Test: Basic of Mathematics - 1 - JEE MCQ

Consider the equation of the line joining (0,14) and (5,0). Applying intercept form gives us

Now it has been shaded away from the origin, hence the inequality has to be false at x,y=(0,0). At x,y=(0,0) we get
14x + 5y = 0. Therefore for false inequality, 14x + 5y ≥ 70
Similarly the equation of the line joining (0,14) and (19,14), is given by y=14.
Since it is shaded towards the origin, hence y≤14.
The equation of the line joining (5,0) and (19,14) is given by

xy − 14x − 5y + 70 = xy − 19y
14x − 14y = 70 or x − y = 5. Since it is shaded towards the origin, x − y ≤ 5
Hence the shaded area is given by 14x + 5y ≥70, y ≤ 14, x − y ≤ 5

Let log₁₆x=t
Hence, t²−t + log₁₆k = 0 and change k > 0
The equation has exactly one solution. Hence, the discriminant must be zero




So, number of real values of k is 1 when k = 2.

We have

Now two cases arise :
Case I : When , i.e., .

Then
and
or and
and

or and

We have, 37−(3x + 5) ≥ 9x − 8(x − 3) (37 − 3x − 5) ≥ 9x − 8x + 24
⇒ 32 − 3x ≥ x + 24
Transferring the term 24 to L.H.S. and the term (−3x) to R.H.S. 32 − 24 ≥ x + 3x
⇒ 8 ≥ 4x
⇒ 4x ≤ 8
Dividing both sides by 4 ,


Solution set is (−∞, 2].

The log functions are defined if and and
Now the inequality is

The shaded region in the figure lies between x = −3 and x = 3 not including the line x = −3 and x = 3 (lines are dotted). Therefore, −3  <x < 3
⇒ |x| < 3 [∵|x| < a ⇔ −a < x < a]

As x ≠ 4 hence x ∈ (2, 4) ∪ (4, 6)
Since, (4,6) is lying in the interval for all the acceptable values of x ∈ (2, 4)∪ (4, 6) so this will be the best option.
∴ The solution of the equality  is x ∈ (4, 6).

Let Ravi got x marks in third unit test.
∴ Average marks obtained by Ravi
Now, it is given that he wants to obtain an average of at least 60 marks. At least 60 marks means that the marks should be greater than or equal to 60 i.e.


Now, transferring the term 145 to R.H.S.,
i.e. Ravi should get greater than or equal to 35 marks in third unit test to get an average of at least 60 marks.
Minimum marks Ravi should get = 35.

Given and

which is true if
, which is true iff

, which is also

If x cm is the breadth, then 2(3x + x) ≥ 160
⇒x ≥ 20

Let

D = 

We have the following cases:
CASE I When x ∈ Z
In this case, we have

CASE II When this case, we have
x = n + k, where ∈ Z and 0 (∴(x) = n + 1 and (x + 1) = n + 2
Now,

Hence, x∈(−5, −4] ∪ (2, 3]

|x − 1| is the distance of x from 1 . |x − 3| is the distance of x from 3 The point x = 2 is equidistant from 1 and 3 . Hence, the solution consists of all x ≥ 2.

Using A.M./G.M. one can show that

holds. Also, Therefore,

or
or
or
Therefore, holds. Again,

and so on. Adding the inequalities, we get

Therefore, does not hold.

We have

base .


or

The inequality is Dividing the problem into three intervals :
(i) If , then

But , hence no common values

(ii) If , then
But , hence no common values
(iii) If , then

We have
or
or or
or or
Thus, all real numbers which are greater than or equal to 1 is the solution set of the given inequality.

∴ 

Clearly, this is defined for (i)

Thus from (i) and (ii)
The solution set is (−2, 3)