CMAT Functions, Graphs and Statistics Questions

Statement I:

Given arithmetic mean of (7,8,9,a,b,10,8,7) is 9 and mode is 8

7+8+9+a+b+10+8+7 = $$9\times\ 8$$

49+a+b = 72

a+b = 23

It is mentioned that mode is 8, 8 occurred twice and 7 also occurred twice. Therefore, a or b should be 8

If a = 8, b = 15

If b = 8, a = 15

ab = 120

Statement I is true

Statement II:

a + b + ab = 84

1 + a + b + ab = 85

(1+a)(1+b) = 85

85 = 1 $$\times\ $$ 85 or 5 $$\times\ $$ 17

It cannot be 1$$\times\ $$85, as it is mentioned both a and b are positive integers

If 1+a = 5, 1+b = 17

      a =4 and b = 16

If 1+a = 17, 1+b = 5

     a = 16 and b = 4

In both cases, a+b = 20

Statement II is true

Answer is option A.

Graph A and D represent the function of x, because a vertical line is drawn in (a) meets the graph at only one point i.e., for one x, in the domain there exist only one f(x) in the codomain.

In these type of questions
we go through the options :
Now when x=6 or -6 the value of y is 0
so let us put value of y=0 in all options and check which option satisfies x=6 and -6
Putting x=6 and y=0 in option A
we get |6+0|-|6-0| = 0
But RHS =6 so this option is discarded
Now putting in option B
|6-0|+|6+0| =12
But RHS is 10
so this option is also discarded .
Now option C we get
|6|-0 =6
so we can say graph can be of |x|-|y|=6
Now satisfying option D
we get |6+0|-|6-0|=0
So we can say 6,0 and -6,0 satisfies option D as well
Now you have to eliminate between C and D
If you observe |x+y|-|x-y|=0 will pass through origin
and the graph here is not passing through origin so you can eliminate option D
Therefore the graph will be |x|-|y|=6

The question appeared in CMAT 2021 Slot 2 and the question is incorrect . Marks were awarded to students.

As per the empirical rule, approximately 99.7% of the data in a bell-shaped distribution lies within three standard deviations of the mean, or $$\mu \pm 3\sigma$$.  $$\mu \pm 6\sigma$$ will have approximately 99.99967% of the data. Hence, option C.

Given f(x) = $$x^3 - 4x + p$$ and  f(0) and f(1) are of opposite signs.

$$\Rightarrow f(0)\times f(1) < 0$$

$$f(0) = 0^3 - 4*0 + p = p$$

$$f(1) = 1^3 - 4*1 + p = p-3$$

$$\Rightarrow (p)(p-3) < 0$$

$$\Rightarrow 0 < p < 3$$.

Let say f(x) = $$x^3 + 2x^2 + bx + 68$$ and given (x+4) is a factor of f(x).

$$\Rightarrow f(x) = (x+4)\times k$$.................(1)

where 'k' is the quotient when f(x) is divided by (x+4).

Substituting 'x= -4' in the equation (1), we get

f(-4) = 0

$$\Rightarrow -4^3 + 2 (-4)^2 + b (-4) + 68 = 0$$

$$\Rightarrow -64 + 32 - 4b + 68 = 0$$

$$\Rightarrow 4b = 36$$

$$\Rightarrow b = 9$$.

Given      0<a<1..............................................(1)

Let us go with the verification of options

Option (i):

squaring on both sides of (1), we get

0<$$a^2$$<1

Subtracting '1' in the above equation, we get

-1<$$a^2$$-1<0....................(2)

$$\Rightarrow a^2$$-1< 0, Hence option (i) is false.

Option (ii):

Squaring on both sides of (1), we get

0<$$a^2$$<1

Adding '1' on both sides of above equation, we get

1<$$a^2$$+1<2....................(3)

$$\Rightarrow a^2$$+1>0, Hence Option (ii) is true.

Option (iii):

As 0<a<1, $$a^2<a$$.............(4)

Subtracting 'a' on both sides of (4), we get

$$a^2-a<0$$, Hence Option (iii) is also false.

Modulus(2a-3) = 3a+2, which can also be written as |2a-3| = 3a+2.

Modulus function  f(x) = |x| is defined as

|x| = x, for x>0,

|x| = -x, for x<0

So, we get two cases here,

Case (1):

2a-3>0

$$\Rightarrow a>\frac{3}{2}$$........................(1)

|2a-3| = 2a-3 = 3a+2

$$\Rightarrow a = -5$$

But from equation (1), The above value of a = -5 is not a possible case.

Case (2):

2a-3<0

$$\Rightarrow a<\frac{3}{2}$$....................(2)

|2a-3| = -(2a-3) = 3a+2

$$\Rightarrow a = \frac{1}{5}$$

This is a possible case, hence the value of $$a = \frac{1}{5}$$