SRMJEEE Maths Mock Test - 1 - JEE MCQ

Given adj
Consider,



.

Consider the given matrix:

Apply 

Therefore, Δ (2a) - Δ (a)  will be completely divisible by (2k + 3a).
Hence, this is the required solution.

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)

Probability to getting heads in both the trials

Let end points be 

∴ Length of focal chord = PQ

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.

Given, x=t²+2t−1 ...(i)

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

This is an equation of a parabola

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

Let AB is the height of Ram and CD be the height of tower. AE = y (ram's shadow) and AC = x (the distance of the Ram from the tower at time t). Then,
 is the rate at which Ram is moving away from the post.
is the rate at which Ram's shadow is increasing.
The tip of the shadow (E) is at a distance x + y from the tower.
So, is the rate at which the tip of the shadow is moving
So at time t,

So,

And

Thus,
The length of the shadow is increasing at the rate of 1.18 m/sec and tip of the shadow is moving away from the tower at the rate of 3.48 m/sec.
Hence, this is required solution.

Required probability = 

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
 

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.

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

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

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.

= cos^⁻1(√(3/7)

The given series is in GP because this series satisfies GP conditions.
The sum of n terms,

Since

And,

So,

Thus,

Hence, this is required solution.

Operating C₃ - C₂ and C₂ - C₁

Apply R₃ - R₂, R₂ - R₁

 Determinent = -2.

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

Step 1: Write the given line in parametric form

Given line:
(x + 3)/5 = (y - 1)/2 = (z + 4)/3 = t

This gives the parametric equations:

• x = -3 + 5t

• y = 1 + 2t

• z = -4 + 3t

So, every point on the line can be written as:
Q(t) = (-3 + 5t, 1 + 2t, -4 + 3t)

Step 2: Identify the direction vector of the line

From the parametric equations, the direction vector of the line is:
d = ⟨5, 2, 3⟩

Step 3: Form vector PQ(t) from external point P(2, 4, 6) to Q(t)

PQ(t) = P - Q(t) =
= ⟨2 - (-3 + 5t), 4 - (1 + 2t), 6 - (-4 + 3t)⟩
= ⟨5 - 5t, 3 - 2t, 10 - 3t⟩

Step 4: Use perpendicularity condition

Since the foot of the perpendicular lies on the line, PQ(t) must be perpendicular to the direction vector d.

So, take the dot product of PQ(t) and d and set it equal to zero:

(5 - 5t)(5) + (3 - 2t)(2) + (10 - 3t)(3) = 0

Now compute:

25 - 25t + 6 - 4t + 30 - 9t = 0
61 - 38t = 0
=> t = 61 / 38

Step 5: Find the coordinates of the foot Q

Substitute t = 61/38 into the parametric equations:

x = -3 + 5t = -3 + 305/38 = ( -114 + 305 ) / 38 = 191 / 38
y = 1 + 2t = 1 + 122/38 = ( 38 + 122 ) / 38 = 160 / 38 = 80 / 19
z = -4 + 3t = -4 + 183/38 = ( -152 + 183 ) / 38 = 31 / 38

Final Answer: Coordinates of the foot of the perpendicular

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

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

Converting cos and sin into tan, we get,



So, for and for