BITSAT Mathematics Test - 8 - JEE MCQ

Given:

Solving equations,
(1) and (2), we get: 
d = −1, a = n + m − 1
p th term:

Thus, the p th term is n + m − p.

p = Probability of success (s) = 2/3 = 1/3
q = Probability of failure 
The probability that success occurs in even number of trials.

Given:

Given: 

Given: 

Either cos⁡ x − 2 = 0 or 2 cos ⁡x + 1 = 0
But cos⁡ x − 2 = 0
i.e. cos⁡ x = 2, which is not possible. Now, from 2 cos⁡ x + 1 = 0, we get:

Therefore, general solution of the equation is:

Given:
For ΔABC we have a = 18, b = 24 and c = 30
Here, we have to find the value of sin⁡(A/2).
As we know that, if a,b and c are the sides of the ΔABC then 2S = a+b+c
⇒ 2S = 18 + 24 + 30
⇒ 2S = 72
⇒ S = 36
As we know that 

Given:

According to the question,
The cross product of two vectors  is given by 
On solving, we get:

Given: 


⇒ 4 x 1 = 4

Given: 

We have,
x² = 4y
x = 4y − 2
On solving both the equations simultaneously, we get the points of intersection 

Required area

In first quadrant sin x and cos x meet at x = π/4.
The required area is as shown in the diagram.
So, the required area 

Given:

And, second curve is 

This is symmetrical about y-axis and y = 0 is the asymptote for the curve
Points of intersection for the curve is 

To find the bounded area, we have to plot all of the four curves

Here, the limits of x will be from 0 to 1.
So, area will be given by

Given

Clearly t can be any real number

So,

This is the circle with centre at origin an radius is unity. 
∴ Required area = π(1)² = π sq. units.

The x−coordinate of A is 1/a.
According to the given condition, 

Shaded portion is required Area.

The general term in the expansion of  is given by

As 5 and 2 are relatively prime, T_r+1 will be rational, if  are both integers i.e., if (100−r) is a multiple of 6 and r is a multiple of 8.
As 0 ≤ r ≤ 100, multiples of  8 upto 100 and corresponding value of 100 − r are
r = 0, 8, 16, 24,…., 88, 96
⇒ 100 − r = 100, 92, 84, 76, …, 12, 4
Out of 100 − r, multiples of 6 are 84, 60, 36, 12
∴ There are four rational terms.
Hence, the number of irrational terms is 101 − 4 = 97.

Using the binomial expansion (a + b)ⁿ = ⁿC₀aⁿb⁰ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ... + ⁿC_na⁰bⁿ,
= ¹⁰C₀(1 + x²⁵³)¹⁰(xⁿ)⁰ + ¹⁰C₁(1 + x²⁵³)⁹ (xⁿ)¹ + ¹⁰C₂(1 + x²⁵³)⁸(xⁿ)² + ... + ¹⁰C₁₀(1 + x²⁵³)⁰(xⁿ)¹⁰ 
As 253 = 23 × 11 and 1012 = 253 × 4, also n ≤ 22
⇒ Coefficient of x¹⁰¹² will come only from the first term, i.e. in

The general term in the expansion of 
Hence, the general term in the expansion of 

Since, 10¹² = 253 × 4, hence r = 4
Thus, the required coefficient is = ¹⁰C₄.

We have,

General term of the binomial expansion of (x + y)ⁿ is given by T_r+1 = ⁿC_r(x)^n−r(y)^r.

So, the least positive integral value of x is 8.

Given,
The second term in the expansion of 
Now,

We know that,

Then,

Hence, the required solution is 4.

We can have 17²⁵⁶ = (290 − 1)¹²8
= 1000I + ¹²⁸C₂(290)² − ¹²⁸C₁(290) + 1, where I is an integer
= 1000I + 128(290)(18415 − 1) + 1
= 1000m + 681
∴ A₁ = 6, A₂ = 8, A₃ = 1

= 2k = (even integer)
∴ f + f′ = 1 (Because Both Values are Between 0 & 1 and I is Integer)
Now, (I + f)f′ = (5 + 2√6)ⁿ (5 − 2√6)ⁿ = (25 − 24)ⁿ = 1
⇒ (I + f)(1 − f) = 1

Now, since term with x² exists, so, equating power of x in general term to 2, we get,
∴  3n − 3r − r/2 = 2

Now we know that 0 ≤ r ≤ n and r ∈ W.
will be the least double digit integer when r = 8
∴ n = 10

We have,

T⁷ = 7^th term from the beginning

And,
 term from the end
= (n + 2 − 7)^th term from the beginning
= (n − 5)^th term from the beginning