Category: Class 12 Mathematics Objective Questions

250+ TOP MCQs on General and Particular Solutions of Differential Equation | Class 12 Maths

Mathematics Multiple Choice Questions on “General and Particular Solutions of Differential Equation”.

1. Which of the following functions is the solution of the differential equation (frac{dy}{dx})+2y=0?
a) y=-2e-x
b) y=2ex
c) y=e-2x
d) y=e2x
Answer: c
Clarification: Consider the function y=e-2x
Differentiating both sides w.r.t x, we get
(frac{dy}{dx}=-2e^{-2x})
(frac{dy}{dx})=-2y
⇒(frac{dy}{dx})+2y=0.

2. The function y=8 sin⁡2x is a solution of the differential equation (frac{d^2 y}{dx^2})+4y=0.
a) True
b) False
Answer: a
Clarification: The given statement is true.
Consider the function y=8 sin⁡2x
Differentiating w.r.t x, we get
(frac{dy}{dx})=16 cos⁡2x –(1)
Differentiating (1) w.r.t x, we get
(frac{d^2 y}{dx^2})=-32 sin⁡2x
(frac{d^2 y}{dx^2})=-4(8 sin⁡2x )=-4y
⇒(frac{d^2 y}{dx^2})+4y=0.

3. Which of the following functions is a solution for the differential equation xy’-y=0?
a) y=4x
b) y=x2
c) y=-4x
d) y=2x
Answer: d
Clarification: Consider the function y=2x
Differentiating w.r.t x, we get
y’=(frac{dy}{dx})=2
Substituting in the equation xy’-y, we get
xy’-y=x(2)-2x=2x-2x=0
Therefore, the function y=2x is a solution for the differential equation xy’-y=0.

4. Which of the following differential equations has the solution y=3x2?
a) (frac{d^2 y}{dx^2})-6x=0
b) (frac{dy}{dx})-3x=0
c) x (frac{d^2 y}{dx^2})–(frac{dy}{dx})=0
d) (frac{d^2 y}{dx^2}-frac{3dy}{dx})=0
Answer: c
Clarification: Consider the function y=3x2
Differentiating w.r.t x, we get
(frac{dy}{dx})=6x –(1)
Differentiating (1) w.r.t x, we get
(frac{d^2 y}{dx^2})=6
∴(frac{xd^2 y}{dx^2}-frac{6dy}{dx})=6x-6x=0
Hence, the function y=3x2 is a solution for the differential equation x (frac{d^2 y}{dx^2})-6 (frac{dy}{dx})=0.

5. Which of the following functions is a solution for the differential equation y”+6y=0?
a) y=5 cos⁡3x
b) y=5 tan⁡3x
c) y=cos⁡3x
d) y=6 cos⁡3x
Answer: a
Clarification: Consider the function y=5 cos⁡3x
Differentiating w.r.t x, we get
y’=(frac{dy}{dx})=-15 sin⁡3x
Differentiating again w.r.t x, we get
y”=(frac{d^2 y}{dx^2})=-30 cos⁡3x
⇒y”+6y=0.
Hence, the function y=5 cos⁡3x is a solution for the differential equation y”+6y=0.

6. Which of the following functions is a solution for the differential equation (frac{dy}{dx})-14x=0?
a) y=7x2
b) y=7x3
c) y=x7
d) y=14x
Answer: a
Clarification: Consider the function y=7x2
Differentiating w.r.t x, we get
(frac{dy}{dx})=14x
∴(frac{dy}{dx})-14x=0
Hence, the function y=7x2 is a solution for the differential equation (frac{dy}{dx})-14x=0

7. Which of the following differential equations given below has the solution y=log⁡x?
a) (frac{d^2 y}{dx^2})-x=0
b) (frac{d^2 y}{dx^2}+(frac{dy}{dx})^2)=0
c) (frac{d^2 y}{dx^2})–(frac{dy}{dx})=0
d) x (frac{d^2 y}{dx^2})-log⁡x=0
Answer: b
Clarification: Consider the function y=log⁡x
Differentiating w.r.t x, we get
(frac{dy}{dx}=frac{1}{x} )–(1)
Differentiating (1) w.r.t x, we get
(frac{d^2 y}{dx^2}=-frac{1}{x^2} )
∴(frac{d^2 y}{dx^2}+(frac{dy}{dx})^2=-frac{1}{x^2}+(frac{1}{x})^2)
=-(frac{1}{x^2}+frac{1}{x^2})=0.

8. How many arbitrary constants will be there in the general solution of a second order differential equation?
a) 3
b) 4
c) 2
d) 1
Answer: c
Clarification: The number of arbitrary constants in a general solution of a nth order differential equation is n.
Therefore, the number of arbitrary constants in the general solution of a second order D.E is 2.

9. The number of arbitrary constants in a particular solution of a fourth order differential equation is __________________
a) 1
b) 0
c) 4
d) 3
Answer: b
Clarification: The number of arbitrary constants for a particular solution of nth order differential equation is always zero.

10. The function y=3 cos⁡x is a solution of the function (frac{d^2 y}{dx^2}-3frac{dy}{dx})=0.
a) True
b) False
Answer: b
Clarification: The given statement is false.
Given differential equation: (frac{d^2 y}{dx^2})-3 (frac{dy}{dx})=0 –(1)
Consider the function y=3 cos⁡x
Differentiating w.r.t x, we get
(frac{dy}{dx})=-3 sin⁡x
Differentiating again w.r.t x, we get
(frac{d^2 y}{dx^2})=-3 cos⁡x
Substituting the values of (frac{dy}{dx}) and (frac{d^2 y}{dx^2}) in equation (1), we get
(frac{d^2 y}{dx^2})-3 (frac{dy}{dx})=-3 cos⁡x-3(-3 sin⁡x)
=9 sin⁡x-3 cos⁡x≠0.
Hence, y=3 cos⁡x, is not a solution of the equation (frac{d^2 y}{dx^2})-3 (frac{dy}{dx})=0.

250+ TOP MCQs on Three Dimensional Geometry – Angle between Two Planes | Class 12 Maths

Mathematics Multiple Choice Questions on “Three Dimensional Geometry – Angle between Two Planes – 1”.

1. Which of the following is the correct formula for the angle between two planes?
a) cos⁡θ=(left |frac{vec{n_1}.vec{n_2}}{|vec{n_1}||vec{n_2}|}right |)
b) sin⁡θ=(left |frac{vec{n_1}.vec{n_2}}{|vec{n_1}||vec{n_2}|}right |)
c) cos⁡θ=(left |frac{vec{n_1}+vec{n_2}}{|vec{n_1}||vec{n_2}|}right |)
d) sin⁡θ=(left |frac{vec{n_1}+vec{n_2}}{(|vec{n_1}|+|vec{n_2}|}right |)
Answer: a
Clarification: If two planes of the form (vec{r}.vec{n_1}=d_1) and (vec{r}.vec{n_2}=d_2) where (vec{n_1} ,and ,vec{n_2}) are the normals to the plane, then the angle between them is given by
cos⁡θ=(left |frac{vec{n_1}.vec{n_2}}{|vec{n_1}||vec{n_2}|}right |)

2. Which of the following is the correct formula for the angle between two planes (A_1 x+B_1 y+C_1 z+D_1)=0 and (A_2 x+B_2 y+C_2 z+D_2)=0?
a) cos⁡θ=(frac{A_1 B_1 C_1}{A_2 B_2 C_2})
b) cos⁡θ=(left |frac{A_1 A_2+B_1 B_2+C_1 C_2}{sqrt{A_1^2+B_1^2+C_1^2}sqrt{A_2^2+B_2^2+C_2^2}}right |)
c) sin⁡θ=(left |frac{A_1 A_2-B_1 B_2-C_1 C_2}{sqrt{A_1^2+B_1^2+C_1^2}sqrt{A_2^2+B_2^2+C_2^2}}right |)
d) cos⁡θ=(A_1 A_2+B_1 B_2+C_1 C_2)
Answer: b
Clarification: If the planes are in the Cartesian form i.e. (A_1 x+B_1 y+C_1 z+D_1)=0 and (A_2 x+B_2 y+C_2 z+D_2)=0, where (A_1,B_1,C_1 ,and ,A_2,B_2,C_2) are the direction ratios of the planes, then the angle between them is given by
cos⁡θ=(left |frac{A_1 A_2+B_1 B_2+C_1 C_2}{sqrt{A_1^2+B_1^2+C_1^2} sqrt{A_2^2+B_2^2+C_2^2}}right |)

3. Find the angle between two planes (vec{r}.(2hat{i}-hat{j}+hat{k})=3) and (vec{r}.(3hat{i}+2hat{j}-3hat{k}))=5.
a) (cos^{-1}⁡frac{1}{sqrt{22}})
b) (cos^{-1}⁡frac{1}{sqrt{6}})
c) (cos^{-1}⁡frac{1}{sqrt{132}})
d) (cos^{-1}⁡frac{1}{sqrt{13}})
Answer: c
Clarification: Given that, the normal to the planes are (vec{n_1}=2hat{i}-hat{j}+hat{k}) and (vec{n_2}=3hat{i}+2hat{j}-3hat{k})
cos⁡θ=(left |frac{vec{n_1}.vec{n_2}}{|vec{n_1}||vec{n_2}|}right |)
(|vec{n_1}|=sqrt{2^2+(-1)^2+1^2}=sqrt{6})
(|vec{n_2}|=sqrt{3^2+2^2+(-3)^2}=sqrt{22})
(vec{n_1}.vec{n_2})=2(3)-1(2)+1(-3)=6-2-3=1
cos⁡θ=(frac{1}{sqrt{22}.sqrt{6}}=frac{1}{sqrt{132}})
∴θ=(cos^{-1}⁡frac{1}{sqrt{132}}).

4. If the planes (A_1 x+B_1 y+C_1 z+D_1)=0 and (A_2 x+B_2 y+C_2 z+D_2)=0 are at right angles to each other, then which of the following is true?
a) (frac{A_1+B_1+C_1}{A_2+B_2+C_2})=0
b) (A_1+A_2+B_1 +B_2+C_1+C_2)=0
c) (A_1+B_1+C_1=A_2 B_2 C_2)
d) (A_1 A_2+B_1 B_2+C_1 C_2)=0
Answer: d
Clarification: We know that the angle between two planes is given by
cos⁡θ=(left |frac{A_1 A_2+B_1 B_2+C_1 C_2}{sqrt{A_1^2+B_1^2+C_1^2} sqrt{A_2^2+B_2^2+C_2^2}}right |)
Given that, θ=90°
∴cos⁡90°=(left |frac{A_1 A_2+B_1 B_2+C_1 C_2}{sqrt{A_1^2+B_1^2+C_1^2}sqrt{A_2^2+B_2^2+C_2^2}}right |)
0=(left |frac{A_1 A_2+B_1 B_2+C_1 C_2}{sqrt{A_1^2+B_1^2+C_1^2}sqrt{A_2^2+B_2^2+C_2^2}}right |)
⇒(A_1 A_2+B_1 B_2+C_1 C_2)=0.

5. Find the angle between the planes 6x-3y+7z=8 and 2x+3y-2z=5?
a) (cos^{-1}frac{⁡11}{sqrt{98}})
b) (cos^{-1}⁡frac{⁡11}{sqrt{1598}})
c) (cos^{-1}frac{⁡⁡13}{sqrt{198}})
d) (cos^{-1}frac{⁡⁡11}{1598})
Answer: b
Clarification: We know that, the angle between two planes of the form (A_1 x+B_1 y+C_1 z+D_1)=0 and (A_2 x+B_2 y+C_2 z+D_2)=0 is given by
cos⁡θ=(left |frac{A_1 A_2+B_1 B_2+C_1 C_2}{sqrt{A_1^2+B_1^2+C_1^2} sqrt{A_2^2+B_2^2+C_2^2}}right |)
Given that, (A_1=6,B_1=-3,C_1=7) and (A_2=2,B_2=3,C_2=-2)
cos⁡θ=(left |frac{6(2)-3(3)+7(-2)}{|sqrt{6^2+(-3)^2+7^2} sqrt{2^2+3^2+(-2)^2}|}right |)
cos⁡θ=(|frac{-11}{sqrt{94}.sqrt{17}}|)
θ=(cos^{-1}⁡frac{⁡11}{sqrt{1598}}).

6. If two vectors (vec{r}.vec{n_1}=d_1) and (vec{r}.vec{n_2}=d_2) are such that (vec{n_1}.vec{n_2})=0, then which of the following is true?
a) The planes are perpendicular to each other
b) The planes are parallel to each other
c) Depends on the value of the vector
d) The planes are at an angle greater than 90°
Answer: a
Clarification: We know that if the scalar or dot product of vectors is 0, then they are at right angles to each other. Here the dot product of the normal vectors of the plane is 0, i.e. (vec{n_1}.vec{n_2})=0. Hence, the planes will be perpendicular to each other.

7. Find the angle between the two planes 2x+2y+z=2 and x-y+z=1?
a) (cos^{-1}frac{⁡1}{3})
b) (cos^{-1}⁡sqrt{3})
c) (cos^{-1}⁡frac{1}{3})
d) (cos^{-1}⁡frac{1}{3sqrt{3}})
Answer: d
Clarification: The angle between two planes of the form (A_1 x+B_1 y+C_1 z+D_1)=0 and (A_2 x+B_2 y+C_2 z+D_2)=0 is given by
cos⁡θ=(left |frac{A_1 A_2+B_1 B_2+C_1 C_2}{sqrt{A_1^2+B_1^2+C_1^2} sqrt{A_2^2+B_2^2+C_2^2}}right |)
According to the given question, (A_1=2,B_1=2,C_1=1 ,and ,A_2=1,B_2=-1,C_2=1)
cos⁡θ=(left |frac{2(1)+2(-1)+1(1)}{|sqrt{2^2+2^2+1^2} sqrt{1^2+(-1)^2+1^2}|}right |)
cos⁡θ=(left |frac{1}{sqrt{9}.sqrt{3}}right |)
θ=(cos^{-1}⁡frac{1}{3sqrt{3}}).

8. Which of the given set of planes are perpendicular to each other?
a) (vec{r}.(2hat{i}+2hat{j}+hat{k}))=5 and (vec{r}.(hat{i}+2hat{j}+2hat{k}))=5
b) (vec{r}.(hat{i}-2hat{j}+hat{k}))=7 and (vec{r}.(hat{i}+hat{j}+2hat{k}))=2
c) (vec{r}.(2hat{i}-2hat{j}+hat{k}))=4 and (vec{r}.(hat{i}+2hat{j}+2hat{k}))=5
d) (vec{r}.(3hat{i}-2hat{j}+hat{k}))=2 and (vec{r}.(hat{i}+2hat{j}+8hat{k}))=8
Answer: c
Clarification: Consider the set of planes (vec{r}.(2hat{i}-2hat{j}+hat{k})=4 ,and ,vec{r}.(hat{i}+2hat{j}+2hat{k}))=5
For the set of planes to be perpendicular (vec{n_1.}vec{n_2})=0
In the above set of planes, (vec{n_1}=2hat{i}-2hat{j}+hat{k}) and (vec{n_2}=hat{i}+2hat{j}+2hat{k})
∴(vec{n_1}.vec{n_2})=2(1)-2(2)+1(2)=0
Hence, they are perpendicular.

9. Find the angle between the planes (vec{r}.(4hat{i}+hat{j}-2hat{k}))=6 and (vec{r}.(5hat{i}-6hat{j}+hat{k}))=7?
a) (cos^{-1}⁡frac{12}{sqrt{1302}})
b) (cos^{-1}⁡frac{1}{sqrt{1392}})
c) (cos^{-1}frac{⁡23}{sqrt{102}})
d) (cos^{-1}⁡frac{15}{sqrt{134}})
Answer: a
Clarification: Given that, the normal to the planes are (vec{n_1}=4hat{i}+hat{j}-2hat{k} ,and ,vec{n_2}=5hat{i}-6hat{j}+hat{k})
The angle between two planes of the form (vec{r}.vec{n_1}=d_1 ,and ,vec{r}.vec{n_2}=d_2) is given by
cos⁡θ=(left |frac{vec{n_1}.vec{n_2}}{|vec{n_1}||vec{n_2}|}right |)
(|vec{n_1}|=sqrt{4^2+1^2+(-2)^2}=sqrt{21})
(|vec{n_2}|=sqrt{5^2+(-6)^2+1^2}=sqrt{62})
(vec{n_1}.vec{n_2})=4(5)+1(-6)-2(1)=20-6-2=12
cos⁡θ=(frac{12}{sqrt{21}.sqrt{62}}=frac{12}{sqrt{1302}})
∴θ=(cos^{-1}⁡frac{12}{sqrt{1302}}).

10. Which of the following sets of planes are parallel to each other?
a) 2x+3y+4z=8 and 3x+9y+12z=7
b) 2x+3y+4z=2 and 4x+6y+8z=9
c) 3x+2y+4z=0 and 3x+4y+2z=0
d) 2x+4y+8z=9 and 4x+2y+7z=0
Answer: b
Clarification: If two planes are of the form (A_1 x+B_1 y+C_1 z+D_1)=0 and (A_2 x+B_2 y+C_2 z+D_2)=0, then they will be parallel if (frac{A_1}{A_2}=frac{B_1}{B_2}=frac{C_1}{C_2})
Consider the planes 2x+3y+4z=2 and 4x+6y+8z=9
(frac{A_1}{A_2}=frac{2}{4})
(frac{B_1}{B_2}=frac{3}{6})
(frac{C_1}{C_2}=frac{4}{8})
(frac{2}{4}=frac{3}{6}=frac{4}{8}=frac{1}{2})
Hence, the above set of planes are parallel.

250+ TOP MCQs on Calculus Application – Maxima and Minima | Class 12 Maths

Mathematics Aptitude Test for Engineering Entrance Exams on “Calculus Application – Maxima and Minima”.

1. What will be the nature of the equation sin(x + α)/sin(x + β)?
a) Possess only minimum value
b) Possess only maximum value
c) Does not possess a maximum or minimum value
d) Data inadequate

Answer: c
Clarification: Let, y = sin(x + α)/sin(x + β)
Then,
dy/dx = [cos(x + α)sin(x + β) – sin(x + α)cos(x + β)]/sin2(x + β)
= sin(x+β – x-α)/sin2(x + β)
Or sin(β – α)/sin2(x + β)
So, for minimum or maximum value of x we have,
dy/dx = 0
Or sin(β – α)/sin2(x + β) = 0
Or sin(β – α) = 0 ……….(1)
Clearly, equation (1) is independent of x; hence, we cannot have a real value of x as root of equation (1).
Therefore, y has neither a maximum or minimum value.

2. Given, f(x) = x3 – 12x2 + 45x + 8. At which point does f(x) has its maximum?
a) 1
b) 2
c) 3
d) 4

Answer: c
Clarification: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.
f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.
Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.
So, f(x) has maximum at 3.

3. Given, f(x) = x3 – 12x2 + 45x + 8. What is the maximum value of f(x)?
a) 61
b) 62
c) 63
d) 54

Answer: b
Clarification: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.
f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.
Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.
So, f(x) has maximum at 3.
Putting, x = 3 in (1)
Thus, its maximum value is,
f(3) = 33 – 12*32 + 45*3 + 8 = 62.

4. Given, f(x) = x3 – 12x2 + 45x + 8. At which point does f(x) has its minimum?
a) 1
b) 7
c) 3
d) 5

Answer: d
Clarification: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.
f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.
Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.
So, f(x) has minimum at 5.

5. Given, f(x) = x3 – 12x2 + 45x + 8. What is the minimum value of f(x)?
a) -1
b) 0
c) 1
d) Value does not exist

Answer: c
Clarification: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or(x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.
f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.
Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.
So, f(x) has minimum at 5.
Putting, x = 5 in (1)
Thus, its maximum value is,
f(3) = 53 – 12*52 + 45*5 + 8 = 58.

6. At which point does f(x) = |x – 1| has its local minimum?
a) They are unequal
b) They are equal
c) Depend on the numbers
d) Can’t be predicted

Answer: b
Clarification: The given function is f(x) = ∣x − 1∣, x ∈ R.
It is known that a function f is differentiable at point x = c in its domain if both
(limlimits_{h rightarrow 0^-}) hf(c + h) – f(c)
And
(limlimits_{h rightarrow 0^+}) hf(c + h) – f(c) are finite and equal.
To check the differentiability of the function at x = 1,
LHS,
Consider the left hand limit of f at x=1
(limlimits_{h rightarrow 0^-}frac{|1+h-1|-|1-1|}{h})
= (limlimits_{h rightarrow 0^-}frac{|h|}{h})
= (limlimits_{h rightarrow 0^-}frac{-h}{h})
= −1
RHS,
Consider the right hand limit of f at x − 1
(limlimits_{h rightarrow 0^+}frac{|1+h-1|-|1-1|}{h})
= (limlimits_{h rightarrow 0^+}frac{|h|}{h})
= 1
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1.
As, LHS = -1 and RHS = 1, it is clear that, f’(1) < 0 on the left of x = 1 and f’(x) > 0 on the right of the point x = 1.
Hence, f’(x) changes sign, from negative on the left to positive on the right of the point x = 1.
Therefore, f(x) has a local minima at x = 1.

7. A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?
a) 1
b) 2
c) 3
d) 4

Answer: c
Clarification: Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1
So, v = dx/dt = t3/3 – 2t2/ + 3t + 1
Thus, dv/dt = t2 – 4t + 3
And d2v/dt2 = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t2 – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.

8. A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. What is the minimum velocity?
a) 1 cm/sec
b) 2 cm/sec
c) 3 cm/sec
d) 4 cm/sec

Answer: a
Clarification: Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1
So, v = dx/dt = t3/3 – 2t2/ + 3t + 1
Thus, dv/dt = t2 – 4t + 3
And d2v/dt2 = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t2 – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
Putting t = 3 in (1) we get,
33/3 – 2(3)2/ + 3(3) + 1
= 1 cm/sec.

250+ TOP MCQs on Inverse Trigonometry | Class 12 Maths

Mathematics Multiple Choice Questions on “Inverse Trigonometry”.

1. What will be the value of x + y + z if cos-1 x + cos-1 y + cos-1 z = 3π?
a) -1/3
b) 1
c) 3
d) -3
Answer: d
Clarification: The equation is cos-1 x + cos-1 y + cos-1 z = 3π
This means cos-1 x = π, cos-1 y = π and cos-1 z = π
This will be only possible when it is in maxima.
As, cos-1 x = π so, x = cos-1 π = -1 similarly, y = z = -1
Therefore, x + y + z = -1 -1 -1
So, x + y + z = -3.

2. Which value is similar to sin-1sin(6 π/7)?
a) sin-1(π/7)
b) cos-1(π/7)
c) sin-1(2π/7)
d) coses-1(π/7)
Answer: a
Clarification: sin-1sin(6 π/7)
Now, sin(6 π/7) = sin(π – 6 π/7)
= sin(2π + 6 π/7) = sin(π/7)
= sin(3π – 6 π/7) = sin(20π/7)
= sin(-π – 6 π/7) = sin(-15π/7)
= sin(-2π + 6 π/7) = sin(-8π/7)
= sin(-3π – 6 π/7) = sin(-27π/7)
Therefore, sin-1sin(6 π/7) = sin-1(π/7).

3. What is the value of sin-1(-x) for all x belongs to [-1, 1]?
a) -sin-1(x)
b) sin-1(x)
c) 2sin-1(x)
d) sin-1(-x)/2
Answer: a
Clarification: Let, θ = sin-1(-x)
So, -π/2 ≤ θ ≤ π/2
=> -x = sinθ
=> x = -sinθ
=> x = sin(-θ)
Also, -π/2 ≤ -θ ≤ π/2
=> -θ = sin-1(x)
=> θ = -sin-1(x)
So, sin-1(-x) = -sin-1(x)

4. What is the value of sin-1(sin 6)?
a) -2π – 6
b) 2π + 6
c) -2π + 6
d) 2π – 6
Answer: c
Clarification: We know, sin x = sin(π – x)
So, sin 6 = sin(π – 6)
= sin(2π + 6)
= sin(3π – 6)
= sin(-π – 6)
= sin(-2π – 6)
= sin(-3π – 6)
So, sin-1(sin 6) = sin-1(sin (-2π + 6))
= -2π + 6

5. What is the value of cos-1(-x) for all x belongs to [-1, 1]?
a) cos-1(-x)
b) π – cos-1(x)
c) π – cos-1(-x)
d) π + cos-1(x)
Answer: b
Clarification: Let, θ = cos-1(-x)
So, 0 ≤ θ ≤ π
=> -x = cosθ
=> x = -cosθ
=> x = cos(-θ)
Also, -π ≤ -θ ≤ 0
So, 0 ≤ π -θ ≤ π
=> -θ = cos-1(x)
=> θ = -cos-1(x)
So, cos-1(x) = π – θ
θ = π – cos-1(x)
=> cos-1(-x) = π – cos-1(x)

6. The given graph is for which equation?

a) y = sinx
b) y = sin-1x
c) y = cosecx
d) y = secx
Answer: b
Clarification: The following graph represents 2 equations.

The pink curve is the graph of y = sinx
The blue curve is the graph for y = sin-1x
This curve passes through the origin and approaches to infinity in both positive and negative axes.

7. The given graph is for which equation?

a) cosec-1x
b) secx
c) cos-1x
d) cotx
Answer: c
Clarification: There are 2 curves.

The green curve is the graph of y = cosx
The red curve is the graph for y = cos-1x
This curve origin from some point before π/3 and approaches to infinity in both positive y axis by intersecting at a point near 1.5 in y axis.

8. The given graph is for which equation?

a) y = cos-1x
b) y = cot-1x
c) y = cosec-1x
d) y = tan-1x
Answer: d
Clarification: There are 2 curves.

The blue curve is the graph of y = tanx
The red curve is the graph for y = tan-1x
This curve passes through the origin and approaches to infinity in the direction of x axis only.
This graph lies below –x axis and above +x axis.

9. The given graph is for which equation?

a) y = cot-1x
b) y = tan-1x
c) y = cotx
d) y = cosec-1x
Answer: a
Clarification: There are 2 curves.

The black curve is the graph of y = cotx
The red curve is the graph for y = cot-1x
This curve does not pass through the origin but approaches to infinity in the direction of x axis only.
The part of the curve that lies in the (x, y) coordinate gradually meets to the x-axis.
This graph lies above +x axis and –x axis.

10. The given graph is for which equation?

a) y = sinx
b) y = log|sinx|
c) y = |sinx|
d) y = |cosx|
Answer: c
Clarification: The given form of equation can be written as,

The green curve is the graph of y = sinx
The blue curve is the graph for y = |sinx|
As sinx is enclosed by a modulus so the curve that lies in the negative y axis will come to the positive y axis.

250+ TOP MCQs on Area of a Triangle | Class 12 Maths

Mathematics Multiple Choice Questions on “Area of a Triangle”.

1. Which of the following is the formula for finding the area of a triangle with the vertices (x1,y1), (x2,y2), (x3,y3).
a) Δ=(begin{Vmatrix}x_1&y_1&1\x_2&y_2&1\x_3&y_3&1end{Vmatrix})
b) Δ=(frac{1}{2}begin{Vmatrix}x_1&y_1&1\x_2&y_2&0\x_3&y_3&1end{Vmatrix})
c) Δ=(frac{1}{2}begin{Vmatrix}x_1&y_1&1\x_2&1&1\x_3&y_3&1end{Vmatrix})
d) Δ=(frac{1}{2}begin{Vmatrix}x_1&y_1&1\x_2&y_2&1\x_3&y_3&1end{Vmatrix})
Answer: d
Clarification: The area of a triangle with vertices (x1,y1), (x2,y2), (x3,y3) is given by
Δ=(frac{1}{2}begin{Vmatrix}x_1&y_1&1\x_2&y_2&1\x_3&y_3&1end{Vmatrix}).

2. What is the area of the triangle whose vertices are (0,1), (0,2), (1,5)?
a) 1 sq.unit
b) 2 sq.units
c) (frac{1}{3}) sq.units
d) (frac{1}{2}) sq.units
Answer: d
Clarification: The area of the triangle with vertices (0,1), (0,2), (1,5) is given by
Δ=(frac{1}{2}begin{Vmatrix}0&1&1\0&2&1\1&5&1end{Vmatrix})
Expanding along C1, we get
Δ=(frac{1}{2}){(0-0+1(1-2)}=(frac{1}{2}) |-1|=(frac{1}{2}) sq.units.

3. The area of the triangle formed by three collinear points is zero.
a) True
b) False
Answer: a
Clarification: The given statement is true. If the three points are collinear then they will be lying in a single line. Therefore, the area of the triangle formed by collinear points is zero.

4. Find the value of k for which (1,2), (3,0), (2,k) are collinear.
a) 0
b) -1
c) 2
d) 1
Answer: d
Clarification: The area of triangle formed by collinear points is zero.
Δ=(frac{1}{2}) (begin{Vmatrix}1&2&1\3&0&1\2&k&1end{Vmatrix})=0
Expanding along C2, we get
(frac{1}{2}){-2(3-2)+0-k(1-3)}=0
(frac{1}{2}) {-2+2k}=0
∴k=1

5. What is the area of the triangle if the vertices are (0,2), (0, 0), (3, 0)?
a) 1 sq.unit
b) 5 sq.units
c) 2 sq.units
d) 3 sq.units
Answer: d
Clarification: The area of the triangle (0,2), (0, 0), (3, 0) with vertices is given by
Δ=(frac{1}{2}begin{Vmatrix}0&2&1\0&0&1\3&0&1end{Vmatrix})
Expanding along R3, we get
Δ=(frac{1}{2}) {0-0+3(2-0)}
Δ=3 sq.units.

6. Find the equation of the line joining A(2,1) and B(6,3) using determinants.
a) 2y-x=0
b) 2y-x=0
c) y-x=0
d) y-2x=0
Answer: a
Clarification: Let C(x,y) be a point on the line AB. Thus, the points A(2,1), B(6,3), C(x,y) are collinear. Hence, the area of the triangle formed by these points will be 0.
⇒Δ=(frac{1}{2}begin{Vmatrix}2&1&1\6&3&1\x&y&1end{Vmatrix})=0
Expanding along C3, we get
(frac{1}{2}) {1(6y-3x)-1(2y-x)+1(6-6)}=0
(frac{1}{2}) {6y-3x-2y+x}=(frac{1}{2}) {4y-2x}=0
⇒2y-x=0

7. Find the value of k if the area is (frac{7}{2}) sq. units and the vertices are (1,2), (3,5), (k,0).
a) (frac{8}{3})
b) –(frac{8}{3})
c) –(frac{7}{3})
d) –(frac{8}{5})
Answer: b
Clarification: Given that the vertices are (1,2), (3,5), (k,0)
Therefore, the area of the triangle with vertices (1,2), (3,5), (k,0) is given by
Δ=(frac{1}{2}begin{Vmatrix}1&2&1\3&5&1\k&0&1end{Vmatrix})=(frac{7}{2})
Expanding along R3, we get
(frac{1}{2}) {k(2-5)-0+1(5-6)}=(frac{1}{2}) {-3k-1}=(frac{7}{2})
⇒-(frac{1}{2}) (3k+1)=(frac{7}{2})
3k=-8
k=-(frac{8}{3}).

8. Find the area of the triangle with the vertices (2,3), (4,1), (5,0).
a) 3 sq.units
b) 2 sq.units
c) 0
d) 1 sq.unit
Answer: c
Clarification: The area of the triangle with vertices (2,3), (4,1), (5,0) is given by
Δ=(frac{1}{2}begin{Vmatrix}2&3&1\4&1&1\5&0&1end{Vmatrix})
Applying R2→R2-R3
Δ=(frac{1}{2}begin{Vmatrix}2&3&1\-1&1&0\5&0&1end{Vmatrix})
Expanding along R2, we get
Δ=(frac{1}{2}) {-(-1)(3-0)+1(2-5)}
Δ=(frac{1}{2}) (0-0)=0.

9. Find the equation of the line joining A(5,1), B(4,0) using determinants.
a) 4x-y=4
b) x-4y=4
c) x-y=4
d) x-y=0
Answer: c
Clarification: Let C(x,y) be a point on the line AB. Thus, the points A(5,1), B(4,0), C(x,y) are collinear. Hence, the area of the triangle formed by these points will be 0.
⇒Δ=(frac{1}{2}begin{Vmatrix}5&1&1\4&0&1\x&y&1end{Vmatrix})=0
Applying R1→R1-R2
(frac{1}{2}begin{Vmatrix}1&1&0\4&0&1\x&y&1end{Vmatrix})=0
Expanding along R1, we get
=(frac{1}{2}) {1(0-y)-1(4-x)}=0
=(frac{1}{2}) {-y-4+x}=0
⇒x-y=4.

10. Find the value of k for which the points (3,2), (1,2), (5,k) are collinear.
a) 2
b) 5
c) 4
d) 9
Answer: a
Clarification: Given that the vertices are (3,2), (1,2), (5,k)
Therefore, the area of the triangle with vertices (3,2), (1,2), (5,k) is given by
Δ=(frac{1}{2}) (begin{Vmatrix}3&2&1\1&2&1\5&k&1end{Vmatrix})=0
Applying R1→R1-R2, we get
(frac{1}{2}begin{Vmatrix}2&0&0\1&2&1\5&k&1end{Vmatrix})=0
Expanding along R1, we get
(frac{1}{2}) {2(2-k)-0+0}=0
2-k=0
k=2.

250+ TOP MCQs on Derivatives Application – Approximations | Class 12 Maths

Mathematics Multiple Choice Questions on “Derivatives Application – Approximations”.

1. Find the approximate value of (sqrt{64.3}).
a) 8.0675
b) 8.03465
c) 8.01875
d) 8.0665
Answer: c
Clarification: Let y=(sqrt{x}). Let x=64 and Δx=0.3
Then, Δy=(sqrt{x+Δx}-sqrt{x})
Δy=(sqrt{64.3}-sqrt{64})
(sqrt{64.3})=Δy+8
dy is approximately equal to Δy is equal to:
dy=(frac{dy}{dx})Δx
dy=(frac{1}{2sqrt{x}}).Δx
dy=(frac{1}{2sqrt{64}}) (0.3)
dy=0.3/16=0.01875
∴ The approximate value of (sqrt{64.3}) is 8+0.01875=8.01875

2. Find the approximate value of (sqrt{49.1}).
a) 7.0142
b) 7.087942
c) 7.022
d) 7.00714
Answer: d
Clarification: Let y=(sqrt{49.1}). Let x=49 and Δx=0.1
Then, Δy=(sqrt{x+Δx}-sqrt{x})
Δy=(sqrt{49.1}-sqrt{49})
(sqrt{49.1})=Δy+7
dy is approximately equal to Δy is equal to
dy=(frac{dy}{dx})Δx
dy=(frac{1}{(2sqrt{x})}).Δx
dy=(frac{1}{(2sqrt{49})}) (0.1)
dy=0.1/14=0.00714
∴ The approximate value of (sqrt{49.1}) is 7+0.00714=7.00714

3. Find the approximate value of f(5.03), where f(x)=4x2-7x+2.
a) 67.99
b) 56.99
c) 67.66
d) 78.09
Answer: a
Clarification: Let x=5 and Δx=0.03
Then, f(x+Δx)=4(x+Δx)2-7(x+Δx)+2
Δy=f(x+Δx)-f(x)
∴f(x+Δx)=Δy+f(x)
Δy=f’ (x)Δx
⇒f(x+Δx)=f(x)+f’ (x)Δx
f(5.03)=(4(5)2-7(5)+2)+(8(5)-7)(0.03) (∵f’ (x)=8x-7)
f(5.03)=(100-35+2)+(40-7)(0.03)
f(5.03)=67+33(0.03)
f(5.03)=67+0.99=67.99

4. Find the approximate value of (sqrt{11}).
a) 3.34
b) 3.934
c) 3.0034
d) 3.544
Answer: a
Clarification: Let y=(sqrt{x}). Let x=9 and Δx=2
Then, Δy=(sqrt{x+Δx}-sqrt{x})
Δy=(sqrt{11}-sqrt{9})
(sqrt{11})=Δy+3
dy is approximately equal to Δy is equal to
dy=(frac{dy}{dx})Δx
dy=(frac{1}{(2sqrt{x})}).Δx
dy=(frac{1}{(2sqrt{9})}(2))
dy=2/6=0.34
∴ The approximate value of (sqrt{11}) is 3+0.34=3.34.

5. What will be the approximate change in the surface area of a cube of side xm caused by increasing the side by 2%.
a) 0.24x
b) 2.4x2
c) 0.4x2
d) 0.24x2
Answer: d
Clarification: Let the edge of the cube be x. Given that dx or Δx is equal to 0.02x(2%).
The surface area of the cube is A=6x2
Differentiating w.r.t x, we get
(frac{dA}{dx})=12x
dA=((frac{dA}{dx}))Δx=12x(0.02x)=0.24x2
Hence, the approximate change in volume is 0.24x2.

6. Find the approximate value of f(4.04), where f(x)=7x3+6x2-4x+3.
a) 346.2
b) 544.345
c) 546.2
d) 534.2
Answer: c
Clarification: Let x=4 and Δx=0.04
Then, f(x+Δx)=7(x+Δx)3+7(x+Δx)2-4(x+Δx)+3
Δy=f(x+Δx)-f(x)
∴f(x+Δx)=Δy+f(x)
Δy=f'(x)Δx
⇒f(x+Δx)=f(x)+f’ (x)Δx
Here, f'(x)=21x2+12x-4
f(4.04)=(7(4)3+6(4)2-4(4)+3)+(21(4)2+12(4)-4)(0.04)
f(4.04)=(448+96-16+3)+(336+48-4)(0.04)
f(4.04)=531+380(0.04)=546.2

7. Find the approximate value of (127)1/3.
a) 5.0267
b) 2.0267
c) 8.0267
d) 5.04
Answer: a
Clarification: Let y=(x)1/3. Let x=125 and Δx=2
Then, Δy=(x+Δx)1/3-x1/3
Δy=(127)1/3-(125)1/3
(127)1/3=Δy+5
dy is approximately equal to Δy is equal to
dy=(frac{dy}{dx})Δx
dy=(frac{1}{3x^{2/3}}).Δx
dy=(frac{1}{3×125^{2/3}} (2))
dy=2/75=0.0267
∴ The approximate value of (127)1/3 is 5+0.0267=5.0267

8. Find the approximate change in the volume of cube of side xm caused by increasing the side by 6%.
a) 0.18x
b) 0.18x3
c) 0.18x2
d) 1.8x3
Answer: b
Clarification: We know that the volume V of a cube is given by
V=x3
Differentiating w.r.t x, we get
(frac{dV}{dx})=3x2
dV=((frac{dV}{dx}))Δx=3x2 Δx
dV=3x2 (6x/100)=0.18x3
Therefore, the approximate change in volume is 0.18x3.

9. Find the approximate value of (82)1/4.
a) 3.025
b) 3.05
c) 3.00925
d) 3.07825
Answer: c
Clarification: Let y=x1/4. Let x=81 and Δx=1
Then, Δy=(x+Δx)1/4-x1/4
Δy=821/4-811/4
821/4=Δy+3
dy is approximately equal to Δy is equal to
dy=(frac{dy}{dx})Δx
dy=(frac{1}{(4x^{3/4})}).Δx
dy=(frac{1}{(4×81^{3/4})} (1))
dy=(frac{1}{(4×27)})=0.00925
∴ The approximate value of 821/4 is 3+0.00925=3.00925

10. Find the approximate error in the volume of the sphere if the radius of the sphere is measured to be 6cm with an error of 0.07cm.
a) 10.08π cm3
b) 10.08cm3
c) 10.4πcm3
d) 9.08cm3
Answer: a
Clarification: Let x be the radius of the sphere.
Then, x=6cm and Δx=0.07cm
The volume of a sphere is given by V=(frac{4}{3}) πx3
∴(frac{dV}{dx}=frac{4}{3}) π(3x2)=4πx2
dV=((frac{dV}{dx}))Δx=4πx2 Δx
dV=4×π×62×0.07
dV=10.08π cm3