Mathematics Test - 18

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Mathematics Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

If \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}\) are unit vectors such that the vector \(\overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}} \quad\) is perpendicular to \(7 \overrightarrow{\mathrm{a}}-5 \overrightarrow{\mathrm{b}}\) and \(\overrightarrow{\mathrm{a}}-4 \overrightarrow{\mathrm{b}} \quad\) is
perpendicular to \(7 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}},\) then the angle between \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\) is

A
π / 6

B
π / 2

C
π / 3

D
such \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}\) are non existent

Solution

Let \thetabe the angle between \({ }_{\mathrm{a}} \hat{\mathrm{x}} \overrightarrow{\mathrm{b}}\). \((\vec{a}+3 \vec{b}) \perp(\overrightarrow{7 a}-5 \vec{b})\)
\((\vec{a}+3 \vec{b}) \cdot(7 \vec{a}-5 \vec{b})=0\)
\(7|a|^{2}+16(i .6)-15|b|^{2}=0\)
\(7+16 \cos \theta-15=0\)
\(\cos \theta=\frac{1}{2}\)
\(\theta=\frac{\pi}{3}\)
And, \((\vec{a}-4 \vec{b}) \perp \overrightarrow{7 a}-2 \vec{b}\)
\((\vec{a}-4 \vec{b}) \cdot(7 \vec{a}-2 \vec{b})=0\)
\(7|a|^{2}+8|b|^{2}-30(i, b)=0\)
\(15-30 \cos \theta=0 \cos \theta=\frac{1}{2} \theta=\frac{\pi}{3}\)
Hence, the correct option is C.
Question 2
1 / -0

Let \(\mathrm{a}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}} \& \dot{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \cdot\) Then, the value of \(\lambda\) for which the vector \(\overrightarrow{\mathrm{c}}=\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+(2 \lambda-1 \hat{\mathrm{k}}\) is parallel to the plane
containing \(\overrightarrow{\mathrm{a}} \& \overrightarrow{\mathrm{b}},\) is

A
1

B
0

C
-1

D
2

Solution

We are given that \(\overline{\mathrm{c}}\) is parallel to the plane containing \(\bar{a}\) and \(\bar{b}\) The plane containing vectors a and b can be \((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})\). Then, \(\overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=0\)
\(\Rightarrow[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=0\)
\(\Rightarrow\left|\begin{array}{ccc}2 & 3 & -1 \\ 1 & -2 & 3 \\ \lambda & 1 & 2 \lambda-1\end{array}\right|=0\)
\(\Rightarrow 2(-4 \lambda+2-3)-3(2 \lambda-1-3 \lambda)-1(1+2 \lambda)=0\)
\(\Rightarrow 2(-4 \lambda-1)-3(-\lambda-1)-1-2 \lambda=0\)
\(\Rightarrow-8 \lambda-2+3 \lambda+3-1-2 \lambda=0\)
\(\Rightarrow-7 \lambda=0\)
\(\Rightarrow \lambda=0\)
Hence, the correct option is B.
Question 3
1 / -0

The plane passing through the point (−2, −2, 2) and containing the line joining the points,,,,
(1, 1, 1) and (1, −1, 2) makes intercepts on the coordinates axes the sum of whose lengths is

A
3

B
4/3

C
4

D
16/3

Solution

Equation of any plane passing through (−2, −2, 2) is
a(x + 2) + b(y + 2) + c(z – 2) = 0
Since it contains the line joining the points (1, 1, 1) and (1, −1, 2), it contains these points as well
so that
3a + 3b – c = 0
and 3a + b + 0 = 0
Solving we get
\(\frac{a}{1}=\frac{b}{-3}=\frac{c}{0}\)
and thus the equation of the plane is \((x+2)-3(y+2)=0\)
\(x+3 y-4=0\)
\(\frac{x}{4}+\frac{y}{4 / 3}=1\)
Intercepts on axes = 4, 4/3
The required sum= 4+(4/3)=16/3
Hence, the correct option is D.
Question 4
1 / -0

All the values of x satisfying the equation \(\left|\begin{array}{lll}x+1 & x+2 & 1 \\ x+2 & x+4 & 2 \\ x+3 & x+6 & 3\end{array}\right|\)= 0 are given by,

A
x = 0, 1

B
x = −1 or 1

C
\(x \in(-\infty, \infty)\)

D
None of these

Solution

Using \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\)we get \(\left|\begin{array}{lll}x+1 & 1 & 1 \\ x+2 & 2 & 2 \\ x+3 & 3 & 3\end{array}\right|\) =0
If any two columns of the determinant is identical, then the determinant is always zero.
Thus, all real values of x are possible.
Hence, the correct option is C.
Question 5
1 / -0

The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy = 0 is

A
log (2x + y – 1) = C + x + y

B
log (4x + 2y – 1) = C + 2x + y

C
log (2x + y + 1) + x + y = C

D
log (2x + y – 1) + x + 2y = C

Solution

Put \(2 x+y=x \Rightarrow 2+\frac{d y}{d x}=\frac{d X}{d x}\). Therefore, the given equation is reduced to \(\frac{d X}{d x}-2=-\frac{X+1}{2 X-1} \Rightarrow \frac{d X}{d x}=\frac{3(X-1)}{2 X-1}\)
\(\Rightarrow \frac{2 \mathrm{X}-1}{3(\mathrm{X}-1)} \mathrm{dX}=\mathrm{dx} \Rightarrow \frac{1}{3}\left[2+\frac{1}{\mathrm{X}-1}\right] \mathrm{dX}=\mathrm{dx}\)
\(\frac{1}{3}[2 \mathrm{X}+\log (\mathrm{X}-1)]=\times+\) constant
\(2(2 x+y)+\log (2 x+y-1)=3 x+\) constant
\(x+2 y+\log (2 x+y-1)=C\)
Hence, the correct option is D.
Question 6
1 / -0

One angle of a triangle is 2x/3 grad, another is 3x/2 degrees, whilst the third is πx/75 radians. Express all angles in degress.

A
Hence three angles of the triangle are 43^o,30^o ,30^o

B
Hence three angles of the triangle are 24^o,60^o, 96^o

C
Hence three angles of the triangle are 74^o,27^o, 98^o

D
Hence three angles of the triangle are 30^o, 60^o, 90^o

Solution

\(\frac{2}{3} x^{g}=\frac{2}{3} x\left(\frac{9}{10}\right)=\frac{3}{5} x^{0}\)
And \(\frac{\pi x^{c}}{75}=\frac{\pi x}{75} \times \frac{180^{\circ}}{\pi}=\frac{12 x^{\circ}}{5}\)
But \(\frac{3}{5} x^{0}+\frac{3}{2} x^{0}+\frac{12}{5} x^{0}=180^{\circ}\)
\(\therefore 6 x^{0}+15 x^{0}+24 x^{0}=1800\)
\(45 x^{0}=1800\)
\(x=40^{\circ}\)
Hence, the correct option is B.
Question 7
1 / -0

If \(\tan x=\frac{4}{3},\) then the value of \(\sqrt{\frac{(1-\sin x)(1+\sin x)}{(1+\cos x)(1-\cos x)}}\) is

A
9/16

B
3/4

C
4/3

D
16/9

Solution

\(\sqrt{\frac{(1-\sin x)(1+\sin x)}{(1+\cos x)(1-\cos x)}}\)
\(=\sqrt{\frac{1-\sin ^{2} x}{1-\cos ^{2} x}} \ldots \ldots\left(\because(a-b)(a+b)=a^{2}-b^{2}\right)\)
\(=\sqrt{\frac{\cos ^{2} x}{\sin ^{2} x}} \ldots \ldots\left(\because \sin ^{2} x+\cos ^{2} x=1\right)\)
\(=\frac{\cos x}{\sin x}\)
\(=\cot \theta\)
\(=\frac{1}{\tan \theta}\)
\(=\frac{1}{\frac{4}{3}}=\frac{3}{4}\)
Hence, the correct option is B.
Question 8
1 / -0

\(\mathrm{f} y=\log \left(\frac{\sqrt{(x+1)}-1}{\sqrt{(x+1)}+1}\right)+\frac{\sqrt{x}}{\sqrt{(x+1)}},\) then by using
substitution \(x=\tan ^{2} \theta, y\) reduces to

A
log tan²⁽θ/4)+sinθ

B
log tan²(θ/2)+sinθ

C
log tan²θ+sinθ

D
log tan²(θ/2)+sin(θ/2)

Solution

Given \(y=\log \left(\frac{\sqrt{(x+1)}-1}{\sqrt{(x+1)}+1}\right)+\frac{\sqrt{x}}{\sqrt{(x+1)}}\)
Using substitution, \(x=\tan ^{2} \theta,\) we get
\(\therefore y=\log \left(\frac{\sec \theta-1}{\sec \theta+1}\right)+\frac{\tan \theta}{\sec \theta}\)
\(=\log \left(\frac{1-\cos \theta}{1+\cos \theta}\right)+\sin \theta\)
\(=\log \left(\tan ^{2} \frac{\theta}{2}\right)+\sin \theta\)
Hence, the correct option is B.
Question 9
1 / -0

Find the value of sin∠M in the right angled triangle shown in the figure.

A
10/12

B
12/10

C
10/√44

D
√44/12

Solution

In a right-angled triangle with angle \(L=90^{\circ}\) Using Pythagorean theorem \((M K)^{2}=(M L)^{2}+(K L)^{2}\)
\((K L)^{2}=(M K)^{2}-(M L)^{2}\)
\(K L=\sqrt{(M K)^{2}-(M L)^{2}}\)
Put the value from the figure, we get \(K L=\sqrt{12^{2}-10^{2}}\)
\(K L=\sqrt{144-100}\)
\(K L=\sqrt{44}\)
\(\sin M=\frac{\text {Perpendicular}}{\text {Hypotenuse}}=\frac{K L}{M K}=\frac{\sqrt{44}}{12}\)
Hence, the correct option is D.
Question 10
1 / -0

(cosec θ−sinθ)(secθ−cosθ)(tanθ+cotθ)=.......

A
0

B
1

C
-1

D
None of these

Solution

\((\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)\)
\(=\left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{1}{\cos \theta}-\cos \theta\right)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)\)
using basic identities \(=\left(\frac{1-\sin ^{2} \theta}{\sin \theta}\right)\left(\frac{1-\cos ^{2} \theta}{\cos \theta}\right)\left(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cdot \cos \theta}\right)\)
\(=\frac{\cos ^{2} \theta}{\sin \theta} \times \frac{\sin ^{2} \theta}{\cos \theta} \times \frac{1}{\sin \theta \cdot \cos \theta}=1 \times 1 \times 1=1\)
Hence, the correct option is B.

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Mathematics Test - 18

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Mathematics Test - 18

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Question 1

π / 6

π / 2

π / 3

such \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}\) are non existent

Solution

Question 2

1

0

-1

2

Solution

Question 3

3

4/3

4

16/3

Solution

Question 4

x = 0, 1

x = −1 or 1

\(x \in(-\infty, \infty)\)

None of these

Solution

Question 5

log (2x + y – 1) = C + x + y

log (4x + 2y – 1) = C + 2x + y

log (2x + y + 1) + x + y = C

log (2x + y – 1) + x + 2y = C

Solution

Question 6

Hence three angles of the triangle are 43o,30o ,30o

Hence three angles of the triangle are 24o,60o, 96o

Hence three angles of the triangle are 74o,27o, 98o

Hence three angles of the triangle are 30o, 60o, 90o

Solution

Question 7

9/16

3/4

4/3

16/9

Solution

Question 8

log tan2(θ/4)+sinθ

log tan2(θ/2)+sinθ

log tan2θ+sinθ

log tan2(θ/2)+sin(θ/2)

Solution

Question 9

10/12

12/10

10/√44

√44/12

Solution

Question 10

0

1

-1

None of these

Solution

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Hence three angles of the triangle are 43^o,30^o ,30^o

Hence three angles of the triangle are 24^o,60^o, 96^o

Hence three angles of the triangle are 74^o,27^o, 98^o

Hence three angles of the triangle are 30^o, 60^o, 90^o

log tan²⁽θ/4)+sinθ

log tan²(θ/2)+sinθ

log tan²θ+sinθ

log tan²(θ/2)+sin(θ/2)