Mathematics MCQ 1 - IIT JAM MCQ

Note that


, hence minimum value is 1 and it is attained at Z = 0, 1/2

Equation is x² - |x| - 6 = 0
Case I: x > 0. then we have
x² - x - 6 = 0 (lx| = x)
⇒ (x-3)(x+2) = 0
⇒ x = 3 is the solution as x > 0.
(So x = -2 can’t be solution)
Case II: x < 0. then we have
x² + x - 6 = 0    (|x| = -x)
⇒ (x + 3)(x - 2) = 0
⇒ x = -3 is the solution as x < 0
(So x = 2 can't be solution)
∴ product of roots = 3. - 3 = -9

Two dice are rolled simultaneously, hence total number of elements in sample space is = 6 * 6 = 36
We have event E is the collection of those elements having sum greater or equal than 10.
i.e. E = {(4,6),(5.6).(6.6).(6.5).(6.4).(5.5)}

Given that f(x) = 2x + 3, g(x) = x²+7
∴   g(f(x)) = g(2x + 3) = (2x + 3)² + 7
= 4x² + 9 + 12x+ 7 = 4x² + 12x + 16
Given that g (f (x)) = 8
⇒ 4x² + 12x + 16 = 8
⇒ 4x² + 12x + 8 = 0
⇒ 4(x² + 3x + 2) = 0
⇒ 4(x + l)(x + 2) = 0
∴ x = -1 and x = -2

Hence limit lies between 0 and 1/2

Method-I: From the figure, it is clear that ABCD fonn a square having each side √2.


Method -II: Area ofABCD = 4 x Area of OBC
  (as equation of CB  is y = -x+ 1)

 (using De'morgans law) 
 
= φ

(d) : For (3, 9) ∈ R, (9, 3) ∉ R 

Therefore,relation is not symmetric which means our choice 

(a) and (b) are out of court. We need to prove reflexivity and transitivity. 

For reflexivity a ∈ R, (a, a) ∈ R which is hold i.e. R is reflexive. Again, 

for transitivity of (a, b) ∈ R , (b, c) ∈ R 

⇒ (a, c) ∈ R 

which is also true in R = {(3, 3)(6, 6), (9, 9), (12, 12), (6,12), (3, 9), (3, 12), (3, 6)}.

Given circle x² + y² - 2x + 4y - 8 = 0
Comparing with the general equation of circle x² + y² + 2gx + 2fy - c = 0, we get g = -1. f = 2, c = -8

For throwing a dice, sample space S = {1,2,3,4,5,6} and
A = {1,3,5}
B = {2,4,6}
C = {4,5,6}
D = {1,2}

Given that f (x) = |x| + |x-1|, then f(1) = 1
Since absolute volue functions are continuous everywhere so f(x) = |x| + |x-1|. being the sum of two continuous function is continuous everywhere. Now we check differentiability at x = 1, we have




Hence Lf'(1) ≠ Rf'(1)
∴ Derivative do not exist at x = 1.

Note that if a. b. c are three consecutive terms

 ......(1)
differentiating both side, we have

i.e.c =yy’ ...(2)
from (1) we have


Squaring both side we have

Hence order of differential equation is 1 and degree is 3.

Let slope of a line making 60° angle with 



So. there will be two lines of such type. One is having slope  and the otehr one is having slope 0. 
Therefore.
Line 1: passing through (3, -2) and slope 

Line 2: passing through (3.-2) and slope 0.

Option (A) and (C) are correct.

In BANANA. Letter A reapets 3 times and N reapets 2 times.
Total number of arrangements of word BANANA is 
Let both N s are appear together, then they are considered is single letter.
In this way total number of arrangements are 
Hence total number of arrangements where N do not appear adjacently is = 60 - 20 = 40

For critical point, we have f'(x) = 0

Note that sign of f' (x) depends on the sign of (1 - 4x).

hence f (x) is increasing when and f (x) is decreasing when 
∴ f (x) is maximum at x = 1/4 

Clearly from options, we have
b = 1. a = -1
Hence (0) is answer

n(E) = 4     n(F) = 2
Then total number of onto functions from E to F are =
Note: If n(A) = n and n(B) = m then total number of onto functions from A to B are

3x + 4y = 9 and 6x + 8y = 15 

We know that the distance between the two parallel lines ax + by = c₁ and ax + by = c₂ is