BITSAT Mathematics Test - 1 - JEE MCQ

Let end points be 

∴ Length of focal chord = PQ

Combined equation of pair of tangents is given by,
SS₁=T²

⇒(y²−4x)((−1)²−4(−2))=(−1⋅y−2(x−2))²
⇒(y²−4x)(9)=(y+2x−4)²
⇒9y²−36x=y²+4x²+16−8y−16+4xy
⇒4x²−8y²+4xy+20x−8y+16=0
⇒2x²−4y²+2xy+10x−4y+8=0

Slope of line
Similarly, slope of line
Similarly, slope of line
We know that the algebraic sum of slopes of all the three concurrent normals of a parabola is equal to zero.

Consider a △ABC whose sides are x=0, y=0 and x+y=2

Therefore, co-ordinates of A, B & C are (0, 2), (0, 0) & (2, 0) respectively.
Since the parabolas touch all the sides, their foci must lie on the circumcircle of the Δ ABC.
We see that Δ ABC is a right angle triangle.
So circumradius
Now, on joining the foci of three parabolas, we get a triangle of maximum area.
Hence, foci must be the vertices of an equilateral triangle inscribed in the circumcircle.

Let side length of equilateral triangle F₁F₂F₃ be a.
From the diagram,
Therefore, required area
 

The given parabola is y2=4ax  ...(i)
Let OA(=l) be the side of equilateral triangle.
​Then OL=lcos30°= √3l/2
and LA=lsin 30°= l/2

∴ The co-ordinates of A are 

⇒  l=8√3a
Hence the length of the side of the triangle = 8√3a units.

Given, y²=−8x
Length of the latus rectum = 4a = 8
⇒a=2
Hence, the equation of the directrix is x=2

for the given parabola , y²=4(x+1)
directrix is x=−2. and  Any point on it is (−2, k)
let  mirror image of (-2,k) in the line x+2y=3 is  (x,y)

From Eqs. (i) and (ii), we get

⇒4y -3x = 16 is the equation of the mirror image of the directrix.

Shortest distance between two curves occurs along the common normal.
Normal to y²=4x at (m², 2m) is
y+mx−2m−m³=0

Both normals are same,  if  −2m−m³=−4m− 
⇒m=0, ±2
So, points will be (4, 4) and (5, 2) or (4,−4) and (5,−2)
Hence, shortest distance will be

[3 (1 + w²)] + w]⁴ = (2w)⁴ = 16w⁴ = 16w 

∴ n = 11

Number of triangles = ¹²C₃ - ⁷C₃
= 220 - 35
= 185

Operating C₃ - C₂ and C₂ - C₁

Apply R₃ - R₂, R₂ - R₁

 Determinent = -2.

Above is skew symmetric deteminent of odd order because
cos (A + B) = - cos C etc.

Probability to getting heads in both the trials

Given: A pair of dice is thrown

Let us first write the all possible events that can occur

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6),

Hence total number of events is 6² = 36

Favorable events i.e. getting the total of numbers on the dice greater than 9 are

(5,5), (5,6), (6,4), (4,6), (6,5) and (6,6),

Hence total number of favorable events i.e. getting the total of numbers on the dice greater than 9 is 6

We know that;

Probability = Number of favorable event
                       Total number of event

Hence probability of getting the total of numbers on the dice greater than 9 is 
 6    =   1  
36        6

Required probability = 

Given, x=t²+2t−1 ...(i)

On putting the value of t in Eq. (i), we get

This is an equation of a parabola

Let there be n men participants. Then, the number of games that the men play between themselves is 2.  ⁿC₂ and the number of games that the men played with the women is 2.(2n)
∴ 2nC₂ − 2⋅2n = 66 (given)
⇒ n (n−1) − 4n − 66 = 0
⇒ n² − 5n − 66 = 0
⇒(n + 5) (n − 11) = 0
⇒ n = 11
∴ Number of participants =11 men+2 women=13

The boys are in majority, if the groups are (4B,3G),(5B,2G),(6B,1G) Total number of combinations
= ⁶C₄× ⁴C₃+ ⁶C₅× ⁴C₂+ ⁶C₆× ⁴C₁
= 15 × 4 + 6 × 6 + 1 × 4 = 100

Rank from ending = Total no of words − Rank from beginning +1
Tota no of words possible u sin gletters of the word HORROR is 
Dictionary rank of the word : arrange in alphabetical order {H,O,O,R,R,R} No of words starting with H O O: 1
No of words starting with HORO : 1
the net word after the above words is HORROR
∴ RANK of the word HORROR from beginning is 3
∴ RANK of the word horror from ending is = 60 − 3 + 1 = 58

Let there are n teams.
Each team play to every other team in  ⁿC₂₃ ways
∴ ⁿC₂=153 (given)

⇒ n(n−1)=306
⇒ n²−n−306=0
⇒ (n−18)(n+17)=0
⇒ n=18 (∵n is never negative)

2 circles can intersect at atmost 2 points. Maximum no. of points can be obtained if no 3 circles intersect at the same point.
no. of possible pair of circles = ⁸C₂
= 28.
max. No. of intersection points = 2 x 28
= 56.

Total number of outcomes when four normal dice are rolled
= 6 × 6 × 6 × 6 = 6⁴ = 1296
Total number of ways in which no dice shows up 2 i.e.
Each of the four dice shows up 1,3,4,5 or  6 as outcomes
=5 × 5 × 5 × 5 = 5⁴ = 625
Hence total number of possible outcomes when no dice shows up 2
=1296 − 625 = 671

Since, telephone number start with 67, so two digits is already fixed
Now, we have to arrangement of three digits from remaining eight digits (i.e., 0,1,2,3,4,5,8,9)

= 8 × 7 × 6
= 336 ways

R₃ cannot remain empty.
Total ways = ⁸C₆×6!=20160
When R₁ is empty =6!=720
When R₂ is empty =6!=720
∴ When no is empty =20160−720×2=18,720

Let A (−2,−6), B(–2,4) and C(1,3) are the vertices of the △ABC
Now using the distance formula, we get 

(AB)² = (BC)² + (CA)²
so the △ABC is a right angle triangle, right angle at C and we know that in right-angle triangle orthocenter is the point where right angle formed,
Therefore orthocenter is C(1, 3)