Mathematics Test

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Mathematics Test - 1

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Question 1
3 / -1

If for the non-singular matrix \(A, A^{2}=I\), then find \(A^{-1}\).

A
A

B
3

C
2

D
1

Solution

Given, \(A^{2}=I\)
Since, \(A\) is non-singular matrix.
\(\therefore|A| \neq 0\), so, \(A^{-1}\) exists.
Multiplying by \(A^{-1}\) on both sides, we get:
\(\Rightarrow A^{-1}\left(A^{2}\right)=A^{-1}(I)\)
\(\Rightarrow \left(A^{-1} A\right) A=A^{-1}\)
\( \Rightarrow I A=A^{-1} \quad\left(\because A^{-1} A=I\right)\)
\(\therefore A^{-1}=A \quad(\because I A=A)\)
Question 2
3 / -1

Let P(x₁, y₁) and Q(x₂, y₂), y₁ < 0, y₂ < 0 be the end points of the latus rectum of the ellipse x₂ + 4y₂ = 4. The equation(s) of parabolas with latus rectum PQ is/are-

A
\(x^{2}+2 \sqrt{3 y}=3+\sqrt{3}\)

B
\(x^{2}-2 \sqrt{3} y=3+\sqrt{3}\)

C

D
B and C are Correct

Solution

\(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)

\(b^{2}=a^{2}\left(1-e^{2}\right)\)

\(\Rightarrow \mathrm{e}=\frac{\sqrt{3}}{2}\)

\(\Rightarrow \mathrm{P}\left(\sqrt{3},-\frac{1}{2}\right)\) and \(\mathrm{Q}\left(-\sqrt{3},-\frac{1}{2}\right)\) (given \(\mathrm{y}_{1}\) and \(\mathrm{y}_{2}\) less than \(\left.0\right)\)

Co-ordinates of mid-point of \(\mathrm{PQ}\) are \(\mathbf{R} \equiv\left(0,-\frac{1}{2}\right)\)

\(\mathrm{PQ}=2 \sqrt{3}=\) length of latus rectum.

\(\Rightarrow\) Two parabolas are possible whose vertices are \(\left(0,-\frac{\sqrt{3}}{2}-\frac{1}{2}\right)\) and \(\left(0, \frac{\sqrt{3}}{2}-\frac{1}{2}\right)\)

Hence, the equations of the parabolas are \(x^{2}-2 \sqrt{3} y=3+\sqrt{3}\) and \(x^{2}+2 \sqrt{3} y=3-\sqrt{3}\)

Hence options B is the correct answer.
Question 3
3 / -1

Let \(a, b, c\) are the Arithmetic means between two numbers such that \(a+b+c=15\) and \(p, q, r\) be Harmonic mean between same numbers such that \(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{5}{3}\) then the numbers are

A
12, 1

B
8, 2

C
9, 1

D
6, 4

Solution

Let numbers be \(x, y\) \(3\left(\frac{x+y}{2}\right)=a+b+c=15\)
\(x+y=10\)
\(\frac{3}{2}\left(\frac{1}{x}+\frac{1}{y}\right)=\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right)=\frac{5}{3}\)
\(\frac{1}{x}+\frac{1}{y}=\frac{10}{9}\)
\(x=9, y=1\)
Hence, The correct answer is C.
Question 4
3 / -1

The number of values of \(\lambda\) for which the given system of equations has No solution
\(x+2 y+3 z=5\)
\(4 x+6 y+2 \lambda z=21\)
\(x+\lambda y+3 z=3\)

A
2

B
0

C
1

D
Infinite

Solution

When \(\Delta=\left|\begin{array}{ccc}1 & 2 & 3 \\ 4 & 6 & 2 \lambda \\ 1 & \lambda & 3\end{array}\right|=0\)
there can be two possibilities
(i) No solution
(ii) Infinite solution
But for Infinite solution
\(\Delta 1\) is the determinant obtained by replacing the third column of \(\Delta\) by the elements of the constant matrix.
\(\Delta_{1}=\left|\begin{array}{lll}1 & 2 & 5 \\ 4 & 6 & 21 \\ 1 & \lambda & 3\end{array}\right|\)
Should also be zero
So, for no solution condition will be,
\(\Delta=0 \quad\) and \(\quad \Delta_{1} \neq 0\)
\(\Delta=0 \Rightarrow\left|\begin{array}{ccc}1 & 2 & 3 \\ 4 & 6 & 2 \lambda \\ 1 & 2 & 3\end{array}\right|=0\)
\(\Rightarrow 1\left(18-2 \lambda^{2}\right)-2(12-2 \lambda)+3(4 \lambda-6)=0\)
\(\Rightarrow \quad \lambda^{2}-8 \lambda+12=0\)
\(\lambda=2,6\)
\(\Delta_{1}=0 \Rightarrow\left|\begin{array}{lll}1 & 2 & 5 \\ 4 & 6 & 21 \\ 1 & 2 & 3\end{array}\right|=0\)
\(\Rightarrow 1(18-21 \lambda)-2(12-21)+5(4 \lambda-6)=0\)
\(\Rightarrow \quad-\lambda,+6=0\)
\(2=6\)
So, \(\lambda\) can have only one value
Hence, The correct answer is C.
Question 5
3 / -1

In a triangle ΔPQR, cos(P - R) cos(Q) + cos(2Q) = 0

Which of the following options is/are correct?

A
sin² P + sin² R = 2 sin²Q

B
p² and q² are in AP.

C
sin P sin R = sin² Q

D
A and B are correct

Solution

Consider the expression. \(\cos (\boldsymbol{P}-\boldsymbol{R}) \cos (Q)+\cos (2 Q)=0\)

\(\cos (P-R) \cos \left\lceil 180^{\circ}-(P+R)\right\rceil+1-2 \sin ^{2} Q=0\)

\(-\cos (P-R) \cos (P+R)+1-2 \sin ^{2} Q=0\)

\(-\cos ^{2} P+\sin ^{2} R+1-2 \sin ^{2} Q=0\)

\(-\left(1-\sin ^{2} P\right)+\sin ^{2} R+1-2 \sin ^{2} Q=0\)

\(\sin ^{2} P+\sin ^{2} R=2 \sin ^{2} Q\)

We know that

\(\frac{\sin P}{p}=\frac{\sin Q}{q}=\frac{\sin R}{r}=k\)

Therefore \(\sin P=p k, \sin Q=q k, \sin R=r k\)

\(p^{2} k^{2}+r^{2} k^{2}=2 q^{2} k^{2}\)

\(p^{2}+r^{2}=2 q^{2}\)

Therefore

\(p^{2}, q^{2}, r^{2}\) are in \(\mathrm{AP}\)

Hence options A is the correct answer.
Question 6
3 / -1

If \(p \rightarrow(p \wedge \sim q)\)
is false, then truth value of \(p\) and \(q\) are respectively

A
T, F

B
T, T

C
F, T

D
F, F

Solution

Truth table:
So, when \(p \rightarrow(p \wedge \sim q)\)
is false, \(\mathrm{p}\) and \(\mathrm{q}\) both are true 'T
Hence, The correct answer is B.
Question 7
3 / -1

If \((\mathrm{n}+1) !=12 \times(\mathrm{n}-1) !\), then the value of \(\mathrm{n}\) is?

A
1

B
2

C
3

D
4

Solution

We have:
\((n+1) !=12 \times(n-1) !\)
\(\Rightarrow(n+1) \times n \times(n-1) !=12 \times(n-1) !\)
\(\Rightarrow({n}+1) \times {n}=12\)
\(\Rightarrow {n}^{2}+{n}-12=0\)
\(\Rightarrow {n}^{2}+4 {n}-3 {n}-12=0\)
\(\Rightarrow n(n+4)-3(n+4)=0\)
\(\Rightarrow(n+4)(n-3)=0\)
\(\Rightarrow {n}=-4,3\)
Since, \(n\) has to be a natural number, \(n=3\)
Question 8
3 / -1

Find the equation of ellipse whose eccentricity is \(\frac{1}{2}\), length of latus rectum is 4 and the center is \((0,0)\).

A
\(9 \mathrm{x}^{2}+12 \mathrm{y}^{2}=64\)

B
\(9 \mathrm{x}^{2}+12 \mathrm{y}^{2}=16\)

C
\(12 \mathrm{x}^{2}+9 \mathrm{y}^{2}=64\)

D
\(12 \mathrm{x}^{2}+9 \mathrm{y}^{2}=16\)

Solution

Given eccentricity \((\mathrm{e})=\frac{1}{2}\)
\( \frac{\sqrt{a^{2}-b^{2}}}{a}=\frac{1}{2}\)[\(\because\)Eccentricity \(=\frac{\sqrt{\left({a}^{2}-\mathrm{b}^{2}\right)}}{{a}}\)]
\(\Rightarrow {a}^{2}-\mathrm{b}^{2}=\frac{{a}^{2}}{4}\)
\(\Rightarrow \mathrm{b}^{2}=\frac{3 a^{2}}{4}\)...(i)
Also Latus rectum length \(=4\)
\( \frac{2 \mathrm{~b}^{2}}{{a}}=4\) [\(\because\) Length of latus rectum \(=\frac{2 b^{2}}{a}\)]
\(\Rightarrow \mathrm{b}^{2}=\mathrm{2 a}\)...(ii)
Comparing (i) and (ii)
\( \frac{3 {a}^{2}}{4}=2 {a}\)
\(\Rightarrow {a}=\frac{8}{3}\)
So,
\(\mathrm{b}^{2}=2 {a}=\frac{16}{3}\)
And
\({a}^{2}=\frac{64}{9}\)
The equation of ellipse
\(\frac{\mathrm{x}^{2}}{{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\)
\(\Rightarrow 9 \mathrm{x}^{2}+12 \mathrm{y}^{2}=64\)
Question 9
3 / -1

If the mean and standard deviation of 10 observations are 20 and 2 respectively. The sum of squares of all the observations is—

A
2020

B
2000

C
4040

D
4000

Solution

\(\operatorname{Mean} x=\frac{\sum_{i=1}^{n} x_{i}}{n}\)
\(\Rightarrow 20=\frac{\sum_{i=1}^{10} x_{1}}{10}\)
\(\Rightarrow \sum_{i=1}^{10} x_{i}=200\)
\((S \text { tandard deviation })^{2}=\frac{\sum_{i=1}^{n}\left(x_{i}\right)^{2}}{n}-(x)^{2}\)
\(\Rightarrow(2)^{2}+400=\frac{1}{10}\left[\sum_{i=1}^{10}\left(x_{i}\right)^{2}\right]\)
\(\Rightarrow 4040=\sum_{i=1}^{10} x_{i}^{2}\)
\(\Rightarrow \sum_{i=1}^{10} x_{i}^{2}=4040\)
Hence, The correct answer is C.
Question 10
3 / -1

The internal bisector of ∠A of a triangle ABC meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and side AB at F. If a, b and c represent the sides of ΔABC, then-

A
AE is HM of b and c.

B
\(A D=\frac{2 b c}{b+c} \cos \frac{A}{2}\)

C
\(E F=\frac{4 b c}{b+c} \sin \frac{A}{2}\)

D
All are Correct

Solution

Given, internal angle bisector of A meets BC at D. we know length of AD is given by, \(A D=\frac{2 b c}{b+c} \cos \frac{A}{2}\) Given,

\(\triangle\) ADE is right angled triangle with right angle at \(D\).

\(\angle D=90^{\circ}, \quad \angle A E D=90-\frac{A}{2}\)

Using sine rule in \(\triangle\) ADE, we get

\(\frac{A D}{\sin \angle A E D}=\frac{A E}{\sin \angle A D E}=\frac{D E}{\sin \angle \frac{A}{2}}\)

\(\frac{A D}{\cos \frac{A}{2}}=A E=\frac{D E}{\sin \frac{A}{2}}\)

Substitute the value of AD in above equation, we get

\(A E=\frac{\frac{2 k}{b+c} \cos \frac{A}{2}}{\cos \frac{A}{2}}=\frac{2 b c}{b+c}\)

\(\mathrm{SO}, \mathrm{AE}\) is harmonic mean of \(\mathrm{b}\) and \(\mathrm{c}\). Consider \(\triangle\) ADE and \(\triangle\) ADF \(,\) we know that \(\angle A D E=\angle A D F=90^{\circ}\)

\(\angle D A E=\angle D A F=\angle \frac{A}{2}\)

\(\therefore \angle A E D=\angle A F D=90-\angle \frac{A}{2}\)

Hence \(\triangle\) AEF is isosceles triangle.

\(E F=2 D E=2 A E \sin \frac{A}{2}\)

\(E F=2 \times \frac{2 b c}{b+c} \sin \frac{A}{2}=\frac{4 b c}{b+c} \sin \frac{A}{2}\)

Hence options D is the correct answer.
Question 11
3 / -1

Find the equation(s) of common tangent(s) to y = x² and y = - x² + 4x - 4.

A
y = 4(x - 1)

B
y = -4(x - 1)

C
y = 5

D
y = -30x - 50

Solution

The equation of tangent to \(y=x^{2}\)

\(y=m x-\frac{m^{2}}{4}\)

Putting in \(y=-x^{2}+4 x-4\)

\(m x-\frac{m^{2}}{4}=-x^{2}+4 x-4\)

\(\Rightarrow x^{2}+x(m-4)+4-\frac{m^{2}}{4}=0\)

\(D=0\)

Now, \((m-4)^{2}-\left(16-m^{2}\right)=0\)

\(\Rightarrow 2 m(m-4)=0\)

\(\Rightarrow \mathbf{m}=0,4\)

\(y=0\) and \(y=4(x-1)\) are the required tangents. Correct answers: \(y=4(x-1), y=0\)

Hence option A is the correct answer.
Question 12
3 / -1

The value of \({ }^{21} \mathrm{C}_{1}+{ }^{21} \mathrm{C}_{2}+\ldots .21 \mathrm{C}_{10}\) is \(-\)

A
\(2^{21}\)

B
\(2^{20}\)

C
\(2^{21}-1\)

D
\(2^{20}-1\)

Solution

\({ }^{21} \mathrm{C}_{1}+{ }^{21} \mathrm{C}_{2}+{ }^{21} \mathrm{C}_{3}+\ldots .^{21} \mathrm{C}_{10}\)
\(=\frac{1}{2}\left\{2\left({ }^{21} \mathrm{C}_{1}+{ }^{21} \mathrm{C}_{2}+{ }^{21} \mathrm{C}_{3}+\ldots+{ }^{21} \mathrm{C}_{10}\right)\right\}\)
Now,
\({ }^{21} \mathrm{c}_{1}={ }^{21} \mathrm{C}_{20}\)
\({ }^{21} \mathrm{C}_{2}={ }^{21} \mathrm{C}_{19}\)
\({ }^{21} C_{10}={ }^{21} C_{11}\)
\(\mathrm{So},{ }^{21} \mathrm{C}_{1}+{ }^{21} \mathrm{C}_{2}+{ }^{21} \mathrm{C}_{3}+\ldots .^{21} \mathrm{C}_{10}\)
\(=\frac{1}{2}\left\{{ }^{21} \mathrm{C}_{1}+{ }^{21} \mathrm{C}_{2}+\ldots{ }_{10}^{2 \mathrm{C}}+{ }^{21} \mathrm{C}_{11}+{ }^{21} \mathrm{C}_{12}+\ldots .^{21} \mathrm{c}_{20}\right\}\)
\(=\frac{1}{2}\left\{\left(21 \mathrm{C}_{0}+{ }^{21} \mathrm{C}_{1}+{ }^{21} \mathrm{C}_{2}+\ldots . .^{21} \mathrm{C}_{20}+{ }^{2 \mathrm{c}} \mathrm{C}_{21}\right)-\left(21 \mathrm{C}_{0}+{ }^{21} \mathrm{C}_{21}\right)\right\}\)
\(=\frac{1}{2}\left\{2^{21}-2\right\}\)
\(\because\left({ }^{n} \mathrm{C}_{0}+{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{2}+\ldots \ldots{ }^{n} \mathrm{C}_{n-1}+{ }^{n} \mathrm{C}_{n}=2^{n}\right)\)
\(=2^{20}-1\)
Hence, The correct answer is D.
Question 13
3 / -1

The angles of elevation of top of a pole from two points \(A\) and \(B\) on the horizontal line lying on opposite side of the pole are observed to be \(30^{\circ}\) and \(60^{\circ}\) If \(A B=100 \mathrm{m}\) then height of the pole is

A
\(20 \sqrt{3} m\)

B
\(15 \sqrt{3} m\)

C
\(10 \sqrt{3} m\)

D
\(25 \sqrt{3} m\)

Solution

Let the height of pole be \(P Q=h\) m. And \(Q B=d\) be the distance between \(B\) and foot of pole. \(\tan 60^{\circ}=\frac{h}{d}\)
\(\Rightarrow d=\frac{h}{\sqrt{3}}\)
\(\tan 30^{\circ}=\frac{h}{100-d}\)
\(\Rightarrow-\frac{h}{\sqrt{3}}+100=\sqrt{3} h\)
\(\Rightarrow 100=h\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)\)
\(\Rightarrow h=25 \sqrt{3} \mathrm{m}\)
Hence, The correct answer is D.
Question 14
3 / -1

\(\int \frac{x^{2}-1}{\left(x^{2}+1\right)\left(x^{4}+1\right)^{1 / 2}} d x\)

A
\(\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)\)

B
\(\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)\)

C
\(\tan ^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)\)

D
\( \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)\)

Solution

\(I=\int \frac{x^{2}-1}{\left(x^{2}+1\right)\left(x^{4}+1\right)^{1 / 2}} d x\)
\(I=\int \frac{1-1 / x^{2}}{\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}\right)^{1 / 2}} d x\)
\(I=\int \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^{2}-2}} d x\)
Let \(x+\frac{1}{x}=t\)
\(\quad=d\left(x+\frac{1}{x}\right)=d t\)
\(\quad=\left(1-\frac{1}{x^{2}}\right) d x=d t\)
\(I=\int \frac{d t}{t \sqrt{t^{2}-2}}=\int \frac{d t}{t \sqrt{t^{2}-(\sqrt{2})^{2}}}\)
\(=\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{t}{\sqrt{2}}\right)\left\{\because \int \frac{d x}{x \sqrt{x^{2}-a^{2}}}=\frac{1}{2} \sec ^{-1} \frac{x}{2}\right)\)
\(=\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)\)
Hence, The correct answer is A.
Question 15
3 / -1

Find the number of four digit numbers that are less than 2000 that can be formed with the digits 1,2,3,4 such that each digit can be repeated any number of times

A
24

B
25

C
64

D
65

Solution

Given that the numbers should be less than 2000 , so the first digit can only be filled by 1 The other three digits can take any of the four given values in 4 ways each
\(\therefore\) Required number of 4 -digit numbers \(=1 \times 4^{3}=64\)
Hence, The correct answer is C.

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Mathematics Test - 1

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Mathematics Test - 1

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Question 1

A

3

2

1

Solution

Question 2

\(x^{2}+2 \sqrt{3 y}=3+\sqrt{3}\)

\(x^{2}-2 \sqrt{3} y=3+\sqrt{3}\)

B and C are Correct

Solution

Question 3

12, 1

8, 2

9, 1

6, 4

Solution

Question 4

2

0

1

Infinite

Solution

Question 5

sin2 P + sin2 R = 2 sin2Q

p2 and q2 are in AP.

sin P sin R = sin2 Q

A and B are correct

Solution

Question 6

T, F

T, T

F, T

F, F

Solution

Question 7

1

2

3

4

Solution

Question 8

\(9 \mathrm{x}^{2}+12 \mathrm{y}^{2}=64\)

\(9 \mathrm{x}^{2}+12 \mathrm{y}^{2}=16\)

\(12 \mathrm{x}^{2}+9 \mathrm{y}^{2}=64\)

\(12 \mathrm{x}^{2}+9 \mathrm{y}^{2}=16\)

Solution

Question 9

2020

2000

4040

4000

Solution

Question 10

AE is HM of b and c.

\(A D=\frac{2 b c}{b+c} \cos \frac{A}{2}\)

\(E F=\frac{4 b c}{b+c} \sin \frac{A}{2}\)

All are Correct

Solution

Question 11

y = 4(x - 1)

y = -4(x - 1)

y = 5

y = -30x - 50

Solution

Question 12

\(2^{21}\)

\(2^{20}\)

\(2^{21}-1\)

\(2^{20}-1\)

Solution

Question 13

sin² P + sin² R = 2 sin²Q

p² and q² are in AP.

sin P sin R = sin² Q