Mathematics Test

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Mathematics Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

The angle between the tangents from any point on the circle x²+y²=r² to the circle x²+y²=r²sin²α is

A
α

B
2α

C
α / 2

D
4α

Solution

From the above figure, \(\mathrm{In} \triangle P A O\)
\(\sin \theta=\frac{r \sin \alpha}{r}\)
or, \(\sin \theta=\sin \alpha\)
or, \(\boldsymbol{\theta}=\alpha\)
Therefore, angle between tangents \(=2 \theta=2 \alpha\)
Hence, B is the correct option.
Question 2
1 / -0

lf t is parameter, A=( a sec t , b tan t ) and B =( − a tan t , b sec t ) , O = ( 0 , 0 ) then the locus of the centroid of ΔOAB is

A
9xy=ab

B
xy=9ab

C
x²−9y²=a²−b²

D
None of these

Solution

Let the centroid be G,
Then
\(G=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)
Here
\(\left(x_{1}, y_{1}\right)=(\)asect,btant\()\)
\(\left(x_{2}, y_{2}\right)=(-a t a n t, b s e c t)\)
\(\left(x_{3}, y_{3}\right)=(0,0)\)
Hence
\(G=\left(\frac{\text {asect}-\text {atant}}{3}, \frac{\text {btant}+\text {bsect}}{3}\right)\)
Or
\(G(x, y)=\left(\frac{\text {asect}-\text {atant}}{3}, \frac{\text {btant}+\text {bsect}}{3}\right)\)
Therefore
\(x=\frac{\text {asect}-\text {atant}}{3}\)
Or
\(3 x=\operatorname{asect}-\)atant\(. .\) (i)
And
\(y=\frac{\text {bsect}+\text {btant}}{3}\)
Or
\(3 y=\) bsect \(+\) btant \(\ldots\) (ii)
Hence
\(3 x \times 3 y=(\)asect\(-\)atant\()(\)bsect\(+\)btant\()\)
\(9 x y=a b(\operatorname{sect}-\operatorname{tant})(\operatorname{sect}+\tan t)\)
\(9 x y=a b\left(\sec ^{2} t-\tan ^{2} t\right)\)
\(9 x y=a b(1)\)
Or
\(9 x y=a b\)
Question 3
1 / -0

The locus of a point which is collinear with the points (3,4) and (−4,3) is

A
2x+3y−12=0

B
2x+3y+12=0

C
x+y+12=0

D
x−7y+25=0

Solution

Let \((h, k)\) be any arbitrary point on the locus. It is given that \((3,4),(-4,3),(h, k)\) are collinear. Hence, area of the triangle formed by these points will be 0 \(\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|=0\)
Condition of col-linearity of three points \(\left(x_{1}, y_{1}\right)\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)\)
\(\left|\begin{array}{ccc}3 & 4 & 1 \\ -4 & 3 & 1 \\ h & k & 1\end{array}\right|=0\)
\(\therefore h(4-3)-k(4+3)+1(9+16)=0\)
\(\Longrightarrow h-7 k+25=0\)
\(\Longrightarrow x-7 y+25=0\) is the required equation of locus. Hence, option D is correct.
Question 4
1 / -0

The greatest distance of the point (10,7) from the circle x²+y²−4x−2y−20=0 is:

A
5

B
15

C
10

D
20

Solution

\(P(10,7)\) is the given point and circle is \(x^{2}+y^{2}-4 x-2 y-20=0\) Centre of circle \(C\) is (2,1) and radius \(r=\sqrt{4+1+20}=\sqrt{25}=5\)
Greatest distance from \(P\) to the circle is \(|C P+r|\) \(C P=\sqrt{64+36}=10\)
\(\therefore\) Greatest distance of \(P\) from circle is 15 Hence, option B.
Question 5
1 / -0

The circles x²+y²+4x−2y+4=0 and x²+y²−2x−4y−20=0

A
intersect each other.

B
touch each other externally.

C
touch each other internally.

D
are such that one circle lies entirely inside another circle.

Solution

\(x^{2}+y^{2}+4 x-2 y+4=0 \quad[\) From point (1)\(]\)
\(C_{1} \equiv(-2,1)\)
\(r_{1}=1\)
and \(x^{2}+y^{2}-2 x-4 y-20=0\)
\(C_{2} \equiv(1,2)\)
\(r_{2}=5\)
\(\therefore\) distance bet \(^{n}\) centre \(=\sqrt{(3)^{2}+(1)^{2}}\)
\(=\sqrt{10}\)
\(r_{1}+r_{2}=6>\sqrt{10} \ldots \ldots(1)\)
and \(r_{2}-r_{1}=4>\sqrt{10} \ldots . .(2)\)
From ( 1 ) and (2) Given two circles are such that the circle lies entirely inside another.
Question 6
1 / -0

lf a line passes through the point P(1,−2) and cuts the circle x²+y²−x−y=0 at A and B, then the maximum value of PA+PB is

A
√26

B
8

C
√8

D
2√8

Solution

Given equation of the circle: \(x^{2}+y^{2}-x-y=0\) For maximum value of \(P A+P B\), the line \(P A B\) must pass through centre of the circle. \(A C=O C=B C=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}\)
\(=\frac{1}{\sqrt{2}}\)
and \(P C=\sqrt{\left(1-\frac{1}{2}\right)^{2}+\left(-2-\frac{1}{2}\right)^{2}}\)
\(=\frac{\sqrt{26}}{2}\)
\(\mathrm{Now}, P A=P C-A C=\frac{\sqrt{26}}{2}-\frac{1}{\sqrt{2}}\)
and \(P B=P C+B C=\frac{\sqrt{26}}{2}+\frac{1}{\sqrt{2}}\)
\(\mathrm{Now}, P A+P B=\frac{\sqrt{26}}{2}-\frac{1}{\sqrt{2}}+\frac{\sqrt{26}}{2}+\frac{1}{\sqrt{2}}=\sqrt{26}\)
Hence, option \(\mathrm{A}\) is the correct option.
Question 7
1 / -0

The edges of a triangular board are 6cm,8cm and 10cm. The cost of painting it at the rate of 9 paise per cm² is

A
Rs. 2.00

B
Rs. 2.16

C
Rs. 2.48

D
Rs. 3.00

Solution

\(s=\frac{a+b+c}{2}=\frac{6+8+10}{2}=12\)
By Heron's formula, Area of the triangle \(=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{12(12-6)(12-8)(12-10)}\)
\(=\sqrt{12(6)(4)(2)}=24 \mathrm{cm}^{2}\)
cost of painting \(=9 \times 24\) paise \(=216\) paise \(=\) Rs. 2.16
Question 8
1 / -0

The equation of the circle which touches x-axis at (0,0) and touches the line 3x+4y−5=0 is

A
x²+y²−4y=0

B
x²+y²−10y=0

C
x²+y²+10x=0

D
x²+y²+10y=0

Solution

Equation of circle touching x-axis at \((0,0),\) means centre of circle lie on Y-axis i.e. \((0, k)\)
\((x-0)^{2}+(y-k)^{2}=k^{2}\)
\(S: x^{2}+y^{2}-2 k y=0 \ldots \ldots(1)\)
Circle \(S\) touches \(3 x+4 y-5=0\) \(\therefore k=\left|\frac{4 k-5}{5}\right|\)
\(5 k=4 k-5\)
\(k=-5\)
\(\therefore\) Equation of circle is \(x^{2}+y^{2}-(-5) \times 2 y=0\)
\(\Rightarrow x^{2}+y^{2}+10 y=0\)
Question 9
1 / -0

lf the point (k+1,k) lies inside the region bounded by the curve x=√25−y2 and y-axis, then k belongs to the interval

A
(−1,3)

B
(−4,3)

C
(−∞,−4)∪(3,∞)

D
(−4,−3)

Solution

\(x=\sqrt{25-y^{2}}\)
\(x^{2}+y^{2}=25\) is a equation of hemicircle with \(x \geqslant 0\) \(\therefore(k+1)^{2}+k^{2}-25<0\)
\(2 k^{2}+2 k-24<0\)
\(k^{2}+k-12<0\)
\((k+4)(k-3)<0\)
\(k \in(-4,3)\)
and as known \(x \geqslant 0\) \(\Rightarrow k+1>0\)
\(k>-1 \quad--(2)\)
from ( 1\()\) \& (2) \(k \in(-1,3)\)
Question 10
1 / -0

The length of the normal from pole on the line r cos (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 \(p=5\)

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Mathematics Test - 36

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Mathematics Test - 36

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SHARING IS CARING

Question 1

α

2α

α / 2

4α

Solution

Question 2

9xy=ab

xy=9ab

x2−9y2=a2−b2

None of these

Solution

Question 3

2x+3y−12=0

2x+3y+12=0

x+y+12=0

x−7y+25=0

Solution

Question 4

5

15

10

20

Solution

Question 5

intersect each other.

touch each other externally.

touch each other internally.

are such that one circle lies entirely inside another circle.

Solution

Question 6

√26

8

√8

2√8

Solution

Question 7

Rs. 2.00

Rs. 2.16

Rs. 2.48

Rs. 3.00

Solution

Question 8

x2+y2−4y=0

x2+y2−10y=0

x2+y2+10x=0

x2+y2+10y=0

Solution

Question 9

(−1,3)

(−4,3)

(−∞,−4)∪(3,∞)

(−4,−3)

Solution

Question 10

3

4

5

10

Solution

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x²−9y²=a²−b²

x²+y²−4y=0

x²+y²−10y=0

x²+y²+10x=0

x²+y²+10y=0