XAT Functions, Graphs and Statistics Questions

$$f(x) = ax+b$$.
$$f(f(x)) = f(ax+b) = a(ax+b)+b = a^2x + ab+b$$
We have been given that $$f(f(x)=9x+8$$
$$9x+8=a^2x+ab+b$$
Equating the co-efficient of $$x$$, we get, $$a^2=9$$. Therefore, $$a$$ can be $$3$$ or $$-3$$. But, it has been given that $$a$$ is a positive real number. 
Equating the constants, we get, $$ab+b=8$$
If we substitute $$a=3$$, we get, $$3b+b=8$$
$$4b=8$$
$$b=2$$
Therefore, $$a+b=3+2=5$$
Therefore, option C is the right answer. 

Expression : $$f(x) = 2x + f(x - 1)$$ and $$f(1) = 1$$

Putting, $$x = 2$$

=> $$f(2) = 2 \times 2 + f(1) = 4 + 1 = 5$$

and $$f(3) = 2 \times 3 + f(2) = 6 + 5 = 11$$

Similarly, $$f(4) = 2 \times 4 + f(3) = 8 + 11 = 19$$

The pattern followed is : $$f(n) = n^2 + (n - 1)$$

Now, $$n = 31$$

$$\therefore f(31) = 31^2 + (31 - 1) = 961 + 30 = 991$$

When x  = -3, y = -10
This is satisfied only in option D.

Hence, option D is the correct answer.

$$f(x^2 - 1) = x^4 - 7x^2 + k_1$$

Put $$x^2 = 1$$ to make it 0

=> $$f(0) = (1)^2 - 7(1) + k_1 = k_1 - 6$$ --------(i)

Also, $$f(x^3 - 2) = x^6 - 9x^3 +k_2$$

Put $$x^3 = 2$$

=> $$f(0) = (2)^2 - 9(2) + k_2 = k_2 - 14$$ -----------(ii)

Equating (i) & (ii), we get :

=> $$k_1 - 6 = k_2 - 14$$

=> $$k_2 - k_1 = 14 - 6 = 8$$

Expression : $$f (x + a) = f (a \times x)$$

Also, $$f(1) = 4$$

Now, $$f(1003) = f(1002 + 1) = f(1002 \times 1) = f(1002)$$

Similarly, $$f(1002) = f(1001) = f(1000) = ......... = f(1) = 4$$

$$\therefore f(1003) = k = 4$$

Mean of the six numbers = 15
So, the sum of the numbers = 15 * 6 = 90
As the median is 18, the mean of middle two numbers must be 18 and thus, their sum must be 36.
Also, the mode is a number less than 18. So, the mode must be appearing as the first and the second number of the six given integers, when arranged in ascending order. To maximize the largest integers, the mode must be equal to 1.
Therefore, out of the six integers, two are 1 and 1.
For the middle two numbers whose sum is 36, we cannot have 18 and 18 because then we will have two modes which is inappropriate as per the question.
So, the middle numbers must be 17 and 19.
The fifth integer can be 20.
Maximum possible value of the largest integer = 90 - (1 + 1 + 17 + 19 + 20) = 32
Hence, option D is the correct answer.

It has been given that f(f(x)) = 15.
From the graph, we can see that the value of f(4) = 15 and f(12) = 15
Therefore, f(x) can be 4 or 12.

When f(x) = 4, x can take 4 values
When f(x) = 12, x can take 3 values.
Therefore, there are 4+3 = 7 solutions in total.
Therefore, option C is the right answer.

A logarithm function of the form $$log_a b$$ is true, iff $$b > 0$$

and $$c = a^b$$ can be written as = $$log_a c = b$$

We have, $$f(x) =log_{7}({ log_{3}(log_{5}(20x-x^{2}-91 )))}$$

=> $$log_{3}(log_{5}(20x - x^{2} - 91 )) > 0$$

=> $$log_{5}(20x - x^{2} - 91 ) > (3)^0$$

=> $$log_{5}(20x - x^{2} - 91 ) > 1$$

=> $$20x - x^{2} - 91 > (5)^1$$

=> $$x^2 - 20x + 96 < 0$$

=> $$(x - 8) (x - 12) < 0$$

=> $$8 < x < 12$$

$$\therefore$$ Domain of $$f(x)$$ = $$(8,12)$$

Let us draw the diagram first,

Let L$$_{1}$$: 2x+ 3y -1 = 0

L$$_{2}$$: x+ 2y -3 = 0

L$$_{3}$$: 5x-6y- 1 = 0

With respect to L$$_{1}$$, we can see that the point $$(a, a^{2})$$ lies within the triangle and (0, 0) are opposite side. Therefore,

 L$$_{(a, a^2)}$$* L$$_{(0, 0)}$$ < 0

$$\Rightarrow$$ $$(2a+ 3a^2-1)(-1)<0$$

$$\Rightarrow$$ $$(3a^2+2a-1)>0$$

$$\Rightarrow$$ a < -1 or a > $$\frac{1}{3}$$   ... (1)

With respect to L$$_{2}$$, we can see that the point $$(a, a^{2})$$ lies within the triangle and (0, 0) are on the same side. Therefore,

 L$$_{(a, a^2)}$$* L$$_{(0, 0)}$$ > 0

$$\Rightarrow$$ $$(a+ 2a^2-3)(-3)>0$$

$$\Rightarrow$$ $$(2a^2+a-3)<0$$

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

With respect to L$$_{3}$$, we can see that the point $$(a, a^{2})$$ lies within the triangle and (0, 0) are on the same side. Therefore,

 L$$_{(a, a^2)}$$* L$$_{(0, 0)}$$ > 0

$$\Rightarrow$$ $$(5a- 6a^2-1)(-1)>0$$

$$\Rightarrow$$ $$(6a^2-5a+1)>0$$

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

From equation (1), (2) and (3) we can say that 

a $$\epsilon$$ (-3/2,-1) ⋃ (1/2, 1). Hence, option C is the correct answer.

Expression : $$f(x + y) = f(x).f(y)$$

and $$f(1) = 2$$

Putting, $$x=y=1$$, => $$f(1 + 1) = f(1).f(1)$$

=> $$f(2) = 2 \times 2 = 4$$

If $$x = 2 , y = 1$$ => $$f(3) = f(2).f(1)$$

=> $$f(3) = 4 \times 2 = 8$$

Similarly, $$f(4) = f(3).f(1) = 8 \times 2 = 16$$

The pattern followed : $$f(n) = 2^n$$

Now,  $$\sum_{x=1}^nf(x) = 1022$$

= $$f(1) + f(2) + f(3) + ............+ f(n) = 1022$$

=> $$2^1 + 2^2 + 2^3 + ......... + 2^n = 1022$$

The series is a G.P. with first term, $$a = 2$$ and common ratio, $$r = 2$$

=> Sum of G.P. = $$\frac{a (r^n - 1)}{r - 1}$$

=> $$\frac{2 (2^n - 1)}{2 - 1} = 1022$$

=> $$2^n - 1 = \frac{1022}{2} = 511$$

=> $$2^n = 511 + 1 = 512 = 2^9$$

Comparing both sides, we get : $$n = 9$$

$$x_1 , x_2 , x_3,......., x_n$$ are in A.P. Let the first term be $$a$$ and common difference be $$d$$

Also, $$x_2 = a + d = 2$$

and $$x_{24} = a + 23d = 68$$

Solving above equations, we get : $$a = -1$$ and $$d = 3$$

=> $$x_8 = a + 7d = -1 + 7(3) = 20$$

To find y coordinate, we will use $$y = px + q$$

$$\because A_2 (2, -2)$$ and $$A_{24} (68, 31)$$

Substituting in above equation, => $$-2 = 2p + q$$

and $$31 = 68p + q$$

Solving above equations, we get : $$p = \frac{1}{2}$$ and $$q = -3$$

$$\therefore y_8 = p x_8 + q = \frac{1}{2} (20) + (-3) = 7$$ 

$$x_1 , x_2 , x_3,......., x_n$$ are in A.P. Let the first term be $$a$$ and common difference be $$d$$

Also, $$x_2 = a + d = 2$$

and $$x_{24} = a + 23d = 68$$

Solving above equations, we get : $$a = -1$$ and $$d = 3$$

=> x coordinates = {$$x_1 , x_2 , x_3, x_4 , x_5 ,.......$$} = {$${-1 , 2 , 5 , 8 , 11 , 14 ,......}$$}        -------------(i)

To find y coordinates, we will use $$y = px + q$$

$$\because A_2 (2, -2)$$ and $$A_{24} (68, 31)$$

Substituting in above equation, => $$-2 = 2p + q$$

and $$31 = 68p + q$$

Solving above equations, we get : $$p = \frac{1}{2}$$ and $$q = -3$$

=> $$y_1 = px_1 + q = \frac{1}{2}(-1) + (-3) = -3.5$$

Similarly, y coordinates = {$$y_1 , y_2 , y_3, y_4 , y_5 ,.......$$} = {$${-3.5 , -2 , \frac{-1}{2} , 1 ,......}$$}         -------(ii)

From (i) and (ii),

=> $$A_1 = (-1 , -3.5)$$

$$A_2 = (2 , -2)$$

$$A_3 = (5 , \frac{-1}{2})$$

$$A_4 = (8 , 1)$$

For the points to lie in the first quadrant, the coordinates of both $$(x_k , y_k)$$ must be positive.

Since, $$d$$ is positive and $$y$$ is a linear relation in $$x$$, the corresponding coordinates of $$A_k$$ i.e. $$A_5$$ onwards will be increasing.

Thus, only for $$A_1 , A_2$$ and $$A_3$$ we do not have both coordinates positive.

$$\therefore$$ Only 3 points do not lie in the first quadrant.