Test: 35 Year JEE Previous Year Questions: Miscellaneous (Sets, Relations, Statistics & Mathematical Reasoning) - 2 - JEE MCQ

As per question if point B moves s distance in t time then point A moves (s + n) distance in time (t + m) after which both have same velocity v.
Then using equation v = u + at we get

....(1)

Using equation v² = u²+ 2 , as we get

v² = 2 f (s + n) = 2 f's ....(2)

Also for point B using the eqn we get

Substituting values of t and s from equations (1) and (2) in the above relation, we get

Let A and B be displaced by a distance x then Change in moment of (A + B) = applied moments
⇒ (A + B) x x = H ⇒ x 

We know that for positive real numbers x₁, x₂, ...., x_n, A.M. of  kth powers of x'_i^s ≥ kth the power of A.M. of x'_i^s

⇒ n ≥ 16 . So only possible value for n = 18

u cos 60° = v cos 30° (as horizontal component of velocity remains the same)

Let F be the larger force

Given R = 

Resolving F in horizontal and vertical direction

R =F cosθ ⇒ cosθ =

F' = Fsinθ =

If we consider unit vectors in the direction AB and AC respectively, then  as per quesiton, forces along AB and AC respectively are

∴ Their resultant along AD 

∴ Magnitude of resultant is

But from figure ΔABC ~ΔDBA

∴ The required magnitude of resultant becomes 

Clearly ( x, x) ∈ R"x ∈ W . So R is relexive.
Let ( x, y) ∈ R ,  then ( y, x) ∈ R as x and y have at least one letter in common. So, R is symmetric.
But R is not transitive for example Let x = INDIA, y = BOMBAY and z = JOKER then ( x, y) ∈R (A is common) an d ( y, z) ∈ R (O is common) but ( x, z) ∉ R . (as no letter is common)

   (Here deviations are taken from the mean).
Since A and B both have 100 consecutive integers, therefore both have same standard deviation and hence the variance.

                  (As    is same in both the cases)

For two velocities u and u at an angle θ to each other the resultant is given by

Now in second case, the new resultant AE (i.e., R") bisects ∠CAB , therefore using angle bisector theorem in ΔABC , we get

Using   and 400 = 

Subtracting, we get 400 = 8g + 4gt  ⇒ t = 8 sec

 = 320

∴ Desired height = 320 + 400 = 720 m

Given : Force P = Pn, Q = 3n, resultant R = 7n & P' = Pn,

Q' = (–3)n, R' =

We know that R² = P² + Q² + 2PQ cosα
⇒ (7)² = P² + (3)² + 2 × P × 3 cosα
⇒ 49 = P² + 9 + 6P cosα
⇒ 40 = P² + 6P cosα .....(i)

And= P² + (–3)² + 2P × –3 cosα
⇒ 19 = P² + 9 – 6P cosα
⇒ 10 = P² – 6P cosα.....(ii)
Adding (i) and (ii)  
50 = 2P² ⇒ P² = 25 ⇒ P = 5n.

Let B be the top of the wall whose coordinates will be (a, b). Range (R) = c

B lies on the trajectory

⇒ 

⇒

⇒

The angle of projection,

Let the number of boys be x and that of girls be y.
⇒ 52x + 42y = 50(x + y) ⇒ 52x – 50x = 50y – 42y

⇒ 2x = 8y ⇒  and 

Required % of boys = 

∵ 13²  = 5 ² + 12² ⇒ AB² = AC² + BC²
⇒ ∠ACB = 90°
Q m is mid point of the hypotenuse AB, therefore MA = MB = MC ⇒ ∠A = ∠ACM =θ
Applying  Lami’s theorem at C, we get

⇒ T₁ = 13 sinθ and T₂ = 13 cosθ

⇒ T₁ = 5 kg and T₂ = 12 kg

Mean  of a, b, 8, 5, 10 is 6

= ⇒ a + b = 7 ...(i)

Variance of  a, b, 8, 5, 10 is 6.80

⇒ a² – 12a + 36 + (1 –a )² + 21= 34   [using eq. (i)]
⇒ 2a² –14a + 24 = 0 ⇒ a² – 7a + 12 = 0
⇒ a = 3  or   4      ⇒ b = 4   or   3
∴ The possible values of a and b are a = 3 and b =  4 or, a = 4 and b = 3

(None) p :  x is an irrational number
q : y is a transcendental number
r : x is a rational number iff y is a transcendental number. clearly r :~p ⇿q
Let us use truth table to check the equivalence of ‘r’ and ‘q or p’; ‘r’ and ~ (p ⇿~ q)

From columns (1), (2) and (3), we observe, none of the these statements are equivalent to each other.
∴ Statement 1as well as statement 2 both are false.
∴ None of the options is correct.

Let us make the truth table for the given statements, as follows :

From table we observe p → (q→p) is equivalent to p→(pνq)

The truth table for the logical  statements, involved in statement 1, is as follows :

We observe the columns for ~ (p ⇿ ~q) and p ⇿ q are identical, therefore ~(p ⇿ ~q) is equivalent to p ⇿ q
But ~ (p ⇿ ~q) is not a tautology as all entries in its column are not T.
∴ Statement-1 is true but statement-2 is false.

For the numbers 2, 4, 6, 8, ......., 2n

And Var = 

∴ Statement-1 is false.  Clearly, statement - 2 is true.

Let x∈ A and x ∈ B ⇔ x ∈A∪B
⇔ x ∈ A∪C (∵ A ∪ B = A∪C) ⇔ x∈C
∴ B = C.
Let x ∈A and x ∈ B ⇔ x ∈ A∩B
⇔ x ∈ A∩C (∵A ∩ B = A∩C)
⇔ x∈C ∴ B=C

Mean = 

∵ Mean deviation from the mean = 255

P : there is a rational number x ∈S such that x > 0
~ P : Every rational number  x ∈S satisfies x ≤ 0

x Ry need not implies yRx ∴ R is not symmetric and hence not an equivalence relation

Given qm = pn  

 qm = pn, ps= rq

⇒ ms = rn (transitive).

S is an equivalence relation.

Variance of combined data

x – y is an integer.
∵  x – x = 0 is an integer ⇒ A is reflexive.
Let x – y is an integer ⇒  y – x is an integer
⇒  A is symmetric Let x – y, y – z are integers
⇒  x – y + y – z is also an integer
⇒  x – z is an integer ⇒  A is transitive
∴  A is an equivalence relation.
Hence statement 1 is true.
Also B can be considered as

xBy if a rational number

is a rational number

But    a rational number need not imply   a rational number because

is rational ⇒   is not rational

∴  B is not an equivalence relation.