Test: Functions- 1


10 Questions MCQ Test UPSC CSAT Preparation | Test: Functions- 1


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QUESTION: 1

f (x*y) = f(x) + f(y)

f(2401) = 10/3 and f(729) = 18

then f(343 * √3) = ?

Solution:

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)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

QUESTION: 2

If f(x) = 2x+2 what is f(f(3))? 

Solution:

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

QUESTION: 3

A certain function f satisfies the equation f(x)+2*f(6-x) = x for all real numbers x. The value of f(1) is

Solution:

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.

QUESTION: 4

If f(x) = x3 - 4x + p, and f(0) and f(1) are of opposite sign, then which of the following is necessarily true ?

Solution:

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.

QUESTION: 5

Let f(x) be a function satisfying f(x) f(y) = f (xy) for all real x, y. If f (2) = 4, then what is the value of f (1/2)?

Solution:

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.

QUESTION: 6

For two positive integers a and b define the function h(a,b) as the greatest common factor (G.C.F) of a, b. Let A be a set of n positive integers. G(A), the G.C.F of the elements of set A is computed by repeatedly using the function h. The minimum number of times h is required to be used to compute G is 

Solution:

Let A = { a1, a2, a3, a4, ................, a}
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.

QUESTION: 7

Let f (x) = max (2x + 1, 3 − 4x), where x is any real number. Then the minimum possible value of f(x) is :-

Solution:

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.

QUESTION: 8

A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f(x) at x = 10 ?

Solution:

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.

QUESTION: 9

A function f(x) satisfies f(1) = 3600, and f(1) + f(2) +...+ f(n) = n² f(n), for all positive integers n > 1. What is the value of f(9) ?

Solution:

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.

QUESTION: 10

Let g (x) be a function such that g (x + 1) + g (x – l) = g (x) for every real x. Then for what value of p is the relation g (x + p) = g (x) necessarily true for every real x ?

Solution:

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.

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