FUNCTIONS MCQ LIST

 Question 1:

If $f\left(x\right)=x^3-\frac{1}{x^3},$ then $f\left(x\right)+f\left(\frac{1}{x}\right)=?$

[1] 0

[2] 1

[3] -1

[4] 2

Answer & Solution

Option # 1

We have, $f\left(x\right)=x^3-\frac{1}{x^3},f\left(\frac{1}{x}\right)=\frac{1}{x^3}-x^3,f\left(x\right)+f\left(\frac{1}{x}\right)=0$

---

Question 2:
Let $f(x+\frac{1}{x})=x^2+1/x^2(x\ne 0),$ then $f\left(x\right)=?$

[1] x2

[2] x2 – 1

[3] x2 – 2

[4] None of these

Answer & Solution

Option # 3

Let $z=x+1/x,$ then

$f\left(z\right)=f(x+1/x)=x^2+1/x^2$

$=(x+1/x{)}^2-2=z^2-2.$ Hence $f\left(x\right)=x^2-2$

---

Question 3:
If f(x) is a polynomial satisfying f(x) f(1/x) = f(x) + f(1/x) and f(3) = 28, then f(4) = ?

[1] 63

[2] 65

[3] 17

[4] None of these

Answer & Solution

Option # 2

Any polynomial satisfying the functional equation

f(x).f(1/x) = f(x) + f(1/x) is of the form $1+x^n$ or $1-x^n$

If $28=f\left(3\right)=-3^n+1$ then $3^n=-27$, which is not possible for any n

Hence $28=f\left(3\right)=3^n+1\Rightarrow 3^n=27\Rightarrow n=3.$ Thus $f\left(x\right)=4^3+1=65$

---

Question 4:
For a real number x,

let f(x) = 1/(1 + x) if x is nonnegative

             = 1 + x, if x is negative

$f^n\left(x\right)=f\left(f^{n-1}\right(x\left)\right),n=2,3\dots$

What is the value of the product $f\left(2\right)f^2\left(2\right)f^3\left(2\right)f^4\left(2\right)f^5\left(2\right)$?

[1] 1/3

[2] 3

[3] 1/18

[4] None of These

Answer & Solution

Option # 3

$f\left(2\right)=\frac{1}{3},f^2\left(2\right)=\frac{3}{4},f^3\left(2\right)=\frac{4}{7},f^4\left(2\right)=\frac{7}{11},f^5\left(2\right)=\frac{11}{18}$

---

Question 5:
The domain of the function $f\left(x\right)=\sqrt{(2-2x-x^2)}$ is

[1] $-\sqrt{3}\le x\le \sqrt{3}$

[2] $-1-\sqrt{3}\le x\le -1+\sqrt{3}$

[3] $-2\le x\le 2$

[4] $-2+\sqrt{3}\le x\le -2-\sqrt{3}$

Answer & Solution

Option # 2

We must have $2-2x-x^2\ge 0\Rightarrow x^2+2x-2\le 0$

$\Rightarrow (x+1{)}^2-3\le 0$ or $-\sqrt{3}\le (x+1)\le \sqrt{3}$ or $-1-\sqrt{3}\le x\le -1+\sqrt{3}$

---

Question 6:
$f\left(x\right)=\sqrt{\left(\frac{(x+1)(x-3)}{(x-2)}\right)}$ is a real value function in the domain

[1] $(-\infty ,-1]\cup [3,\infty )$

[2] $(-\infty ,-1]\cup [2,3]$

[3] $[-1,2)\cup \left[3\infty \right)$

[4] None of these

Answer & Solution

Option # 3

$\sqrt{\left\{g\right(x\left)\right\}}$ is real if $g\left(x\right)\ge 0$

so we should have

$(x+1)(x-3)/(x-2)\ge 0$

Which hold in the domain

$[-1,2)\cup [3,\infty )$

---

Question 7:
The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained at x = ?

[1] 2

[2] 4

[3] 8

[4] 10

Answer & Solution

Option # 4

In all such questions, the minimum value will be obtained at one of the critical points. So check the value of f(x) at x = 10, 2 and 4 and see which gives the least value. Here f(10) = 2, f(2) = 6, f(4) = 8 and f(8) = 4. The minimum occurs at x = 10

---

Question 8:
[ x ] denotes the greatest integer less than or equal to X. If X is a positive integer and $\left[\frac{X}{5}\right]-\left[\frac{X}{7}\right]=1$. If the minimum value of X is a and the maximum value is b, then a + b = ?

[1] 40

[2] 33

[3] 35

[4] 34

Answer & Solution

Option # 4

X = 5 (minimum) and X = 29 (maximum). Sum = 34.

---

Question 9:
If $0<x<1000,$ and $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{5}\right]=\frac{31x}{30}$ where [x] denotes the greatest integer less than or equal to x, then the number of possible values of x will be

[1] 30

[2] 33

[3] 43

[4] 45

Answer & Solution

Option # 2

x must be a multiple of 30. So there are 33 solutions.

---

Question 10:
The maximum possible value of $y=min(\frac{1}{2}-\frac{3x^2}{4},\frac{5x^2}{4})$ for the range $0<x<1$ is

[1] 1/3

[2] 1/2

[3] 5/27

[4] 5/16

Answer & Solution

Option # 4

As x changes from 0 to 1, as long as x is less than 1/2, (1/2 – 3x2 /4) is greater than 5x2/4 and is the value of y. As x becomes greater than 1/2, 5x 2 /4 is greater than (1/2 – 3x2 /4), x become equal to $\frac{1}{2},\frac{5x^2}{4}$ and $(\frac{1}{2}-\frac{3x^2}{4})$ are equal to $\frac{5}{16}$