Answer: b
Clarification: Let r be the radius and h be the height of the cylinder. Then,
(frac{dr}{dt})=2 cm/s
The area of the cylinder is given by A=2πrh
(frac{dA}{dt})=2πh((frac{dr}{dt}))=4πh=4π(10)=40π cm/s.

6. If the circumference of the circle is changing at the rate of 5 cm/s then what will be rate of change of area of the circle if the radius is 6cm.
a) 20 cm²/s
b) 40 cm²/s
c) 70 cm²/s
d) 30 cm²/s
Answer: d
Clarification: The circumference of the circle is given by C=2πr, where r is the radius of the circle.
∴(frac{dC}{dt})=2π.(frac{dr}{dt})=5 cm/s
(frac{dr}{dt})=5/2π cm/s
(frac{dA}{dt})|_r=6=2πr.(frac{dr}{dt})=2πr.(frac{5}{2 pi})=5r=5(6)=30 cm²/s.

7. The total cost N(x) in rupees, associated with the production of x units of an item is given by N(x)=0.06x³-0.01x²+10x-43. Find the marginal cost when 5 units are produced.
a) Rs. 1.44
b) Rs. 144.00
c) Rs. 14.4
d) Rs. 56.2
Answer: b
Clarification: The marginal cost is given by the rate of change of revenue.
Hence, (frac{dN(x)}{dt})=0.18x²-0.02x+10.
∴(frac{dN(x)}{dt})|__x=5=0.18(5)²-0.02(5)+10
=4.5-0.1+10
=Rs. 14.4

8. The length of the rectangle is changing at a rate of 4 cm/s and the area is changing at the rate of 8 cm/s. What will be the rate of change of width if the length is 4cm and the width is 1 cm.
a) 5 cm/s
b) 6 cm/s
c) 2 cm/s
d) 1 cm/s
Answer: d
Clarification: Let the length be l, width be b and the area be A.
The Area is given by A=lb
(frac{dA}{dt})=l.(frac{db}{dt})+b.(frac{dl}{dt}) -(1)
Given that, (frac{dl}{dt})=4cm/s and (frac{dA}{dt})=8 cm/s
Substituting in the above equation, we get
8=l.(frac{db}{dt})+4b
Given that, l=4 cm and b=1 cm
∴8=4((frac{db}{dt}))+4(1)
8=4((frac{db}{dt}))+4
(frac{db}{dt})=1 cm/s.

9. For which of the values of x, the rate of increase of the function y=3x²-2x+7 is 4 times the rate of increase of x?
a) -1
b) (frac{1}{3})
c) 1
d) 0
Answer: c
Clarification: Given that, (frac{dy}{dt}=4.frac{dx}{dt})
y=3x²-2x+7
(frac{dy}{dt})=(6x-2) (frac{dx}{dt})
4.(frac{dx}{dt})=(6x-2) (frac{dx}{dt})
4=6x-2
6x=6
⇒x=1

10. The volume of a cube of edge x is increasing at a rate of 12 cm/s. Find the rate of change of edge of the cube when the edge is 6 cm.
a) (frac{1}{8})
b) (frac{2}{9})
c) –(frac{1}{9})
d) (frac{1}{9})
Answer: d
Clarification: Let the volume of cube be V.
V=x³
(frac{dV}{dt})=3x² (frac{dx}{dt})
12=3x² (frac{dx}{dt})
(frac{dx}{dt}=frac{4}{x^2})
(frac{dx}{dt})|__x=6=(frac{4}{6^2}=frac{4}{36}=frac{1}{9}).

Mathematics for Interviews,

Mathematics Multiple Choice Questions on “Differential Equations Basics-1”.

1. Find the order of the differential equation y’-20y+2=0.
a) 2
b) 8
c) 0
d) 1
Answer: d
Clarification: In the given D. E the highest order derivative is y’. Therefore, the order of the given D.E is 1.

2. Find the order of the D.E (frac{7d^2 y}{dx^2}-frac{6dy}{dx})=1
a) 4
b) 2
c) 3
d) 1
Answer: b
Clarification: The highest order derivative in the given differential equation is (frac{d^2 y}{dx^2}). Therefore, the order of the D.E is 2.

3. Find the order and degree of the differential equation y”’-(4y’)³=0
a) Order -3, Degree-1
b) Order -1, Degree-3
c) Order -2, Degree-1
d) Order -3, Degree-2
Answer: a
Clarification: In the D.E y”’-(4y’)³=0, the highest order derivative is y”’. Therefore, the order of the D.E is 3. Since it is a polynomial equation, the degree will be the power raised to y”’. Therefore, the degree is 1.

4. Find the order of the differential equation (frac{dy}{dx})+15 cos⁡x=0.
a) 4
b) 3
c) 2
d) 1
Answer: d
Clarification: The highest order derivative in the given D.E is (frac{dy}{dx}). Hence, the order of the given differential equation is 1.

5. Find the degree of the differential equation y”-12cosec y=0.
a) 1
b) 2
c) 4
d) Not defined
Answer: a
Clarification: The highest order derivative in this D.E is y”. The given D.E is a polynomial equation in y”. Therefore, the degree of the D.E is the power raised to y” which is 1.

6. Find the order and degree of the differential equation 7y’-3y=0.
a) Order -1, Degree-2
b) Order -2, Degree-3
c) Order -1, Degree-1
d) Order -3, Degree-2
Answer: c
Clarification: In the given D.E, the highest order derivative is y’. Therefore, the order is 1. The given D.E 7y’-3y=0 is a polynomial equation in y’ and the power raised to the highest derivative is 1. Hence, the degree is 1.

7. Find the degree of the D.E (frac{d^2 y}{dx^2}+5 cot⁡(frac{dy}{dx}))=0
a) five
b) three
c) two
d) not defined
Answer: d
Clarification: The given differential equation is not polynomial. Hence, the degree of the differential equation will not be defined.

8. Find the order and degree of the differential equation (y”’)²+7(y’)²-(cos⁡x)²=0
a) Order- 0, Degree-2
b) Order- 3, Degree-2
c) Order- 3, Degree-3
d) Order- 1, Degree-2
Answer: c
Clarification: The highest order derivative in the equation is y”’. Hence, the order is 3. The equation is polynomial in y”’. Therefore, the degree of the D.E will be the power of the derivative y”’ i.e. 2.

9. Find the degree of the D.E ((frac{d^2 y}{dx^2}))-3 tan⁡x=0.
a) 2
b) 1
c) 3
d) 4
Answer: b
Clarification: The given equation is a polynomial differential equation. Therefore, the degree of the equation will the power of the highest derivative (frac{d^2 y}{dx^2}) i.e. 1.

10. Find the order of the differential equation (frac{9d^2 y}{dx^2}-frac{7dy}{dx}+y^6=0)
a) 3
b) 1
c) 4
d) 2
Answer: d
Clarification: The highest derivative of the given D.E is (frac{d^2 y}{dx^2}). Therefore, the order is 2.

Mathematics Written Test Questions and Answers for Class 12 on “Three Dimensional Geometry – Shortest Distance between Two Lines”.

1. Which of the below given is the correct formula for the distance between two skew lines l₁ and l₂?
a) d=(left |frac{(vec{b_1}×vec{b_2}).(a_2-a_1)}{|vec{b_1}×vec{b_2}|}right |)
b) 2d=(left |frac{(vec{b_1}-vec{b_2}).(a_2-a_1)}{|vec{b_1}-vec{b_2}|}right |)
c) d=(left |frac{(vec{b_1}×vec{b_2}).(a_2.a_1)}{3|vec{b_1}×vec{b_2}|}right |)
d) d²=(left |frac{(vec{b_1}×vec{b_2}).(a_2-a_1)}{|vec{b_1}-vec{b_2}|}right |)
Answer: a
Clarification: The distance between two lines l₁ and l₂ with the equations
(vec{r}=vec{a_1}+λvec{b_1})
(vec{r}=vec{a_2}+μvec{b_2})
Then, the distance between the two lines is given by the formula
d=(left |frac{(vec{b_1}×vec{b_2}).(a_2-a_1)}{|vec{b_1}×vec{b_2}|}right |)

2. Find the shortest distance between two lines l₁ and l₂ whose vector equations is given below.
(vec{r}=3hat{i}-4hat{j}+2hat{k}+λ(4hat{i}+hat{j}+hat{k}))
(vec{r}=5hat{i}+hat{j}-hat{k}+μ(2hat{i}-hat{j}-3hat{k}))
a) (frac{11}{sqrt{12}})
b) (frac{23}{sqrt{10}})
c) (frac{18}{sqrt{10}})
d) (frac{10}{sqrt{11}})
Answer: c
Clarification: The distance between two skew lines is given by
d=(left |frac{(vec{b_1}×vec{b_2}).(a_2-a_1)}{|vec{b_1}×vec{b_2}|}right |)
(vec{r}=3hat{i}-4hat{j}+2hat{k}+λ(4hat{i}+hat{j}+hat{k}))
(vec{r}=5hat{i}+hat{j}-hat{k}+μ(2hat{i}-hat{j}-3hat{k}))
d=(left |frac{((4hat{i}+hat{j}+hat{k})×(2hat{i}-hat{j}-hat{k})).((3hat{i}-4hat{j}+2hat{k})-(2hat{i}-hat{j}-hat{k}))}{|4hat{i}+hat{j}+hat{k})×(2hat{i}-hat{j}-3hat{k})|}right |)
((4hat{i}+hat{j}+hat{k})×(2hat{i}-hat{j}-hat{k})=begin{vmatrix}hat{i}&hat{j}&hat{k}\4&1&1\2&-1&-1end{vmatrix})
=(hat{i}(-1+1)-hat{j}(-4-2)+hat{k}(-4-2))
=(6hat{j}-6hat{k})
d=(left |{6hat{j}-6hat{k}).(hat{i}-3hat{j}+3hat{k})}{|6hat{i}-2hat{k}|}right |)
(left|frac{0-18-18}{sqrt{6^2+2^2}}right |=frac{36}{sqrt{40}}=frac{18}{sqrt{10}})

3. Find the equation between the two parallel lines l₁ and l₂ whose equations is given below.
(vec{r}=3hat{i}+2hat{j}-hat{k}+λ(3hat{i}-2hat{j}+hat{k}))
(vec{r}=2hat{i}-hat{j}+hat{k}+μ(3hat{i}-2hat{j}+hat{k}))
a) (sqrt{frac{172}{14}})
b) (sqrt{frac{145}{14}})
c) (sqrt{frac{171}{14}})
d) (sqrt{frac{171}{134}})
Answer: c
Clarification: The distance between two parallel lines is given by
d=(left |frac{vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
=(left |frac{((3hat{i}-2hat{j}+hat{k})×((3hat{i}+2hat{j}-hat{k})-(2hat{i}-hat{j}+hat{k})))}{|sqrt{3^2+(-2)^2+1^2}|}right |)
=(left |frac{(3hat{i}-2hat{j}+hat{k})×(hat{i}+3hat{j}-2hat{k}))}{sqrt{14}}right |)
(
(3hat{i}-2hat{j}+hat{k})×(hat{i}+3hat{j}-2hat{k})=begin{vmatrix}hat{i}&hat{j}&hat{k}\3&-2&1\1&3&-2end{vmatrix})
=(hat{i}(4-3)-hat{j}(-6-1)+hat{k}(9+2))
=(hat{i}+7hat{j}+11hat{k})
∴d=(frac{|hat{i}+7hat{j}+11hat{k}|}{sqrt{14}}=frac{sqrt{1+49+121}}{sqrt{14}}=sqrt{frac{171}{14}}).

4. Find the shortest distance between the lines given.
l₁:(frac{x-5}{2}=frac{y-2}{5}=frac{z-1}{4})
l₂:(frac{x+4}{3}=frac{y-7}{6}=frac{z-3}{7})
a) (frac{115}{sqrt{134}})
b) (frac{115}{sqrt{184}})
c) (frac{115}{134})
d) (frac{sqrt{115}}{134})
Answer: a
Clarification: The shortest distance between two lines in cartesian form is given by:
l₁:(frac{x-x_1}{a_1}=frac{y-y_1}{b_1}=frac{z-z_1}{c_1})
l₂:(frac{x-x_2}{a_2}=frac{y-y_2}{b_2}=frac{z-z_2}{c_2})
∴d=(left |frac{begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\a_1&b_1&c_1\a_2&b_2&c_2end{vmatrix}}{sqrt{(b_1 c_2-b_2 c_1)^2+(c_1 a_2-c_2 a_1)^2+(a_1 b_2-a_2 b_1)^2}}right |)
d=(left |frac{begin{vmatrix}-9&5&2\2&5&4\3&6&7end{vmatrix}}{sqrt{√(35-24)^2+(12-14)^2+(12-15)^2}}right |)
d=(left |frac{-9(35-24)-5(14-12)+2(12-15)}{sqrt{11^2+2^2+3^2}}right |)
d=(left |frac{-99-10-6}{sqrt{134}}right |)
d=(frac{115}{sqrt{134}}).

5. Which of the following is the correct formula for the distance between the parallel lines l₁ and l₂?
a) d=(left|frac{vec{a_2}+vec{a_1})×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
b) d²=(left|frac{vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
c) 2d=(left|frac{vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
d) d=(left|frac{vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)
Answer: d
Clarification: If l₁ and l₂ are two parallel lines, then they are coplanar and hence can be represented by the following equations
(vec{r}=vec{a_1}+λvec{b})
(vec{r}=vec{a_2}+μvec{b})
Then the distance between the lines is given by
d=(left|frac{vec{b}×(vec{a_2}-vec{a_1})}{|vec{b}|}right |)