Mathematics Test

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Mathematics Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

A ray of light along x+√3y=√3 gets reflected upon reaching x-axis. Equation of the reflected rays is

A
√3 y=x−√3

B
y=√3x−√3

C
√3 y=x−1

D
y=x+√3

Solution

\(x+\sqrt{3} y=\sqrt{3}\)
\(y=\frac{-1}{\sqrt{3}} x+1\)
Hence the line makes an angle of \(-30^{0}\) with respect to \(\mathrm{x}\) axis.
Hence the reflected line will make a \(+30^{0}\) angle with the x axis. Thus the equation of the line will be of the form \(y=\frac{x}{\sqrt{3}}+c \ldots(\) i) where \(c\) is a constant.
Now the reflected line and the original line meet where the original line cuts the x axis. The point is \(x=\sqrt{3}\) Hence substituting the point \((\sqrt{3}, 0)\) in equation i gives us
\(0=1+c\)
\(O r\)
\(c=-1\)
Hence the equation of the reflected line will be \(y=\frac{x}{\sqrt{3}}-1\)
\(\sqrt{3} y-x+\sqrt{3}=0\)
Hence, the correct option is (A)
Question 2
1 / -0

A line segment of 2 units is sliding with its ends on two perpendicular lines. Then the locus of the middle point, is

A
x+2y+1=0

B
x²+y2=1

C
y²=4ax

D
x²=4ay

Solution

Let \((h, k)\) be the middle point of \((0, a)\) and \((b, 0)\) \(\therefore a=2 k\) and \(b=2 h\)
Using distance formula we have \(\sqrt{(2 h)^{2}+(2 k)^{2}}=2\)
Squaring on both the sides, we have \((2 h)^{2}+(2 k)^{2}=4\)
\(h^{2}+k^{2}=1\)
Locus of the point will be \(x^{2}+y^{2}=1\)
Hence, the correct option is (B)
Question 3
1 / -0

The distance between the parallel lines \(3 \cos \theta+4 \sin \theta+\frac{5}{r}=0\) and \(6 \cos \theta+8 \sin \theta+\frac{7}{r}=0\) is

A
3/5

B
3/10

C
3/20

D
6/5

Solution

Given equations of lines \(3 \cos \theta+4 \sin \theta+\frac{5}{r}=0\)
or, \(6 \cos \theta+8 \sin \theta+\frac{10}{r}=0\)
\(\Rightarrow 6 r \cos \theta+8 r \sin \theta+10=0 \ldots . .(i)\)
and \(6 \cos \theta+8 \sin \theta+\frac{7}{r}=0\)
\(\Rightarrow 6 r \cos \theta+8 r \sin \theta+7=0\)
\(d=\frac{|3|}{10}=\frac{3}{10}\)
Hence, the correct option is (B)
Question 4
1 / -0

The length of the normal from pole on the line rcos(θ−π/3)=5 is

A
3

B
4

C
5

D
10

Solution

The equation of line at a distance \(p\) from the normal and the normal make an angle \(\alpha\) with polar axis is \(r \cos (\theta-\alpha)=p\)
Given equation is \(\Rightarrow r \cos \left(\theta-\frac{\pi}{3}\right)=5\)
So, here \(p=5\)
Hence, the correct option is (C)
Question 5
1 / -0

The angle made by the perpendicular of the line r(cosθ+√3sinθ)=5 with the polar axis is

A
π/2

B
π/3

C
π/4

D
π/6

Solution

The equation of line at a distance \(p\) from the normal and the normal make an angle \(\alpha\) with polar axis is \(r \cos (\theta-\alpha)=p\)
Given equation is \(r(\cos \theta+\sqrt{3} \sin \theta)=5\)
\(\Rightarrow r\left(\frac{1}{2} \cos \theta+\frac{\sqrt{3}}{2} \sin \theta\right)=\frac{5}{2}\)
\(\Rightarrow r \cos \left(\theta-\frac{\pi}{3}\right)=\frac{5}{2}\)
So, here \(\alpha=\frac{\pi}{3}\)
Hence, the correct option is (B)
Question 6
1 / -0

The curve with parametric equations x=3(cost+sint), y=4(cost−sint) is

A
Ellipse

B
Parabola

C
Hyperbola

D
Circle

Solution

Given, \(x=3(\cos t+\sin t)\)
\(\Rightarrow \frac{x}{3}=\cos t+\sin t\)
\(\Rightarrow \frac{x^{2}}{9}=\cos ^{2} t+\sin ^{2} t+2 \cos t \ldots \sin t \ldots(\mathrm{i})\)
And \(y=4(\cos t-\sin t)\)
\(\Rightarrow \frac{y}{4}=\cos t-\sin t\)
\(\Rightarrow \frac{y^{2}}{16}=\cos ^{2} t+\sin ^{2}-2 \cos t \sin t \ldots(\) ii \()\)
Adding
(i) and (ii) gives us \(\frac{x^{2}}{4}+\frac{y^{2}}{6}=2\left(\cos ^{2} t+\sin ^{2} t\right)=2\)
Hence, \(\frac{x^{2}}{4}+\frac{y^{2}}{6}=2\)
This represents an ellipse.
Hence, the correct option is (A)
Question 7
1 / -0

The equation of the line passing through \(\left(5, \tan ^{-1} \frac{3}{4}\right)\) and perpendicular to \(\cos \theta+2 \sin \theta=\frac{4}{r}\)

A
2cosθ−sinθ=5/r

B
2cosθ−sinθ=1/r

C
2cosθ−sinθ=2/r

D
2cosθ−sinθ=3/r

Solution

Given point is \(\left(5, \tan ^{-1}\left(\frac{3}{4}\right)\right)\)
Let \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\)
\(\Rightarrow \tan \theta=\frac{3}{4}\)
\(\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}\)
Given equation is \(\cos \theta+2 \sin \theta=\frac{4}{r}\)
Equation of required line perpendicular to given line is \(\cos \left(\frac{\pi}{2}+\theta\right)+2 \sin \left(\frac{\pi}{2}+\theta\right)=\frac{k}{r}\)
\(\Rightarrow-\sin \theta+2 \cos \theta=\frac{k}{r}\)
since, it passes through \(\left(5, \tan ^{-1} \frac{3}{4}\right)\) \(\Rightarrow k=5\)
So, the required equation is \(\Rightarrow 2 \cos \theta-\sin \theta=\frac{5}{r}\)
\(\Rightarrow 2 r \cos \theta-r \sin \theta=5\)
Hence, the correct option is (A)
Question 8
1 / -0

A straight line L through the point (3,−2) is inclined at an angle 60^o to the line √3x+y=1. lf L also intersects the x−axis, then the equation of L is

A
y+√3x+2−3√3=0

B
y−√3x+2+3√3=0

C
√3y−x+3+2√3=0

D
√3y−x+3-2√3=0

Solution

Equation of given line is \(\sqrt{3} x+y=1\)
or, \(y=-\sqrt{3} x+1\)
So, slope of this line \(=-\sqrt{3}\) Let the slope of the required line be m. \(\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|\)
\(\tan 60^{\circ}=\left|\frac{m-(-\sqrt{3})}{1-\sqrt{3} m}\right|\)
\(\sqrt{3}=\left|\frac{m+\sqrt{3}}{1-\sqrt{3} m}\right|\)
or, \(\frac{m+\sqrt{3}}{1-\sqrt{3} m}=\pm \sqrt{3}\)
Taking \((+) \operatorname{sign}, \frac{m+\sqrt{3}}{1-\sqrt{3} m}=\sqrt{3}\)
\(m+\sqrt{3}=\sqrt{3}-3 m\)
\(m=0\)
Taking \((+) \operatorname{sign}, \frac{m+\sqrt{3}}{1-\sqrt{3} m}=-\sqrt{3}\)
\(m+\sqrt{3}=-\sqrt{3}+3 m\)
\(m=\sqrt{3}\)
since, the required line intersects \(x-\) axis, so \(m\) will be \(\sqrt{3}\). Required line passes through the point (3,-2) , its equation is given by \(y-(-2)=\sqrt{3}(x-3)\)
\(y+2=\sqrt{3} x-3 \sqrt{3}\)
\(y-\sqrt{3} x+2+3 \sqrt{3}=0\)
Hence, the correct option is (B)
Question 9
1 / -0

If the line 2x+3y+12=0 cuts the axes at A and B, then the equation of the perpendicular bisector of AB is

A
3x−2y+5=0

B
3x−2y+7=0

C
3x−2y+9=0

D
3x−2y+8=0

Solution

\(2 x+3 y+12=0\)
At \(x-\) axis, \(y=0\) So, \(2 x+12=0\)
\(\Rightarrow x=-6, A=(-6,0)\)
And at \(y-\) axis, \(x=0\) So, \(3 y+12=0\)
\(\Rightarrow y=-4, B(0,-4)\)
So, mid point of \(A B=\left(\frac{-6}{2}, \frac{-4}{2}\right) \equiv(-3,-2)\)
Slope of \(A B=-\frac{4}{6} \equiv-\frac{2}{3}\)
\(\therefore\) Equation of the perpendicular bisector of \(A B\) is
\(y+2=\frac{3}{2}(x+3)\)
\(\Rightarrow 2 y+4=3 x+9\)
\(\Rightarrow 2 y-3 x-5=0\)
Hence, option \(\mathrm{A}\) is the correct answer.
Question 10
1 / -0

The straight line ax+by+c=0(a,b,c≠0) will pass through the first quadrant if

A
ac>0,bc>0

B
c>0

C
ac>0ac>0 and/or bc>0

D
ac<0ac<0 and/or bc<0

Solution

A line can pass through the first quadrant if
a) \(X\) and \(Y\) intercept are positive
b) Only \(X\) intercept is positive
c) Only \(Y\) intercept is positive
Xintercept\(=\frac{-c}{a}=\frac{-c^{2}}{a c}\)
\(Y\) intercept \(=\frac{-c}{b}=\frac{-c^{2}}{b c}\)
For case -a ) \(a c<0,\) and \(b c<0\) For case \(-\mathrm{b}\) ) \(a c<0,\) and \(b c>0\) For case - \(c\) ) \(a c>0\) and \(b c<0\)
Hence, the common intersection is \(a c<0\) or \(b c<0\) or both \(a c<0\) and \(b c<0\)
Hence, Option \(D\) is correct.
Hence, the correct option is (D)

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Mathematics Test - 22

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Mathematics Test - 22

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SHARING IS CARING

Question 1

√3 y=x−√3

y=√3x−√3

√3 y=x−1

y=x+√3

Solution

Question 2

x+2y+1=0

x2+y2=1

y2=4ax

x2=4ay

Solution

Question 3

3/5

3/10

3/20

6/5

Solution

Question 4

3

4

5

10

Solution

Question 5

π/2

π/3

π/4

π/6

Solution

Question 6

Ellipse

Parabola

Hyperbola

Circle

Solution

Question 7

2cosθ−sinθ=5/r

2cosθ−sinθ=1/r

2cosθ−sinθ=2/r

2cosθ−sinθ=3/r

Solution

Question 8

y+√3x+2−3√3=0

y−√3x+2+3√3=0

√3y−x+3+2√3=0

√3y−x+3-2√3=0

Solution

Question 9

3x−2y+5=0

3x−2y+7=0

3x−2y+9=0

3x−2y+8=0

Solution

Question 10

ac>0,bc>0

c>0

ac>0ac>0 and/or bc>0

ac<0ac<0 and/or bc<0

Solution

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x²+y2=1

y²=4ax

x²=4ay