Mathematics Test

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Mathematics Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The line 2x+3y=6, 2x+3y=8 cut the X-axis at A, B respectively. A line L=0 drawn through the point (2,2) meets the X-axis at C in such a way that abscissa of A, B,C are in arithmetic progression. Then, the equation of the line L is

A
2x+3y=10

B
3x+2y=10

C
2x−3y=10

D
3x−2y=10

Solution

Equation of line drawn through (2,2) is given by \(y-2=m(x-2)\)
\(y=m x+2-2 m\)
at x-axis, \(y=0, x=\frac{2(m-1)}{m} c=\left(\frac{2(m-1)}{m}, 0\right)\)
as given abscissae of \(A, B, C\) are in A.P. \(\therefore 2 \times 4=3+\frac{2(m-1)}{m}\) [abscissa \(=\mathrm{x}\) - co ordinate \(]\) \(8=3+\frac{(m-1)}{m} 2 \Rightarrow 5 m=2 m-2\)
\(3 m=-2\)
So, equation of line is \(\Rightarrow m=\frac{-2}{3}\) \(y=\frac{-2}{3} x+2+\frac{2 \times 2}{3}\)
\(=\frac{-2}{3} x+\frac{10}{3}\)
\(\Rightarrow 3 y+2 x=10\)
Hence, the correct option is (A)
Question 2
1 / -0

Let aa and bb be non-zero real numbers. Then, the equation (ax²+by²+c)(x²−5xy+6y²)=0 represents

A
four straight lines, when c=0 and a,b are of the same sign

B
two straight lines and a circle, when a=b, and c is of sign opposite to that of a

C
two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a

D
a circle and an ellipse, when a and b are of the same sign and cc is of sign opposite to that of a

Solution

\(\left(a x^{2}+b y^{2}+c\right)\left(x^{2}-5 x y+6 y^{2}\right)=0\)
\(\Rightarrow a x^{2}+b y^{2}+c=0\) or \(x^{2}-5 x y+6 y^{2}=0\)
\(\Rightarrow x^{2}+y^{2}=\left(-\frac{c}{a}\right)\) if \(a=b, x-2 y=0\) and \(x-3 y=0\)
Hence the given equation represents two straight lines and a circle, when \(a=b\) and \(c\) is of sign opposite to that of \(a\).
Hence, the correct option is (B)
Question 3
1 / -0

Find the equation of a straight line parallel to 2x+3y+11=0 and which is such that the sum of its intercepts on the axes is 15.

A
2x+3y=15

B
3x+2y+11=0

C
2x−3y=10

D
2x+3y=18

Solution

Line parallel to \(2 x+3 y+11=0\) is \(2 x+3 y+c=0\)
So, x-intercept \(=\frac{-c}{2}(\) Put \(y=0)\)
\(y\) -intercept \(=\frac{-c}{3}(\) Put \(x=0)\)
As given \(\left(\frac{-c}{2}\right)+\left(\frac{-c}{3}\right)=15\)
\(\frac{-5 c}{6}=15\)
\(c=-18\)
From ( 1 ),
\(\therefore\) equation of that line is \(2 x+3 y-18=0\)
Hence, the correct option is (D)
Question 4
1 / -0

If the slope of the line \(\left(\frac{1}{a}+\frac{k}{b}\right) x+\left(\frac{1}{b}+\frac{k}{a}\right) y-(1+k)=0\) is \(-1,\) then the value
of \(k\) is

A
2

B
-1

C
1

D
-2

Solution

Slope of given line is \(-1 .\) So, \(\left(\frac{1}{a}+\frac{k}{b}\right) x+\left(\frac{1}{b}+\frac{k}{a}\right) y-(1+k)=0\)
\(\Rightarrow-1=\frac{-\left(\frac{1}{a}+\frac{k}{b}\right)}{\left(\frac{1}{b}+\frac{k}{a}\right)}=m\)
\(\Rightarrow \frac{1}{b}+\frac{k}{a}=\frac{1}{a}+\frac{k}{b}\)
\(\Rightarrow k\left(\frac{1}{a}-\frac{1}{b}\right)=\left(\frac{1}{a}-\frac{1}{b}\right)\)
\(\Rightarrow k=1\)
Hence, the correct option is (C)
Question 5
1 / -0

The perpendicular bisector of the line segment joining P(1,4) and Q(k,3) has y-intercept −4, then a possible value of k is

A
−4

B
1

C
2

D
-2

Solution

The mid point of the line joining of the given two points is \(\left(\frac{k+1}{2}, \frac{7}{2}\right)\) and the slope of this line is \(\frac{1}{1-k}\)
\(\therefore\) The slope of the perpendicular bisector line is \(k-1 .\) Let the equation of this line be \(y=(k-1) x+c \rightarrow(1)\) \(y-\) intercept is \(-4 \rightarrow(2) \quad\) (given) From ( 1 ) and (2) , equation of the perpendicular bisector is \(y=(k-1) x-4\) This line passes through the point \(\left(\frac{k+1}{2}, \frac{7}{2}\right)\) On substituting this point in above line equation, we get \(k=\pm 4\) So, the possible value of \(k\) is -4
Hence, the correct option is (A)
Question 6
1 / -0

The acute angle between the lines lx+my=l+m,l(x−y)+m(x+y)=2m is

A
π/4

B
π/6

C
π/2

D
π/3

Solution

The mid point of the line joining of the given two points is \(\left(\frac{k+1}{2}, \frac{7}{2}\right)\) and the slope of this line is \(\frac{1}{1-k}\)
\(\therefore\) The slope of the perpendicular bisector line is \(k-1 .\) Let the equation of this line be \(y=(k-1) x+c \rightarrow(1)\) \(y\) - intercept is \(-4 \rightarrow(2)\) (given) From ( 1 ) and (2) , equation of the perpendicular bisector is \(y=(k-1) x-4\) This line passes through the point \(\left(\frac{k+1}{2}, \frac{7}{2}\right)\) On substituting this point in above line equation, we get \(k=\pm 4\) So, the possible value of \(k\) is -4
Hence, the correct option is (A)
Question 7
1 / -0

The line 4y−3x+48=0 cuts the curve y²=64x in A and B. If AB subtends an angle θ at the origin, then tanθ=

A
20/9

B
10/9

C
5/9

D
40/9

Solution

\(4 y-3 x+48=0\)
\(\frac{3 x-4 y}{48}=1\)
\(\therefore y^{2}-64 x\left(\frac{3 x-4 y}{48}\right)=0\) [By homogenisation. \(]\)
\(y^{2}-\frac{4}{3} x(3 x-4 y)=0\)
\(y^{2}-4 x^{2}+\frac{16}{3} y x=0\)
\(3 y^{2}-12 x^{2}+16 x y=0\)
\(\therefore \tan \theta=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\)
\(=\left|\frac{2 \sqrt{64+36}}{(3-12)}\right|\)
\(=\left|\frac{20}{-9}\right|\)
\(\tan \theta=\frac{20}{9}\) is a angle subtended by \(A B\) at origin.
Hence, the correct option is (A)
Question 8
1 / -0

The angle made by the line whose intercepts are 3 and 3√3 on X and Y axis respectively with X-axis is .....

A
30^o

B
120^o

C
60^o

D
90^o

Solution

Any straight line which makes an angle \(\alpha\) with the positive direction of \(X-\) axis will have their slope as \(\tan \alpha=-\frac{b}{a},\) where \(a\) and \(b\) are the intercepts w.r.t. \(X\) and \(Y\) axes Here, \(a=3, b=3 \sqrt{3}\)
\(\Longrightarrow \tan \alpha=-\frac{3 \sqrt{3}}{3}=-\sqrt{3}\)
\(\Longrightarrow \alpha=\tan ^{-1}(-\sqrt{3})\)
\(\therefore \alpha=120^{\circ}\)
Hence, option B is correct.
Hence, the correct option is (B)
Question 9
1 / -0

lf the pair of lines 2x²+3xy+y²=0 makes angles θ₁and θ₂ with x-axis then tan(θ₁−θ₂)=

A
1

B
1/2

C
1/3

D
1/4

Solution

\(2 x^{2}+3 x y+y^{2}=0\)
Let \(t=\frac{x}{y}\)
\(2 t^{2}+3 t+1=0\)
\(\Rightarrow t=\frac{-3 \pm \sqrt{9-8}}{4}\)
\(\Rightarrow t=\frac{-3 \pm 1}{4}\)
\(\Rightarrow t=-1, \frac{-1}{2}\)
\(\therefore y=-x\) and \(y=-2 x\)
\(\tan \theta_{1}=-1\) tan \(\theta_{2}=-2\)
\(\therefore \tan \left(\theta_{1}-\theta_{2}\right)=\frac{\tan \theta_{1}-\tan \theta_{2}}{1+\tan \theta_{1} \tan \theta_{2}}\)
\(\Rightarrow \tan \left(\theta_{1}-\theta_{2}\right)=\frac{-1+2}{1+2}=\frac{1}{3}\)
Hence, the correct option is (C)
Question 10
1 / -0

If the two pairs of lines 3x²−5xy +2y²=0 and 6x²−xy+ky²=0 have one line in common, then k²+7k−10 =

A
0

B
-20

C
-10

D
20

Solution

\(3 x^{2}-5 x y+2 y^{2}=0\)
\(t=\frac{y}{x}\)
\(2 t^{2}-5 t+3=0\)
\(t=\frac{5 \pm \sqrt{25-24}}{4}\)
\(t=\frac{5 \pm 1}{4}\)
\(t=\frac{3}{2}, 1\)
given \(3 x^{2}-5 x y+2 y^{2}=0\) and \(6 x^{2}-x y+k y^{2}=0\) have one line in common So, If \(t=\frac{3}{2}\)
\(k t^{2}-1 t+6=0\)
\(k \frac{9}{4}-\frac{3}{2}+6=0\)
\(k=-2\)
If, \(t=1, k-1+6=0\)
\(k=-5 \quad \ldots-(2)\)
When \(k=-2, \quad k^{2}+7 k-10=4-14-10\)
\(=-20\)
When, \(k=-5 \quad k^{2}+7 k-10=25-35-10\)
\(k^{2}+7 k-10=-20\)
Hence, the correct option is (B)

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 21

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Mathematics Test - 21

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SHARING IS CARING

Question 1

2x+3y=10

3x+2y=10

2x−3y=10

3x−2y=10

Solution

Question 2

four straight lines, when c=0 and a,b are of the same sign

two straight lines and a circle, when a=b, and c is of sign opposite to that of a

two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a

a circle and an ellipse, when a and b are of the same sign and cc is of sign opposite to that of a

Solution

Question 3

2x+3y=15

3x+2y+11=0

2x−3y=10

2x+3y=18

Solution

Question 4

2

-1

1

-2

Solution

Question 5

−4

1

2

-2

Solution

Question 6

π/4

π/6

π/2

π/3

Solution

Question 7

20/9

10/9

5/9

40/9

Solution

Question 8

30o

120o

60o

90o

Solution

Question 9

1

1/2

1/3

1/4

Solution

Question 10

0

-20

-10

20

Solution

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30^o

120^o

60^o

90^o