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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(3^{6}) = 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(7^{3}) + 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) = x^{3} - 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) = 0^{3}– 4(0) + p = p.

=> f(1) = 1^{3} - 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 = { a_{1}, a_{2}, a_{3}, a_{4}, ................, 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 a_{1} and a_{2} and the next number a_{3}. 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 ax^{2} + bx + c. So, f(x) = ax^{2} + bx + c

At x = 0, the value of function is 1.

**x = 0, f(x) = 1**

ax^{2} + 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 (ax^{2} + 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 -2x^{2} + 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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