BITSAT Mathematics Test - 4 - JEE MCQ

cos 7θ + cos θ = cos (8θ - θ) + cos θ.
= cos (π - θ) + cos θ
= - cosθ + cosθ = 0.

sec² (tan^-12) + cosec² (cot^-13) = {1 + tan² (tan^-1 2)} + {1 + cot² (cot^-13)}
= 1 + {tan (tan^-12)}² + 1 + {cot (cot^-13)}².
= 1 + 2² + 1 + 3² = 15

which is of the form 

andtherefore, does not exist.

⇔ 

Since, the given spheres are concentric and are of different radii, hence they do not have any point in common.

Here, first slot can be filled in 3 ways.
Second slot can be filled in 3 ways and so on.
Therefore, required number of ways = 3 x 3 x 3 x 3... (24 times) - 3²⁴
Hence, this is the required solution.

LHS = = (x + 1)(x + 2) - (x - 3)(x - 1) = x² + 3x + 2 - (x² - 4x + 3)
= 7x - 1
RHS = = 4(3) - 1(-1) = 12 + 1 = 13
LHS = RHS
7x - 1 = 13 ⇒ x = 2

Given: n(P) = 400, n(Q) = 600 and n(P ∩ Q) = 200
We know that,

Also,

Hence, this is the required solution.

Given,
2 sin θ < 1
Sin θ < 1/2
Here,

So, 
Hence, this is required solution.

Let the vertex of the parabola be the point (h,k) and length of its latus rectum be 4a.
Since its axis is parallel to y - axis, its equation can be written as
(x−h)² = 4a(y−k) ..... (1)
It passes through the given points (0,4),(1,9) and (−2,6)
∴(0−h)²=4a(4−k)
⇒h²=4a(4−k) ...... (2)

(1−h)²=4a(9−k)
⇒1−2h+h²=4a(9−k) ...... (3)

(−2−h)²=4a(6−k)
⇒4+4h+h²=4a(6−k) ...... (4)
Subtracting (2),(3) and (3),(4) we have
1−2h=20a ..... (5) and 3+6h=−12a i.e. 1+2h=−4a ..... (6)
Then solving (5) and (6), we get
a= 1/8, and h=−3/4​
Substituting in any of the equations (2),(3) and (4), we get
k= 8/23
Substituting in (1), the equation of parabola is

Since OA has projections on the coordinate axes, then

Let X divide AB in the ratio λ : 1.
Then, the position vector of X is

Therefore,

Therefore, X divides AB in the proportion 3 : 2.
Hence, this is the required solution.

We know that,

From question,

Thus,

Hence, this is the required solution.

Given,

So,

Let x = vy

Putting these values, we get

Let log v = t
So,

Returning the value of t,

Thus,

Hence, this is the required solution.

Since,

Hence, this is the required solution.

Given,


Multiplying both sides by in equation (i), we get

Thus, 
Hence, this is required solution.

Given that: Cost function C (x) = 26,000+30x and revenue function R(x) = 43x
Now for profit P(x), R(x)>C(x)
⇒ 43x > 26000 + 30x
⇒ 43x − 30x > 26000
⇒ 13x > 26000
⇒ x > 2000
Hence, number of cassettes to be manufactured for some profit must be more than 2000.

Let T be the temperature of the body at time t and T_m=20∘C (the temperature of the air)
We have,

where k is the constant of proportionally and t is the time.
Thus,

The solution of differential equation Is

Let height of the reflection of cloud in the lake = H
The height of the cloud above the lake = H - 2500
Given point is high above the sea level = h = 2500 m

From the above figure, 

Substitute the value h = 2500 and cot 15° = 2 + √3 in equation (1), we get

Height of the cloud H - 2500 = 2500√3m.

Using A.M > G.M.

So minimum value of 

AM > GM

25sin²θ+16cosec²θ≥40

Let the GP be a, ar, ar²,  ar³, ........arⁿ⁻¹
Where a  = first term and r = Common ratio
According to question
We have t1+t2=12 ⇒a+ar=12  ...(i)
t3+t4=48 ⇒ ar2 + ar3 = 48     .....(ii)
Divide the equations (i) & (ii)

But the terms are alternately positive and negative,
∴ r = -2
Now using equation (i) 

Let the common ratio be r

On multiplying Eqs. (i) and (ii), we get

From the property of H.P., we have

So comparing the above equation we get that he line  passes through the point (1,-2)

Therefore,