Mathematics Test - 4

Let marks required be x  
= (45 + 62 + x)/3 >60
45+62+ x >60*3
107 + x >180  
x >180 - 107
x >73

So, 73 is the minimum score required in the third attempt to qualify.

The given inequality is 5x –3 <7
=>5x –3 + 3 <7 + 3                [3 is added both sides]
=>5x <10
=>x <10/5
=>x <2
When x is a real number, the solutions of the given inequality are given by x <2, i.e., all real numbers x which are less than 2.

x² + y² + 6x −6y + 5 = 0
Center O = (-3, 3)
radius r = √{(-3)² + (3)² - 5} 
= √{9 + 9 - 5} 
= √13
Since diameter of the circle passes through the center of the circle.
So (-3, 3) satisfies the equation 2x –y + λ= 0
=>-3*2 - 3 + λ= 0
=>-6 - 3 + λ= 0
=>-9 + λ= 0
=>λ= 9

Given equation of circles are  

S₁  = x²  + y²  + 2x + 6y = 0 ..................1

S₂  = x²  + y²  −4x −2y −6 = 0 ............2

Subtract equation 1 - equation 2, we get

S₁  - S₂  = 6x + 8y + 6 = 0

=> 6x + 8y + 6 = 0 ............3

Cente of the circle S₂  is (2, 1)

Now, the length of the perpendicular from the center (2, 1) of of the circle 2 upon the common chord 3 is

l = (6*2 + 8*1 + 6)/√{6²  + 8²  }

=> l = (12 + 8 + 6)/√{36 + 64}

=> l = 26/√100

=>l = 26/10

=>l = 13/5

Radius of the circle 2 is

r = √{2²  + 1²  - (-6)}

=> r = √{4 + 1  + 6}

=> r = √11

Now length of common chord = 2 √{r²  - l²  }

 = 2 √{(√11)²  - (13/5)²  }

= 2 √{11 - 169/25}

= 2 √{(275 - 169)/25}

= 2 √{106/25} unit

Given equation of circles are
x² +y² +6x+6y=0....(i)
and x² +y² −12x −12y=0....(ii)
Here, g1 = 3,f2 = 3, g2 = −6 and f₂ = −6
∴Centres of circles are C₁ (−3,−3) and C₂ (6,6) respectively and radii are r₁ = 3 √2 and r₂ = 6 √2 respectively.
Now, C₁ C₂ = √[(6+3)² + (6+3)² ]
= 9 √2
and r₁ + r₂
​= 3(2)^1/2 + 6(2)^1/2
= 9(2)^1/2
​⇒C₁ C₂ = r₁ + r₂
∴Both circles touch each other externally.

Ellipse passing through O (0, 0) and having foci P (3, 3) and Q (-4, 4) ,
Then  

is an ellipse, whose focus is (2, -3) , directrix 3x - 4y + 7 = 0 and eccentricity is 1/2.
Length of from focus to directrix is  



So length of major axis is 20/3

Any point on the ellipse is (2 cos θ, √3 sin θ). The focus on the positive x-axis is (1,0). 

Given that (2cos θ-1)² + 3sin²  θ = 25/16
⇒ cos  θ = 3/4