VITEEE Maths Test - 7 - JEE MCQ

Using A.M > G.M.

So minimum value of 

Equations are
3x−4y+7z=0  …(i)
2x−y−2z=0    …(ii)
3x³ − y³ + z³ = 18  …(iii)
From (i) and (ii) and applying cross-multiplication we get,

Putting these values in (iii) we get,
3(3k)³−(4k)³+k³ = 18
⇒18k³=18
∴ k=1
So,
x=3×1=3
y=4×1=4
z=1
Hence x=3, y=4, z=1.

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.

AM > GM

25sin²θ+16cosec²θ≥40

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.

Now multiplying the first fraction by a in the numerator and denominator and likewise second by b,  third by c  and adding as above, we have each of the above ratio equal to

Hence, 

Since,

Hence, this is the required solution.

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.

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 

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

Let the common ratio be r

Let x be added to each term of the ratio 5:37
Now the given ratio becomes (5 + x) : (x + 37)
Then, 
⇒3x + 15 = x + 37
⇒x = 11
Hence, the required term, which needs to be added, is 11.

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

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 the mean proportional be x.
Then we can say
6 : x :: x : 24

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.

Required two digit numbers are 12, 19, ...,96 which leave a remainder 5 5 when they are divided by 7.
Here, a = 12, d = 7, l = 96
∴l = a + (n − 1)d
⇒96=12+7(n−1)
⇒ n=13
∴ 

andtherefore, does not exist.

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) 

Since, T_p=q=a+(p−1)d  ...(i)
and T_q=p=a+(q−1)d   ..(ii)
On solving Eqs. (i) and (ii), we get
d=−1 and  a=p+q−1
∴ T₁₀=a+(10−1)d=p+q−10