CAT 2000 Question Paper

In this graph, f(-1) = 1 and f(1) = 2 and since the two lines pass origin, their values increase linearly.

3f(x) = 6f(-x)

In this graph, we can see that the graph on LHS of y axis is symmetrical to the graph on RHS of y axis about origin.

Hence f(x) = -f(-x)

After first instruction: A = 5-2 = 3L. C = 2L. In the third instruction, A yields 2L to C. It means the capacity of A before third operation was 3L. It means that C's water was either drained or transferred to B. Option B is correct.

We know that after 3 operations A contain 1 ltr and 4 th operation is Drain(a) , so 1 ltr is drained and only 4 ltrs remain in all there cylinders. Also after 6 operations A should have all the 4 ltrs in it. hence the other 2 vessels must contain 0 ltrs because there is only one drain operation in the entire sequence.

f(2) = 1/3

$$f^2(2)$$ = f(1/3) = 3/4

$$f^3(2)$$ = f(3/4) = 4/7

$$f^4(2)$$ = f(4/7) = 7/11

$$f^5(2)$$ = f(7/11) = 11/18

So, product = 1/18

f(-2) = -1

$$f^2(-2)$$ = f(f(-2)) = f(-1) = 0

$$f^3(-2)$$ = f(f(-2)) = f(0) = 1

So, sum = 0

Case 1: $$a_1=0$$
So, D equals $$0.0a_20a_20a_2...$$
So, 100D equals $$a_2.0a_20a_2...$$
So, 99D equals $$a_2$$

Case 2: $$a_2=0$$
So, D equals $$0.a_10a_10a_1...$$
So, 100D equals $$a_10.a_10a_1....$$
So, 99D equals $$a_10$$

So, in both the cases, 99D is an integer. From the given options, only option C satisfies this condition (198=2*99) and hence the correct answer is C. 

y=a+bx is a linear function and clearly the function shown in the table is not linear.
Lets take option B:

Suppose $$y=f(x)=a+bx+cx^2$$
f(1)=4 => a+b+c=4 -- (1)
f(2)=8 => a+2b+4c=8 -- (2)
f(3)=14 => a+3b+9c=14 -- (3)

Let us solve the equations and see if $$f(4), f(5),f(6)$$ also satisfy the given equation.

(2)-(1) => b+3c=4
(3)-(2) => b+5c=6
This gives c=1 and b=1 ==> a=2

So, $$f(x)=2+x+x^2$$
$$f(4)=2+4+4^2 = 22$$
$$f(5)=2+5+5^2 = 32$$
$$f(6)=2+6+6^2 = 44$$

As all the three also satisfy the numbers given in the table, it can be inferred that the relationship between x and y is quadratic and the correct option is option (b)

$$a_2 = 2*1 + 5$$
$$a_3 = 2*(2 + 5) + 5 = 2^2 + 5*2 + 5$$
$$a_4 = 2^3 + 5*2^2 + 5*2 + 5$$ 
...
$$a_{100} = 2^{99} + 5*(2^{98} + 2^{97} + ... + 1)$$
$$= 2^{99} + 5*1*\dfrac{(2^{99} - 1)}{(2-1)} = 2^{99} + 5*2^{99} - 5 = 6*2^{99} - 5$$

$$\dfrac{1}{(2^2-1)}+\dfrac{1}{(4^2-1)}+\dfrac{1}{(6^2-1)}+...+\dfrac{1}{(20^2-1)}$$ 

= $$\dfrac{1}{(2+1)(2-1)}+\dfrac{1}{(4+1)(4-1)}+..+\dfrac{1}{(20+1)(20-1)}$$

= 1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/(19*21) 

=1/2 * ( 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... +1/19 - 1/21)

=1/2 * (1 - 1/21) = 10/21