Solved Examples

1. Given that f(x) = x×$\times$f(x - 1) for any natural number x. If f(x + 2) = 20*f(x), then what is the value of x?
a. 2           b. 4            c. 5            d. 1          e. None of these
Given:f(x) = x×$\times$f(x - 1) ....(a)
and f(x + 2) = 20×$\times$f(x) ....(b)
Substitute x = x + 1 in (a)
Then, f(x + 1) = (x + 1)×$\times$f(x) .........(c)
Substitute x = x + 2 in (a)
Then,  f(x + 2) = (x + 2)×$\times$f(x + 1) ......(d)

From (b), 20×$\times$f(x) = f(x + 2) = (x + 2)×$\times$f(x + 1)= (x + 2)(x + 1)×$\times$f(x)
⇒$\Rightarrow$ 20×$\times$f(x) = (x + 2) (x + 1)×$\times$(f(x)
Cancelling f(x) on both sides and upon simplification we get
or x2$x^2$+ 3x - 18 = 0
or (x + 6)(x - 3) = 0
or x = -6 or 3. Hence the answer is option (e).

2. If 2*f(x) - f(1/x) = x2$x^2$ , then f(x) is?
a. f(x) = 3{2*x2$x^2$ + (1x)2${\left({\displaystyle \frac{1}{x}}\right)}^2$ }
b. 2 f(x) = {2*x2$x^2$+(1x)2${\left({\displaystyle \frac{1}{x}}\right)}^2$ }
c. f(x) = {2*x2$x^2$ +(1x)2${\left({\displaystyle \frac{1}{x}}\right)}^2$ }
d. 3 f(x) = {2*x2$x^2$ +(1x)2${\left({\displaystyle \frac{1}{x}}\right)}^2$ }
e. None of these
Sol: 2*f(x) - f(1/x) = x2$x^2$ ..............(1)
Take x = (1/x),
2*(1/x) - f(x) = (1x)2${\left({\displaystyle \frac{1}{x}}\right)}^2$ ...............(2)
Add: 2 (equation 1) + equation 2
3*f(x) = 2*x2$x^2$+ (1x)2${\left({\displaystyle \frac{1}{x}}\right)}^2$
f(x) = {2*x2$x^2$ + (1x)2${\left({\displaystyle \frac{1}{x}}\right)}^2$}/3

3. A function f(x) is said to reflect onitself if, y = f(x) and x = f(y). Which of the following function reflects on itself?
a. f(x) = x
b. f(x) = (2x + 1) / (x - 2)
c. f(x) = (3x + 2) / (2x - 3)
d. f(x) = (4x + 3) / (3x - 4)
e. All of these
(a) f(x) = x = f(y)
y = x = f(y) Hence (a) is true.
(b) f(x) = 2x+1x−2${\displaystyle \frac{2x+1}{x-2}}$ = y
xy - 2y = 2x + 1
xy - 2x = 1 + 2y
x(y - 2) = 1 + 2y
x = 1+2yy−2${\displaystyle \frac{1+2y}{y-2}}$ Hence (b) is also true.
Similarly if we check for options (c) & (d), they will also hold true, hence t he answer is option (e).

4. If f(x) = 1 - x            if x > 0
         = 1/(1 - x)      if x≤$\le$0.
Also fn(x)=f(fn−1(x))$f^n(x)=f\left(f^{n-1}\left(x\right)\right)$ for n > 1. What is the value of  f101(−101)$f^{101}(-101)$ ?
a. 102
b. 1/102
c. 101/102
d. 1/101
e. None of these
Let x≤$\le$0

f(x) = 11−x${\displaystyle \frac{1}{1-x}}$

f(f(x)) = 1 - 11−x${\displaystyle \frac{1}{1-x}}$   (As 11−x${\displaystyle \frac{1}{1-x}}$ x≥0$x\ge 0$

= −x1−x${\displaystyle \frac{-x}{1-x}}$   (Observe −x1−x${\displaystyle \frac{-x}{1-x}}$ is Positive for x is negitive)

f3(x)$f^3(x)$ = 1−−x1−x=1−x−(−x)1−x=11−x$1-{\displaystyle \frac{-x}{1-x}={\displaystyle \frac{1-x-(-x)}{1-x}={\displaystyle \frac{1}{1-x}}}}$

So f1(x)=f3(x)=f5(x)=.....f101(x)$f^1(x)=f^3(x)=f^5(x)=.....f^{101}(x)$

f101(x)=f1(x)=11−x=11−(−101)=1102$f^{101}(x)=f^1(x)={\displaystyle \frac{1}{1-x}={\displaystyle \frac{1}{1-(-101)}={\displaystyle \frac{1}{102}}}}$

5.  If f(x) = 2x + 3, & g(x) = (x - 3)/2, then what is the value of, fo(fo(go(go(fo(fo(go(go................(fo(fo(go(gof(x) )))).....)))))) ?
a. (x - 3) / 2
b. x
c. 2x - 3
d. 2x + 3
e. None of these
f(x) = 2x + 3
g(x) = (x - 3)/2
gof(x) = 2x+3−32=x${\displaystyle \frac{2x+3-3}{2}=x}$
gogof(x) = x−32${\displaystyle {\displaystyle \frac{x-3}{2}}}$
fogogof(x) = 2(x−32)+3$2\left({\displaystyle \frac{x-3}{2}}\right)+3$ = 2x−6+62${\displaystyle \frac{2x-6+6}{2}}$ = x
fofogogof(x) = 2x + 3
This means, when we apply two times g(x) and two time f(x) on f(x) we get f(x).  This pattern continues, Hence the answer is option (d).

6. For any function Fn(x)=Fn−1(F(x))$F^n(x)=F^{n-1}\left(F\left(x\right)\right)$ if for n > 1 also g(x) = 1/x,  h(x) = x√$\sqrt{x}$   and k(x) = x2$x^2$  then what is the value of g(h3(k2(x)))?$g\left(h^3\left(k^2\left(x\right)\right)\right)?$
a. x√$\sqrt{x}$ 
b. 1/x 
c. x2$x^2$  
d. 1x√${\displaystyle \frac{1}{\sqrt{x}}}$ 
e. None of these
Given Fn(x)=Fn−1(F(x));$F^n\left(x\right)=F^{n-1}\left(F\left(x\right)\right);$ g(x) = 1/x; h(x) = x√$\sqrt{x}$; k(x) = x√$\sqrt{x}$
Then, k(k(x)) = (x2)2=x4${\left(x^2\right)}^2=x^4$
And, h3(k2(x))=h(h(h(k2(x))))=h(h(x2))=h(x)=x√.$h^3\left(k^2\left(x\right)\right)=h\left(h\left(h\left(k^2\left(x\right)\right)\right)\right)=h\left(h\left(x^2\right)\right)=h\left(x\right)=\sqrt{x}.$
And, g(h3(k2(x)))=g(x√)=1/x√.$g\left(h^3\left(k^2\left(x\right)\right)\right)=g\left(\sqrt{x}\right)=1/\sqrt{x}.$
Hence answer is option (d).

7. If f1(x)=f(x)=12x$f^1(x)=f(x)={\displaystyle \frac{1}{2x}}$ and fn(x)=f(fn−1(x)${\mathrm{f}}^{\mathrm{n}}(x)=f(f^{n-1}(x)$, then find f5(x)+f10(x)${\mathrm{f}}^5(x)+f^{10}(x)$ where x = 1
Sol: f(x) = 12x${\displaystyle \frac{1}{2x}}$ and f(f(x)) = f(12x)$f\left({\displaystyle \frac{1}{2x}}\right)$ = 12(12x)${\displaystyle \frac{1}{2\left({\displaystyle \frac{1}{2x}}\right)}}$ = x
Similarly  f3(x)=f(f2(x))=f(x)=12x$f^3(x)=f(f^2(x))=f(x)={\displaystyle \frac{1}{2x}}$ and f4(x)=f(f3(x))=f(12x)=12×12x=x$f^4(x)=f(f^3(x))=f({\displaystyle \frac{1}{2x})={\displaystyle \frac{1}{2\times {\displaystyle \frac{1}{2x}}}=x}}$
Hence fn(x)=12x$f^n(x)={\displaystyle \frac{1}{2x}}$ for n = odd and fn(x)=x$f^n(x)=x$ for n = even
f5(x)+f10(x)=12x+x$f^5(x)+f^{10}(x)={\displaystyle \frac{1}{2x}+x}$
If x = 1,  ⇒12×1+1=32$\Rightarrow {\displaystyle \frac{1}{2\times 1}+1={\displaystyle \frac{3}{2}}}$

8. f1(x)$f^1\left(x\right)$ = 2x - 1 and fn(x)=f1(fn−1(x))$f^n\left(x\right)=f^1\left(f^{n-1}\left(x\right)\right)$  for n ≥$\ge$ 2. Find f5(2)$f^5\left(2\right)$
a. 30 b. 33 c. 31  d. None of these
f1(2)$f^1\left(2\right)$ = 2 x 2 - 1 = 3
f2(2)=f1(f1(2))=f1(3)$f^2\left(2\right)=f^1\left(f^1\left(2\right)\right)=f^1(3)$ = 2 x 3 - 1 = 5
f3(2)=f1(f2(2))=f1(5)$f^3\left(2\right)=f^1\left(f^2\left(2\right)\right)=f^1\left(5\right)$= 2  x 5 - 1 = 9
f4(2)=f1(f3(2))=f1(9)$f^4\left(2\right)=f^1\left(f^3\left(2\right)\right)=f^1\left(9\right)$= 2 x 9 - 1 = 17
f5(2)=f1(f4(2))=f1(17)$f^5\left(2\right)=f^1\left(f^4\left(2\right)\right)=f^1\left(17\right)$= 2 x 17 - 1 = 33
In fact, The general term of the above function is 2n$2^n$+1

9. Let g(x) be a function such that g(x + 1) + g(x - 1) = 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?
a. 5 b. 3    c. 2  d. 6
Sol: g(x + 1) + g(x - 1) = g(x)
g(x + 2) + g(x) = g(x + 1)
Adding these two equations we get
⇒$\Rightarrow$ g(x + 2) + g(x  - 1) = 0
⇒$\Rightarrow$ g(x + 3) + g(x) = 0    (By substituting x = x+1)
⇒$\Rightarrow$ g(x + 4) + g(x + 1) = 0  (By substituting x = x+1)
⇒$\Rightarrow$ g(x + 5) + g(x + 2) = 0  (By substituting x = x+1)
⇒$\Rightarrow$ g(x + 6) + g(x + 3) = 0  (By substituting x = x+1)
⇒$\Rightarrow$ g(x + 6) - g(x) = 0    ( g(x+3) + g(x) = 0  ⇒$\Rightarrow$ g(x+3) = - g(x))
⇒$\Rightarrow$ g(x + 6) = g(x)
So for p = 6 the function is periodic.

10. Find the area of region enclosed by | x | + | y | = 3 
a. 9 units   b. 18 units  c. 12 units           d. 6 units
We have to consider separately the form of the function w.r.t to where x, y lies.
x >0, y >0, x + y = 3
x > 0, y < 0, x - y = 3
x < 0, y >0, x - y =  - 3
x < 0, y < 0, x + y = - 3

After having drawn the above 4 lines on the graph, the graph looks like aboe
Now the area of the graph = 4 right angle triangles.
Area of the right angle triangle = 12×3×3=92${\displaystyle \frac{1}{2}\times 3\times 3=\frac{9}{2}}$
Area of the entire region = 92×4${\displaystyle \frac{9}{2}\times 4}$ = 18 units.

11. For any natural number n, F(n) is the least common multiple of all natural numbers upto.  For how many values of n less than 100, F(n+1) - F(n) = 0?
1.  16 2.  25 3.  49 4.  64
Sol: F(n) is the LCM of al natural numbers upto n.
For example,
F(3) = 6
F(5) = 60
F(6) = 60 etc.
F(n + 1) - F(n) = 0
⇒$\Rightarrow$ F(n + 1) = F(n)
It would hapen only when factors of (n + 1) are already present in the LCM of first n natural numbers.  In other words, Whenever (n + 1) is Prime number or is any Integral power of any prime number, F(n + 1) ≠$\ne$ F(n)
for example, F(31) ≠$\ne$ F(32) as 32 = 25$2^5$ which is, 2 raised to power 5. or F(4) ≠$\ne$ F(5) as 5 is prime so the power of 5 does not contain in F(4).
Between 1 and 100, there are 25 prime numbers. So if (n+1) is one of these number then F(n + 1) ≠$\ne$ F(n).
Also {4, 8, 16, 32, 64, 9, 27, 81, 25, 49} are prime numbers raised to some exponent.  
Hence, none of these 35 numbers from 1 to 99 cannot be the possible values of n. So for 99 - 35 = 64 numbers F(n) = F(n+1)

12. Given that f(1) = 1 and f(2) = 1.  If f(n) = f(n+1) - f(n - 1), then find the value of f(8)−f(7)+f(5)f(7)−f(6)−f(4)${\displaystyle \frac{f(8)-f(7)+f(5)}{f(7)-f(6)-f(4)}}$
f(n) = f(n+1) - f(n - 1)
⇒$\Rightarrow$ f(n - 1) = f(n + 1) = f(1)
 f(8) - f(7) = f(6) and f(7) - f(6) = f(5)
⇒$\Rightarrow$f(n + 1) = f(n) + f(n - 1)
⇒$\Rightarrow$f(3) = f(2) + f(1) = 1 + 1 = 2
⇒$\Rightarrow$f(4) = f(3) + f(2) = 2 + 1 = 3
⇒$\Rightarrow$f(5) = f(4) + f(3) = 3 + 2 = 5
⇒$\Rightarrow$ f(6) = f(5) + f(4) = 5 + 3 = 8
f(8)−f(7)+f(5)f(7)−f(6)−f(4)=f(6)+f(5)f(5)−f(4)=8+55−3=132${\displaystyle \frac{f(8)-f(7)+f(5)}{f(7)-f(6)-f(4)}=\frac{f(6)+f(5)}{f(5)-f(4)}=\frac{8+5}{5-3}=\frac{13}{2}}$

13. If f(x) + f(1 - x) + f(1 + x) + f(2 + x) = 2x for all real value of x.  If f(0) = 1, then find the value of f(4)
1.  4 2.  5 3.  6 4.  3 5.  1
For x = 0, f(0) + f(1) + f(2) = 0 ............. (i)
For x = 1, f(1) + f(0) + f(3) = 2 ............. (ii)
subtracting (i) from (ii), we get f(3) - f(1) = 2 .......... (a)
For x = 2, f(2) + f(-1) + f(3) + f(4) = 4      ........  (iii)
For x = -1, f(-1) + f(2) + f(0) + f(1) = -2   .........  (iv)
subtracting from (iii) from (iv), we get
⇒f(0)−f(4)−2=−6$\Rightarrow f(0)-f(4)-2=-6$ [ from (a) we know that f(3) - f(1) = 2]