Mathematics Test

Result

Mathematics Test - 14

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is

A
3100

B
3300

C
2900

D
1400

Solution

n(A) = 40% of 10,000 = 4,000

n(B) = 20% of 10,000 = 2,000

n(C) = 10% of 10,000 = 1,000

n(A ∩ B) = 5% of 10,000 = 500

n(B ∩ C) = 3% of 10,000 = 300

n(C ∩ A) = 4% of 10,000 = 400

n(A ∩ B ∩ C) = 2% of 10,000 = 200

We want to find the number of families which buy only A = n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B ∩ C)]

=4000 - [500 + 400 - 200] = 4000 - 700 = 3300
Question 2
4 / -1

If the vertices P, Q, R of a triangle PQR are rational points, then which of the following points of the triangle PQR is not always a rational point?

A
Centroid

B
Incentre

C
Circumcentre

D
Orthocentre

Solution

Incentre of is not ΔPQR always a rational point.

Hence option B is correct.
Question 3
4 / -1

\(\int \sqrt{\frac{\cos x-\cos ^{3} x}{1-\cos ^{3} x}} d x\) is equal to

A
(2/3) sin^-1 (cos^3/2 x) + C

B
(3/2) sin^-1(cos^3/2 x) + C

C
(2/3) cos^-1 (cos^3/2 x) + C

D
None of these

Solution

\(\int \sqrt{\frac{\cos x-\cos ^{3} x}{1-\cos ^{3} x}} d x=\int \sqrt{\frac{\cos x}{1-\cos ^{3} x}} \sin x d x\)
Put \(\cos x=t-\sin x d x=d t\)
\(=-\int_{\sqrt{1-t^{3}}}^{t} d t=-\int \frac{\sqrt{t}}{\sqrt{1-\left(t^{3 / 2}\right)^{2}}} d t\)
\(=-(2 / 3) \int \frac{3 / 2 \sqrt{t}}{\sqrt{1-\left(t^{3 / 2}\right)^{2}}} d t=(2 / 3) \cos ^{-1}\left(t^{3 / 2}\right)+C\)
\(=(2 / 3) \cos ^{-1}\left(\cos ^{3 / 2} x\right)+C\)
Question 4
4 / -1

If P(1, 0), Q(– 1, 0) and R(2, 0) are three given points, then the locus of point S, which satisfies the relation (SQ)² + (SR)² = 2(SP)², is (where SQ, SR and SP are line segments)

A
a straight line || to x-axis

B
a circle through the origin

C
a circle with centre at the origin

D
a straight line || to y-axis

Solution

Let the point be (h, k)

So, SQ² = (h + 1)² + k²

SR² = (h - 2)² + k²

SP² = (h - 1)² + k²

Substituting in the given equation and replacing the values of h and k by x and y, we get

x = - 3/2 (A straight line parallel to y-axis).
Question 5
4 / -1

If

L(X, Y) = Square of the larger number,

S(X, Y) = Square of the smaller number,

MODD (X, Y) = Absolute value of the difference (X - Y) and

X * Y = Product of X and Y,

then the value of MODD [{L (1, 2) * S(1, - 3)}, {S(0, 2) * MODD (3, - 4)}] *MODD[{S(3, -1) * L(1, - 2)}, { MODD (1, 2) * L(1, 2)}] will be

A
30

B
108

C
241

D
50

Solution

\(L(1,2) \cdot S(1,-3)=4^{*}(-3)^{2}-4^{*} 9-36\)
S(0.2) \(\cdot M O D D(3,-4)=0 \cdot 7=0\)
\(\Rightarrow \mathrm{MODD}[36, \mathrm{Ol}=36\)
\(S(3-1) \cdot L(1,-2)=1^{2} \cdot 1=1\)
MODD (1, 2) \(^{*} L(1,2)=1^{*} 4-4\)
\(\Rightarrow M O D D|1,4|-3\)
\(\Rightarrow M O D D \| L(1,2) \cdot S(1,-3) I, I S(0,2) \cdot M O D D(3,-4)]\)
\(\because[3]=108\)
Question 6
4 / -1

The number of arbitrary constants in the solution of a differential equation of degree 2 and order 3 is

A
2

B
3

C
2³

D
1

Solution

Note: The number of arbitrary constants in a solution of a differential equation of order n is equal to its order.

Order of differential equation = Number of arbitrary constants = 3
Question 7
4 / -1

If A and B are two given sets, then A ∩(A∩B)^cis equal to

A
A

B
B

C
∅

D
A ∩ (B^c).

Solution

\(\begin{aligned}\left.\mathrm{A} \cap(A \cap B)^{\complement}\right)=& A \cap\left(A^{c} \cup B^{\circ}\right) \\ &=\left(A \cap\left(A^{c}\right) \cup\left(A \cap\left(B^{c}\right)\right.\right.\\ &=\emptyset \cup\left(A \cap\left(B^{c}\right)=A \cap\left(B^{c}\right)\right.\end{aligned}\)
Question 8
4 / -1

Find the period of the given function:

f(x) = tan(x/3) + sin(2x)

A
π/2

B
π

C
3π

D
3π/2

Solution

Given, f(x) = tan(x/3) + sin(2x).

Here, tan(x/3) is periodic with period 3π and sin2x is periodic with period π.
Hence, f(x) will be periodic with period 3π.
Question 9
4 / -1

If f(x) = log (1 + x) - log (1 - x) and log (g(x)) = log (3x + x3) - log (1 + 3x2), then what is the value of f(g(x))?

A
g(f(x))

B
2g(f(x))

C
3g(x)

D
3f(x)

Solution

\(f(g(x))=f\left(\frac{3 x+x^{3}}{1+3 x^{2}}\right)=\log \frac{1+\frac{3 x+x^{2}}{1+3 x^{2}}}{1-\frac{3 x+x^{2}}{1+3 x^{2}}}=\log \frac{1+3 x^{2}+3 x+x^{3}}{1+3 x^{2}-3 x-x^{3}}\)
\(\log \left(\frac{1+x}{1-x}\right)^{3}-3 \log \left(\frac{1+x}{1-x}\right)=3 f(x)\)
Question 10
4 / -1

The distance of the point \((3.8 .2)\) from the line \(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3},\) measured parallel to the plane \(3 x+2 y\) \(-2 z+15=0\) is-

A
1

B
2

C
5

D
7

Solution

Let equation of the line through \(P(3.8,2)\) and parallel to the plane \(3 x+2 y-2 z+15=0\) be
\(\frac{x-3}{A}=\frac{y-8}{B}=\frac{z-2}{C}\)
Then. \(3 A+2 B-2 C=0-(i i)\)
Let line (i) intersect \(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}\) at \(Q\)
then \(\left|\begin{array}{ccc}x_{1}-x_{2} & y_{1}-y_{2} & z_{1}-z_{2} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|=0\)
\(\Rightarrow\left|\begin{array}{ccc}3-1 & 8-3 & 2-2 \\ A & B & C \\ 2 & 4 & 3\end{array}\right|=0\)
\(\Rightarrow 15 \mathrm{A}-6 \mathrm{B}-2 \mathrm{C}=0\)
From equation (ii) and equation (iii)

\(\Rightarrow \frac{A}{2}=\frac{B}{3}=\frac{C}{6}\)
Substituting A, B and C in equation (i)
\(\frac{x-3}{2}=\frac{y-8}{3}=\frac{z-2}{6}\)
This is the equation of line PQ parallel to the plane \(3 x\). \(+2 y-2 z+15=0\)
Now, ' \(Q\) ' is the point where this line cuts the line \(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}\)
So. point ' \(\mathrm{Q}^{\prime}\) will satisfy this line also.
Any point ' \(Q\) ' on line

\(\frac{x-3}{2}=\frac{y-8}{3}=\frac{z-2}{6}=\lambda\)
is \(Q(2 \lambda+3, \quad 3 \lambda+8,6 \lambda+2)\)
When this point also lies on \(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}\)
\(\Rightarrow \frac{(2 \lambda+3)-1}{2}=\frac{3 \lambda+8-3}{4}=\frac{6 \lambda+2-2}{3}\)
\(\Rightarrow \lambda=1\)
Point \(Q=(5.11,8)\)
length of \(P Q=\sqrt{(5-3)^{2}+(11-8)^{2}+(8-2)^{2}}\)
\(=\sqrt{4+9+36}\)
\(=7\) unit

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mathematics Test - 14

Result

Mathematics Test - 14

Score

-

Rank

-

SHARING IS CARING

Question 1

3100

3300

2900

1400

Solution

Question 2

Centroid

Incentre

Circumcentre

Orthocentre

Solution

Question 3

(2/3) sin-1 (cos3/2 x) + C

(3/2) sin-1(cos3/2 x) + C

(2/3) cos-1 (cos3/2 x) + C

None of these

Solution

Question 4

a straight line || to x-axis

a circle through the origin

a circle with centre at the origin

a straight line || to y-axis

Solution

Question 5

30

108

241

50

Solution

Question 6

2

3

23

1

Solution

Question 7

A

B

∅

A ∩ (Bc).

Solution

Question 8

π/2

π

3π

3π/2

Solution

Question 9

g(f(x))

2g(f(x))

3g(x)

3f(x)

Solution

Question 10

1

2

5

7

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

(2/3) sin^-1 (cos^3/2 x) + C

(3/2) sin^-1(cos^3/2 x) + C

(2/3) cos^-1 (cos^3/2 x) + C

2³

A ∩ (B^c).