Functions Test - 1

Given, f(2401) = 10/3

⇒ f(49*49) = 10/3 

⇒ f(49) + f(49) = 10/3 

⇒ 2 * f(49) = 10/3

⇒ 2 * f(7*7) = 10/3 

⇒ 2 * [f(7) + f(7)] = 10/3  

⇒ 4 * f(7) = 10/3

⇒ f(7) = 5/6

Also given that, f(729) = 18

⇒ f(3⁶) = 18  

⇒ f((√3)¹²) = 18

f(√3)¹²) can be re-written as:   f [(√3)⁶ * (√3)⁶)] = f [(√3)⁶]+ f [(√3)⁶]

f [(√3)⁶)] can be re-written as: f [(√3)³ * (√3)³] = f [(√3)³)] + f [(√3)³] 

f((√3)³) can be re-written as: f [(√3)² * √3)] = f((√3)²) + f((√3))

f(√3)² can be re-written as: f(√3 * √3) =  f(√3) + f(√3)

So, f(√3)¹²) can also be expressed as: 

f(√3) + f(√3) + f(√3) + f(√3) + ... upto 12 terms = 12 * f(√3)

So, 12 * f(√3) = 18

⇒ f(√3) = 3/2

So, f(343 * √3) =  f(343) + f(√3)

 = f(7³) + f(√3)

= 3*f(7) + f(√3)

= 3 * 5/6 + 3/2 

= 4

This is very simple :

f(x) = 2x+2 

f(3) = 2*3+2=8

f(f(3))= 2*f(3)+2

       = 2*8+2

       = 18

Hence (A) is the answer

Here's the Solution :

f(1)+2*f(6-1)=1......... (1)

f(5)+2*f(6-5)=5......... (2)

Substituting we have (2) in (1) we have :-

-3f(1)=-9,

Hence answer f(1)=3.

Hence (A) is the correct answer.

We have
=> f(0) = 0³– 4(0) + p = p.
=> f(1) = 1³ - 4(1) + p = p – 3.
If P and P – 3 are of opposite signs then p(p – 3) < 0.
Hence 0 < p < 3.

We need to find f(1/2).
According to the question, f(x) * f(y) = f(xy).
So, f(2) * f(1/2) = f(2*1/2) = f(1).                        ----------------(i)
But we do not know f(1).
Using the same formula, f(1)*f(2) = f(1*2) ;

i.e. f(1)*f(2) = f(2)  f(1) = 1.

Substituting for f(1) in (i), we get f(2) * f(1/2) = 1.
f(2) = 4, so f(1/2) = 1/4.

Let A = { a₁, a₂, a₃, a₄, ................, a_n }
Initially, the function 'h' is applied to a1 and a2 and obtain a G.C.F.

Now, the function 'h' is applied to the HCF of a₁ and a₂ and the next number a₃. This process is continued till the last number.

The final G.C.F is the G.C.F of the set A and the number of times we used the function 'h' is 1 [for a1 & a2 ] + (n-2) [for the rest of the values] i.e n-1.

f(x) is the maximum value of 2x + 1 and 3 – 4x.

Here, we see that in 1 place, x is added to something else, while in the other equation (3 – 4x), x is subtracted from something else.

So, in the 1st equation, as x goes higher, f(x) will become bigger, while in the second case, when x goes bigger, the value of f(x) goes smaller, and it will be vice-versa when x goes lower.

So, we need to find an optimum solution, which we can obtain by equating both the equations.
2x + 1 = 3 – 4x. On solving, we get x = 1/3 and f(x) = 5/3.
If we put any value more than 1/3, 2x + 1 becomes high, and f(x) increases.
If we put any value less than 1/3, 3 – 4x becomes high, and f(x) increases.
So, for x = 1/3, f(x) = 5/3 is the minimum possible value.

Let, the quadratic equation be ax² + bx + c. So, f(x) = ax² + bx + c
At x = 0, the value of function is 1.
x = 0, f(x) = 1
ax² + bx + c = a * 0 + b * 0 + c = c
c = 1.
At x = 1, f(x) = 3
x = 1, f(x) = 3
a *1 + b * 1 + c = 3
Since c = 1, a + b = 2.
Also, we know that f(x) is maximum when x = 1. If f(x) is maximum, (dx/dt)(f(x)) = 0 
Differentiating f(x), we have d/dt (ax² + bx + c) = 2ax + b
At x = 1, 2ax + b = 0.
2a + b = 0.
b = -2a.
Substituting we have a + b = 2, or a + -2a = 2.  a = -2. So, b = 4.

So the equation is -2x² + 4x + 1. 
At x = 10, the value is -2 * 100 + 4 * 10 + 1 i.e -159.

f(1) = 3600 ,  f(1) + f(2) = 4 * f(2).  

f(2) = f(1)/3.   We can write this as f(2) = f(1) * 1/3.

Now, f(1) + f(2) + f(3) = 9 * f(3).

f(3) = f(1) + f(2) / 8. We know f(2) = f(1) / 3.

So, f(3) = f(1) + (f(1)/3) / 8.

We can write this as f(3) = f(1) *1/3 * 2/4. Since we already have f(1) * 1/3, we write it in such a way that the equation has f(1) * 1/3 and then what comes rest, so that we can generalize
Calculating similarly for f(4), we get f(4) = f(1) * 1/3 * 2/4 * 3/5.
So, f(9) will be f(1) * 1/3 * 2/4 * 3/5 * 4/6 * 5/7 *6/8 * 7/9 * 8/10.
=> f(9) = (f(1) * 1 * 2) / (9 *10).
=> f(9) = 3600 * 2/ 9 * 10 = 80.

Given, g (x+1) = g (x) – g (x-1)
Let g (x) = p and g (x-1) = q.
Then,  g (x+1) =  g (x) – g (x-1)  =  p – q
g (x+2) = g (x+1) – g (x)  =  p – q – p = -q
g (x+3) = g (x+2) – g (x+1) = -q-p + q  = -p
g (x+4) = g (x+3) – g (x+2)  = q - p
g (x+5) = g (x+4) – g (x+3)  = q –p +p  = q = g (x-1)
So, we see g (x+5) = g (x-1).

Thus, entry repeats after 6 times i.e p = 6.