Category: Class 12 Mathematics Objective Questions

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

250+ TOP MCQs on Methods of Solving First Order & First Degree Differential Equations | Class 12 Maths

Mathematics Multiple Choice Questions on “Methods of Solving First Order & First Degree Differential Equations”.

1. Find the general solution of the differential equation (frac{dy}{dx}=5x^2+2).
a) 10x3+12x-3y2+C=0
b) 12x-3y2+C=0
c) 10x3+12x-y2+C=0
d) 10x2-3y2+C=0
Answer: a
Clarification: Given that, (frac{dy}{dx}=5x^2+2)
Separating the variables, we get
dy=(5x2+2)dx –(1)
Integrating both sides of (1), we get
(int y ,dy=int 5x^2+2 ,dx)
(frac{y^2}{2}=frac{5x^3}{3}+2x+C_1)
3y2=(10x^3+12x+6C_1)
10x3+12x-3y2+C=0 (where 6C1=C)

2. Find the general solution of the differential equation (frac{dy}{dx}=frac{y-3}{x-3}) (x, y≠3).
a) x-3=0
b) y-3=0
c) y+3=0
d) x-3y=0
Answer: b
Clarification: Given that, (frac{dy}{dx}=frac{y-3}{x-3})
Separating the variables, we get
(frac{dy}{y-3}=frac{dx}{x-3})
log⁡(y-3)=log⁡(x-3)+log⁡C1
log⁡(y-3)-log⁡(x-3)=log⁡C1
(log⁡(frac{y-3}{x-3}))=log⁡C1
(frac{1}{C_1} frac{y-3}{x-3}=0)
y-3=0 is the general solution for the given differential equation.

3. Find the general solution of the differential solution (frac{dy}{dx}=2-x+x^3).
a) x4-2x2-4y+C=0
b) x4-2x2+C=0
c) 2x2+4x-4y+C=0
d) x4-2x2+4x-4y+C=0
Answer: d
Clarification: Given that, (frac{dy}{dx})=2-x+x4
Separating the variables, we get
dy=(2-x+x3)dx
Integrating on both sides, we get
(int dy=int 2-x+x^3 ,dx)
(y=2x-frac{x^2}{2}+frac{x^4}{4}+C_1)
4y=4x-2x2+x4+4C1
∴x4-2x2+4x-4y+4C1=0
x4-2x2+4x-4y+C=0 (where 4C1=C)

4. Find the general solution of the differential equation (frac{dy}{dx}=frac{3 ,sec,⁡y}{2 ,cosec⁡,x}).
a) 3 cos⁡x-2 cos⁡y=C
b) 3 sin⁡x+2 sin⁡y=C
c) 3 cos⁡x+2 tan⁡x=C
d) 3 cos⁡x+2 sin⁡y=C
Answer: d
Clarification: Given that, (frac{dy}{dx}=frac{3 ,sec⁡,y}{2cosec ,x})
(frac{2 ,dy}{sec⁡ ,y}=frac{3dx}{cosec,⁡x})
Separating the variables, we get
2 cos⁡y dy=3 sin⁡x dx
Integrating both sides, we get
∫ 2 cos⁡y dy = ∫ 3 sin⁡x dx
2 sin⁡y=3(-cos⁡x)+C
3 cos⁡x+2 sin⁡y=C

5. Find the general solution of the differential equation (frac{dy}{dx}=frac{2+x^3}{4-y^3}).
a) x3-y3-4y+C=0
b) x4+8x+y4-16y+C=0
c) 2x+y4-4y+C=0
d) x3+2x+C=0
Answer: b
Clarification: Given that, (frac{dy}{dx}=frac{2+x^3}{4-y^3})
Separating the variables, we get
(4-y3)dy=(2+x3)dx
Integrating both sides, we get
(int 4-y^3 ,dy=int 2+x^3 ,dx)
(4y-frac{y^4}{4}=2x+frac{x^4}{4}+C_1)
x4+8x+y4-16y+C=0 (where 4C1=C)

6. Find the general solution of the differential equation (frac{dy}{dx}=3e^x+2)
a) y=3ex+2x+C
b) y=3ex-2x+C
c) y=2ex+3x+C
d) y=2ex-3x+C
Answer: a
Clarification: Given that, (frac{dy}{dx}=3e^x+2)
Separating the variables, we get
dy=(3ex+2)dx
Integrating both sides, we get
(int dy=int (3e^x+2),dx) –(1)
y=3ex+2x+C which is the general solution of the given differential equation.

7. Find the particular solution of the differential equation (frac{dy}{dx})+2x=5 given that y=5, when x=1.
a) y=5x+x2+1
b) y=x-x2+4
c) y=5x-x2+1
d) y=5x-x2
Answer: c
Clarification: Given that, (frac{dy}{dx}+2x=5)
(frac{dy}{dx}=5-2x)
Separating the variables, we get
dy=(5-2x)dx
Integrating both sides, we get
(int dy=int 5-2x ,dx)
y=5x-x2+C –(1)
Given that, y=5, when x=1
⇒5=5(1)-(1)2+C
∴C=1
Substituting value of C to equation (1), we get
y=5x-x2+1 which is the particular solution of the given differential equation.

8. Find the particular solution of the differential equation (frac{dy}{dx}+8x=16x^2+4) given that y=(frac{1}{3}) when x=1.
a) y=(frac{(2x+1)^2}{3})
b) y=(frac{(4x+1)^2}{12})
c) y=(frac{(4x-2)^2}{3})
d) y=(frac{(2x-1)^2}{3})
Answer: d
Clarification: Given that, (frac{dy}{dx}+8x=16x^2+4)
(frac{dy}{dx}=16x^2-8x+4)
(frac{dy}{dx}=(4x-2)^2)
Separating the variables, we get
dy=(4x-2)2 dx
Integrating both sides, we get
(int dy=int (4x-2)^2 ,dx)
y=(frac{(4x-2)^2}{12}+C)
y=(frac{(2x-1)^2}{3}+C) –(1)
Given that, y=1/3 when x=1
Therefore, equation (1) becomes,
(frac{1}{3}=frac{(2(1)-1)^2}{3}+C)
(C=frac{1}{3}-frac{1}{3})=0
Hence, the particular solution for the given differential solution is y=(frac{(2x-1)^2}{3}).

9. Find the particular solution for the differential equation (frac{dy}{dx}=frac{3x^2}{7y}) given that, y=1 when x=1.
a) 7x2=2y3+5
b) 7x3=2y2+5
c) 7y2=2x3+5
d) 2y2=5x3+6
Answer: c
Clarification: Given that, (frac{dy}{dx}=frac{3x^2}{7y})
Separating the variables, we get
7y dy=3x2 dx
Integrating both sides, we get
(7int y ,dy=3int x^2 ,dx)
(frac{7y^2}{2}=3(frac{x^3}{3})+C)
(frac{7y^2}{2}=x^3)+C –(1)
Given that y=1, when x=1
Substituting the values in equation (1), we get
(frac{7(1)^2}{2}=(1)^3+C)
(C=frac{7}{2}-1=frac{5}{2})
Hence, the particular solution of the given differential equation is:
(frac{7y^2}{2}=x^3+frac{5}{2})
⇒7y2=2x3+5

10. Find the particular solution of the differential equation (frac{dy}{dx}=frac{9y ,log⁡x}{5x ,log⁡y}).
a) (log⁡y)2+(log⁡x)2=0
b) (log⁡y)2-(log⁡x)2=0
c) log⁡y-log⁡x=0
d) 2 log⁡x+log⁡y=0
Answer: b
Clarification: (frac{dy}{dx}=frac{9y ,log⁡x}{5x ,log⁡y})
Separating the variables, we get
(frac{5 ,log⁡y}{y} dy=frac{9 ,log⁡x}{x} dx)
Integrating both sides, we get
(5int frac{log⁡y}{y} ,dy=9int frac{log⁡x}{x} ,dx) –(1)
First, for integrating (frac{log⁡y}{y})
Let log⁡y=t
Differentiating w.r.t y, we get
(frac{1}{y}) dy=dt
∴(int frac{log⁡y}{y} ,dy=int t ,dt)
=(frac{t^2}{2}=frac{log⁡y^2}{2})
Similarly integrating (frac{log⁡x}{x})
Let log⁡x=t
Differentiating w.r.t x, we get
(frac{1}{x}) dx=dt
∴(int frac{log⁡x}{x} dy=int t ,dt)
=(frac{t^2}{2}=frac{(log⁡x)^2}{2})
Hence, equation (1), becomes
(frac{(log⁡y)^2}{2}=frac{(log⁡x)^2}{2}+C) –(2)
Given y=2, we get x=2
Substituting the values in equation (2), we get
(frac{(log⁡y)^2}{2}=frac{(log⁡x)^2}{2}+C)
C=0
Therefore, the equation becomes ((log⁡y)^2=(log⁡x^2))
∴(log⁡y)2-(log⁡x)2=0.

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

Mathematics Online Quiz for Engineering Entrance Exams on “Three Dimensional Geometry – Angle between Two Planes – 2”.

1. _____ is the angle between the normals to two planes.
a) Normal between the planes
b) The angle between the planes
c) Tangent between the planes
d) Distance between the planes
Answer: b
Clarification: The angle between the normals to two planes is called the angle between the planes. A trigonometric identity, cosine is used to find the angle called ‘θ’ between two planes.

2. If θ is the angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c21z + d2 = 0 then
cos θ=(frac {a1a2.b1b2.c1c2}{sqrt {a1^2+b1^2+c1^2} sqrt {a2^2+b2^2+c2^2 }}).
a) True
b) False
Answer: b
Clarification: The formula to find angle between the normal of two planes is
cos θ=(frac {a1a2+b1b2+c1c2}{sqrt {a1^2+b1^2+c1^2} sqrt{a2^2+b2^2+c2^2 }}) not cos θ=(frac {a1a2.b1b2.c1c2}{sqrt {a1^2+b1^2+c1^2} sqrt {a2^2+b2^2+c2^2 }}) because the numerator should contain sum of co-efficients not their product.

3. What is the formula to find the angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0?
a) cos θ=(frac {a1a2+b1b2+c1c2}{sqrt {a1^2+b1^2+c1^2} sqrt {a2^2+b2^2+c^2 }})
b) sec θ=(frac {a1a2+b1b2+c1c2}{sqrt {a1^2+b1^2+c1^2} sqrt{a2^2+b2^2+c2^2 }})
c) cos θ=(frac {a1a2.b1b2.c1c2}{sqrt {a1^2+b1^2+c1^2} sqrt{a2^2+b2^2+c2^2 }})
d) cot θ=(frac {a1a2+b1b2+c1c2}{sqrt {a1^2+b1^2+c1^2} sqrt{a2^2+b2^2+c2^2 }})
Answer: a
Clarification: The formula to find the angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is cos θ=(frac {a1a2+b1b2+c1c2}{sqrt {a1^2+b1^2+c1^2} sqrt {a2^2+b2^2+c2^2 }}). θ is the angle between the normal of two planes.

4. Which trigonometric function is used to find the angle between two planes?
a) Tangent
b) Cosecant
c) Secant
d) Sine
Answer: b
Clarification: The symbol ‘θ’ represents the angle between two planes. A trigonometric function called cosine is used the find the angle i.e.; θ between the normal of two planes.

5. Find s for the given planes 2x + 2y + sz + 2 = 0 and 3x + y + z – 2 = 0, if they are perpendicular to each other.
a) 21
b) – 7
c) 12
d) – 8
Answer: d
Clarification: If their normals are perpendicular to each other then a1a2 + b1b2 + c1c2 = 0.
2(3) + 2(1) + s(1) = 0
s(1) = – 8
k = – 8

6. What is the relation between the the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c21z + d2 = 0, if their normal are parallel to each other?
a) (frac {a1}{b1} = frac{a2}{c1} = frac{c2}{b2})
b) (frac {a1}{a2} = frac{b1}{c2} = frac{c1}{b2})
c) (frac {a1}{a2} = frac{b1}{b2} = frac{c1}{c2})
d) (frac {c1}{a2} = frac{b1}{b2} = frac{a1}{c2})
Answer: c
Clarification: Relation between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c21z + d2 = 0, if their normal are parallel to each other is a1 : b1 : c1 = a2 : b2 : c2 ⇒ (frac {a1}{a2} = frac{b1}{b2} = frac{c1}{c2}).

7. What is the relation between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c21z + d2 = 0, if their normal are perpendicular to each other?
a) a1a2 . b1b2 . c1c2 = 0
b) a1a2 + b1b2 + c1c2 = 0
c) a1a2 + b1b2 – c1c2 = 0
d) a1a2 + b1b2 – c1c2 = 0
Answer: b
Clarification: θ = 90 degrees ⇒ cos θ
a1a2 + b1b2 – c1c2 = 0
Relation between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c21z + d2 = 0, if their normal are perpendicular to each other is a1a2 + b1b2 + c1c2 = 0.

8. _____ planes have an angle 90 degrees between them.
a) Orthogonal
b) Tangential
c) Normal
d) Parallel
Answer: a
Clarification: The planes which are perpendicular to each other i.e.; having an angle 90 degrees between them are called orthogonal planes. Hence, Orthogonal planes have an angle 90 degrees between them.

9. The condition a1a2 + b1b2 + c1c2 = 0 is for the planes whose normals are _____ to each other.
a) integral
b) parallel
c) perpendicular
d) concentric
Answer: c
Clarification: θ = 90 degrees ⇒ cos θ
a1a2 + b1b2 – c1c2 = 0
Relation between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c21z + d2 = 0, if their normal are perpendicular to each other is a1a2 + b1b2 + c1c2 = 0.

10. The condition (frac {a1}{a2} = frac{b1}{b2} = frac{c1}{c2}) is for the planes whose normals are _____ to each other.
a) perpendicular
b) parallel
c) differential
d) tangential
Answer: a
Clarification: Relation between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c21z + d2 = 0, if their normal are parallel to each other is a1 : b1 : c1 = a2 : b2 : c2 ⇒ (frac {a1}{a2} = frac{b1}{b2} = frac{c1}{c2}).

11. Find the angle between 2x + 3y – 2z + 4 = 0 and 4x + 3y + 2z + 2 = 0.
a) 38.2
b) 19.64
c) 89.21
d) 54.54
Answer: d
Clarification: Angle between two planes cos cos θ=(frac {a1a2+b1b2+c1c2}{sqrt {a1^2+b1^2+c1^2} sqrt {a2^2+b2^2+c2^2 }})
cos θ = 0.58
θ = cos-1(0.58)
θ = 54.54

12. Find the angle between x + 2y + 7z + 2 = 0 and 4x + 4y + z + 2 = 0.
a) 69.69
b) 84.32
c) 63.25
d) 83.25
Answer: c
Clarification: Angle between two planes cos θ=(frac {a1a2+b1b2+c1c2}{sqrt {a1^2+b1^2+c1^2} sqrt {a2^2+b2^2+c2^2 }})
cos θ = 0.45
θ = 63.25

13. The planes 5x + y + 3z + 1 = 0 and x + y – kz + 6 = 0 are orthogonal, find k.
a) 4
b) 2
c) 6
d) 8
Answer: b
Clarification: Relation between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c21z + d2 = 0, if their normal are perpendicular to each other is a1a2 + b1b2 + c1c2 = 0.
5(1) + 1(1) + 3(-k) = 0
-3k = -6
K = 2

14. Find the angle between the planes 5x + y + 3z + 1 = 0 and x + y – 2z + 6 = 0.
a) 30.82
b) 34.91
c) 11.23
d) 7.54
Answer: b
Clarification: Angle between two planes cos θ=(frac {a1a2+b1b2+c1c2}{sqrt {a1^2+b1^2+c1^2} sqrt {a2^2+b2^2+c2^2 }})
cos θ = 0.82
θ = 34.91

15. Find k for the given planes x + 2y + kz + 2 = 0 and 3x + 4y – z + 2 = 0, if they are perpendicular to each other.
a) 21
b) 17
c) 12
d) 11
Answer: d
Clarification: Relation between the the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c21z + d2 = 0, if their normal are perpendicular to each other is a1a2 + b1b2 + c1c2 = 0.
1(3) + 2(4) + k(-1) = 0
k(-1) = -11
k = 11

250+ TOP MCQs on Calculus Application – Acceleration | Class 12 Maths

Mathematics Multiple Choice Questions on “Calculus Application – Acceleration”.

1. A particle is moving in a straight line and its distance s cm from a fixed point in the line after t seconds is given by s = 12t – 15t2 + 4t3. What is the acceleration of the particle after 3 seconds?
a) 41 cm/sec2
b) 42 cm/sec2
c) 43 cm/sec2
d) 44 cm/sec2
Answer: b
Clarification: We have, s = 12t – 15t2 + 4t3 ……….(1)
Differentiating both side of (1) with respect to t we get,
(ds/dt) = 12 – 30t + 12t2
And d2s/dt2 = -30 + 24t
So, acceleration of the particle after 3 seconds is,
[d2s/dt2]t = 3 = – 30 + 24(3)
= 42 cm/sec2.

2. A particle is moving in a straight line is at a distance x from a fixed point in the straight line at time t seconds, where x = 2t3 – 12t + 11. What is the acceleration of the particle at the end of 2 seconds?
a) 22 cm/sec2
b) 24 cm/sec2
c) 26 cm/sec2
d) 28 cm/sec2
Answer: a
Clarification: We have, x = 2t3 – 12t + 11  ……….(1)
Let v and f be the velocity and acceleration respectively of the particle at time t seconds.
Then, v = dx/dt = d(2t3 – 12t + 11)/dt
= 6t2 – 12  ……….(2)
And f = dv/dt = d(6t2 – 12)/dt
= 12t   ……….(3)
Putting the value of t = 2 in (3),
Therefore, the displacement of the particle at the end of 2 seconds,
12t = 12(2)
= 24 cm/sec2

3. A particle is moving in a straight line is at a distance x from a fixed point in the straight line at time t seconds, where x = 2t3 – 12t + 11. What is the average acceleration of the particle at the end of 3 seconds?
a) 28 cm/sec2
b) 30 cm/sec2
c) 32 cm/sec2
d) 26 cm/sec2
Answer: b
Clarification: We have, x = 2t3 – 12t + 11  ……….(1)
Let v and f be the velocity and acceleration respectively of the particle at time t seconds.
Then, v = dx/dt = d(2t3 – 12t + 11)/dt
= 6t2 – 12  ……….(2)
And f = dv/dt = d(6t2 – 12)/dt
= 12t  ……….(3)
Putting the value of t = 2 in (3),
Therefore, the acceleration of the particle at the end of 2 seconds,
12t = 12(2)
= 24 cm/sec2
Now putting the value of t = 2 in (2),
We get the displacement of the particle at the end of 2 seconds,
6t2 – 12 = 6(2)2 – 12
= 12 cm/sec  ……….(4)
And putting the value of t = 3 in (2),
We get the displacement of the particle at the end of 3 seconds,
6t2 – 12 = 6(3)2 – 12
= 42 cm/sec  ……….(5)
Thus, change in velocity is, (5) – (4),
=42 – 12
= 30cm/sec.
Thus, the average acceleration of the particle at the end of 3 seconds is,
= (change of velocity)/time
= (30 cm/sec)/1 sec
= 30 cm/sec2

4. A particle is moving along the straight line OX and its distance x is in metres from O after t seconds from start is given by x = t3 – t2 – 5t. What will be the acceleration of the particle when it is at a distance 28 metres from O?
a) 20 m/sec2
b) 22 m/sec2
c) 24 m/sec2
d) 26 m/sec2
Answer: b
Clarification: We have, x = t3 – t2 – 5t  ……….(1)
When x = 28, then from (1) we get,
t3 – t2 – 5t = 28
Or t3 – t2 – 5t – 28 = 0
Or (t – 4)(t2 + 3t +7) = 0
Thus, t = 4
Let v and f be the velocity and acceleration respectively of the particle at time t seconds. Then,
v = dx/dt = d(t3 – t2 – 5t)/dt
= 3t2 – 2t – 5
And f = dv/dt = d(3t2 – 2t – 5)/dt
= 6t – 2
Therefore, the acceleration of the particle at the end of 4 seconds i.e., when the particle is at a distance of 28 metres from O,
[f]t = 4 = (6*4 – 2) m/sec2
= 22 m/sec2

5. A particle moves in a straight-line OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct2 (a, b > 0). What is the meaning of the constant c?
a) Uniform acceleration
b) Non – uniform acceleration
c) Uniform retardation
d) Non – uniform retardation
Answer: a
Clarification: We have, x = a + bt + ct2  ……….(1)
Let, v and f be the velocity and acceleration of a particle at time t seconds.
Then, v = dx/dt = d(a + bt + ct2)/dt = b + ct  ……….(2)
And f = dv/dt = d(b + ct)/dt = c   ……….(3)
Since f = dv/dt = c, hence, c represents the uniform acceleration of the particle.

6. A particle moves in a straight-line OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct2 (a, b > 0). What will be the nature of motion of the particle when c > 0?
a) Uniform retardation
b) Uniform speed
c) Uniform positive acceleration
d) Uniform velocity
Answer: c
Clarification: We have, x = a + bt + ct2  ……….(1)
Let, v and f be the velocity and acceleration of a particle at time t seconds.
Then, v = dx/dt = d(a + bt + ct2)/dt = b + ct   ……….(2)
And f = dv/dt = d(b + ct)/dt = c   ……….(3)
Clearly, when c > 0,implies f > 0.
Hence in this case the particle moves withan uniform positive acceleration.

7. A particle moves in a straight-line OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct2 (a, b > 0). What will be the nature of motion of the particle when c < 0?
a) Uniform retardation
b) Uniform speed
c) Uniform acceleration
d) Uniform velocity
Answer: a
Clarification: We have, x = a + bt + ct2  ……….(1)
Let, v and f be the velocity and acceleration of a particle at time t seconds.
Then, v = dx/dt = d(a + bt + ct2)/dt = b + ct   ……….(2)
And f = dv/dt = d(b + ct)/dt = c  ……….(3)
Clearly, when c < 0,implies f < 0.
Hence in this case the particle moves withan uniform retardation.

8. A particle moves in a straight line and its velocity v at time t seconds is given byv = 3t2 – 4t + 5 cm/second. What will be the acceleration of the particle during first 3 seconds after the start?
a) 10cm/sec2
b) 12cm/sec2
c) 14cm/sec2
d) 16cm/sec2
Answer: c
Clarification:Let f be the acceleration of the particle in time t seconds. Then,
f = dv/dt = d(3t2 – 4t + 5)/dt
= 6t – 4   ……….(1)
Therefore, the acceleration of the particle at the end of 3 seconds,
= [f]t = 3 = (6*3 – 4) cm/sec2
= 14cm/sec2

9. A particle moves along the straight-line OX, starting from O with a velocity 4 cm/sec. At time t seconds its acceleration is (5 + 6t) cm/sec2. What will be the velocity of the particle from O after 4 seconds?
a) 70 cm/sec
b) 71 cm/sec
c) 72 cm/sec
d) 73 cm/sec
Answer: c
Clarification: Let, vcm/sec be the velocity and x cm be the distance of the particle from O and time t seconds.
Then the velocity of the particle at time t seconds is, v = dx/dt
By the question, dv/dt = 5 + 6t
Or dv = (5 + 6t) dt
Or ∫dv = ∫(5 + 6t) dt
Or v = 5t + 6*(t2)/2 + A   ……….(1)
By question v = 4, when t = 0;
Hence, from (1) we get, A = 4.
Thus, v = dx/dt = 5t + 3(t2) + 4  ……….(2)
Thus, velocity of the particle after 4 seconds,
= [v]t = 4 = (5*4 + 3*42 + 4) [putting t = 4 in (2)]
= 72 cm/sec.

10. A particle moves along the straight-line OX, starting from O with a velocity 4 cm/sec. At time t seconds its acceleration is (5 + 6t) cm/sec2. What will be the distance from O after 4 seconds?
a) 110 cm
b) 120 cm
c) 130 cm
d) 140 cm
Answer: b
Clarification: Let, vcm/sec be the velocity and x cm be the distance of the particle from O and time t seconds.
Then the velocity of the particle and acceleration at time t seconds is, v = dx/dt and dv/dt respectively.
By the question, dv/dt = 5 + 6t
Or dv = (5 + 6t) dt
Or ∫dv = ∫(5 + 6t) dt
Or v = 5t + 6*(t2)/2 + A   ……….(1)
By question v = 4, when t = 0;
Hence, from (1) we get, A = 4.
Thus, v = dx/dt = 5t + 3(t2) + 4   ……….(2)
Or ∫dx = ∫(5t + 3(t2) + 4) dt
Or x = 5t2/2 + t3 + 4t + B   ……….(3)
By question x = 0, when t = 0;
Hence, from (3) we get, B = 0
Thus, x = 5t2/2 + t3 + 4t
Thus, distance of the particle after 4 seconds,
= [x]t = 4 = (5/2*42 + 43 + 4*4)  [putting t = 4 in (4)]
= 120 cm.

11. A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?
a) 1/v = 1/u + kx
b) 1/v = 1/u – 2kx
c) 1/v = 1/u – kx
d) 1/v = 1/u + 2kx
Answer: a
Clarification: Since the particle is moving in a straight line under a retardation kv3, hence, we have,
dv/dt = -kv3  ……….(1)
Or dv/dx*dx/dt = -kv3
Or v(dv/dx) = -kv3  [as, dx/dt = v]
Or ∫v-2 dv = -k∫dx
Or v-2+1/(-2 + 1) = -kx – B, where B is a integration constant
Or 1/v = kx + B   ……….(3)
Given, v = u, when x = 0; hence, from (3) we get, B = 1/u
Thus, putting B = 1/u in (3) we get,
1/v = 1/u + kx.

12. The distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax2 + bx + c (a, b, c are positive constants). If v be the velocity of the particle at the instant, then which one is correct?
a) Moves with retardation 2av2
b) Moves with retardation 2av3
c) Moves with acceleration 2av3
d) Moves with acceleration 2av2
Answer: b
Clarification: We have, t = ax2 + bx + c  ……….(1)
Differentiating both sides of (1) with respect to x we get,
dt/dx = d(ax2 + bx + c)/dx = 2ax + b
Thus, v = velocity of the particle at time t
= dx/dt = 1/(dt/dx) = 1/(2ax + b) = (2ax + b)-1  ……….(2)
Thus, acceleration of the particle at time t is,
= dv/dt = d((2ax + b)-1)/dt
= -1/(2ax + b)2 * 2av
= -v2*2av   [as, v = 1/(2ax + b)]
= -2av3
That is the particle is moving with retardation 2av3.

13. Two straight railway lines meet at right angles. A train starts from the junction along one line and at the same time instant, another train starts towards the junction from a station on the other line and they move at the same uniform velocity.When will they be nearest to each other?
a) When they are equal distance from the junction
b) When they are in unequal distance from the junction
c) When they form a right angle at the junction
d) Data not sufficient
Answer: a
Clarification: Let OX and OY be two straight railway lines and they meet at O at right angles.
One train starts from the junction and moves with uniform velocity u km/hr along the line OY.
And at the same instant, another train starts towards the junction O from station A on the line OX with same uniform velocity u km/hr.
Let C and B be the position of the two trains on lines OY and OX respectively after t hours from the start.
Then OC = AB = ut km. Join BC and let OA = a km and BC = x km.
Then OB = a – ut.
Now, from the right angled triangle BOC we get,
BC2 = OB2 + OC2
Or x2 = (a – ut)2 + (ut)2
Thus, d(x2)/dt = 2(a – ut)(-u) + u2(2t)
And d2(x2)/dt2 = 2u2 + 2u2 = 4u2
For maximum or minimum value of x2(i.e., x) we must have,
d(x2)/dt = 0
Or 2(a – ut)(-u) + u2(2t) = 0
Or 2ut = a   [Since u ≠ 0]
Or t = a/2u
Again at t = a/2u we have, d2(x2)/dt2 = 4u2 > 0
Therefore, x2(i.e., x) is minimum at t = a/2u
Now when t= a/2u, then OC = ut = u(a/2u) = a/2 and OB =a – ut = a – u(a/2u) = a/2 that is at t = a/2u we have, OC = OB.
Therefore, the trains are nearest to each other when they are equally distant from the junction.

14. A particle starts moving from rest with an acceleration in a fixed direction. If its acceleration at time t be(a – bt2),where a and b are positive constants then which one is correct?
a) [v]max = 4a√a/3√b
b) [v]max = 2a√a/3√b
c) [v]max = 2a√a/3√b
d) [v]max = 4a√a/3√b
Answer: b
Clarification: If v be the velocity of the moving particle at time t then its acceleration at time t will be dv/dt. By question,
dv/dt = a – bt2
Integrating we get, v = ∫ a – bt2 dt = at – bt3/3 + k   ……….(1)
where k is constant of integration.
Given, v = 0, when t = 0; hence from (1) we get,
0 = a(0) – b/3(0) + k
Or k = 0
Thus, v = at – bt3/3   ……….(2)
Again, d2v/dt2 = d(a – bt2)/dt = -2bt
Now, for minimum or maximum value of v we have,
dv/dt = 0
Or a – bt2 = 0
Or t2 = a/b
Or t = √a/√b   [Since t > 0 and a, b are positive constants]
At t = √a/√b we have d2v/dt2 = -2b(√a/√b) < 0 [Since t < 0 and a, b are positive constants]
Putting t = √a/√b in (2),
We find, v is maximum at t = √a/√b and the minimum value of v is,
[v]max = 2a√a/3√b.