1015 + Interview Questions in FUNCTIONS IN GENERAL AWARENESS Page 4 InterviewSolution

151.	Let `g(x)=1=x-[x] and f(x)={-1, x < 0 , 0, x=0 and 1, x > 0,` then for all `x, f(g(x))` is equal to(i) x(ii) 1(iii) f(x)(iv) g(x)A. xB. 1C. f(x)D. g(x)
Answer» Correct Answer - B We know that `0 le x-[x] lt 1"for all " x in R` `Rightarrow 1 le 1 +x-[x] lt 2"for all "x in R` `Rightarrow 1 le g(x)lt 2" for all "x in R` `Rightarrow f(g(x))=2" for all "x in R`

Discussion

152.	Let f : R → R : f(x) = x2, g : R → R : g(x) = tan xand h : R → R : h(x) = log x.Find a formula for h o (g o f).Show that [h o (g o f)] \(\sqrt{\frac{\pi}{4}}=0\)
Answer» To find: formula for h o (g o f) To prove: Show that [h o (g o f)] \(\sqrt{\frac{\pi}{4}}=0\) Formula used: f o f = f(f(x)) Given: (i) f : R → R : f(x) = x² (ii) g : R → R : g(x) = tan x (iii) h : R → R : h(x) = log x Solution: We have, h o (g o f) = h o g(f(x)) = h o g(x²) = h(g(x²)) = h (tan x²) = log (tan x²) h o (g o f) = log (tan x²) For, [h o (g o f)] \(\sqrt{\frac{\pi}{4}}=0\) \(=log[tan(\frac{\pi}{4}^2)]\) \(=log[tan\frac{\pi}{4}]\) = log 1 = 0 Hence Proved.

152.

Let f : R → R : f(x) = x2, g : R → R : g(x) = tan xand h : R → R : h(x) = log x.Find a formula for h o (g o f).Show that [h o (g o f)] \(\sqrt{\frac{\pi}{4}}=0\)

Answer»

To find: formula for h o (g o f)

To prove: Show that [h o (g o f)] \(\sqrt{\frac{\pi}{4}}=0\)

Formula used: f o f = f(f(x))

Given: (i) f : R → R : f(x) = x²

(ii) g : R → R : g(x) = tan x

(iii) h : R → R : h(x) = log x

Solution: We have,

h o (g o f) = h o g(f(x)) = h o g(x²)

= h(g(x²)) = h (tan x²)

= log (tan x²)

h o (g o f) = log (tan x²)

For, [h o (g o f)] \(\sqrt{\frac{\pi}{4}}=0\)

\(=log[tan(\frac{\pi}{4}^2)]\)

\(=log[tan\frac{\pi}{4}]\)

= log 1

= 0

Hence Proved.

Discussion

153.	Let f : Z → Z : f(x) = 2x. Find g : Z → Z : g o f = IZ.
Answer» To find: g : Z → Z : g o f = I_Z Formula used: (i) f o g = f(g(x)) (ii) g o f = g(f(x)) Given: (i) g : Z → Z : g o f = I_Z Solution: We have, f(x) = 2x Let f(x) = y ⇒ y = 2x \(\Rightarrow y=\frac{y}{2}\) \(\Rightarrow x=\frac{y}{2}\) let \(g(y)=\frac{y}{2}\) Where g: Z → Z For g o f, ⇒ g(f(x)) ⇒ g(2x) \(\Rightarrow \frac{2x}{2}\) ⇒ x = I_Z Clearly we can see that (g o f) = x = I_Z The required function is g(x) =\( \frac{2x}{2}\)

153.

Let f : Z → Z : f(x) = 2x. Find g : Z → Z : g o f = IZ.

Answer»

To find: g : Z → Z : g o f = I_Z

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) g : Z → Z : g o f = I_Z

Solution: We have,

f(x) = 2x

Let f(x) = y

⇒ y = 2x

\(\Rightarrow y=\frac{y}{2}\)

\(\Rightarrow x=\frac{y}{2}\)

let \(g(y)=\frac{y}{2}\)

Where g: Z → Z

For g o f,

⇒ g(f(x))

⇒ g(2x)

\(\Rightarrow \frac{2x}{2}\)

⇒ x = I_Z

Clearly we can see that (g o f) = x = I_Z

The required function is g(x) =\( \frac{2x}{2}\)

Discussion

154.	Mark (√) against the correct answer in the following: Let \(f(x)=\frac{x}{(x^2-1)}\) Then, dom (f) = ? A. R B. R – {1} C. R – { - 1} D. R – { - 1, 1}
Answer» Correct Answer is (D) R - { - 1,1} \(f(x)=\frac{x}{(x^2-1)}\) x² - 1≠0 x≠(1, - 1) ∴ Dom(f) = R - { - 1,1}

154.

Mark (√) against the correct answer in the following: Let \(f(x)=\frac{x}{(x^2-1)}\) Then, dom (f) = ? A. R B. R – {1} C. R – { - 1} D. R – { - 1, 1}

Answer»

Correct Answer is (D) R - { - 1,1}

\(f(x)=\frac{x}{(x^2-1)}\)

x² - 1≠0

x≠(1, - 1)

∴ Dom(f) = R - { - 1,1}

Discussion

155.	Let `f:R to R` and `g:R to R` be given by `f(x)=3x^(2)+2 and g(x)=3x-1` for all `x to R`. Then,A. `fog(x)=27x^(2)-18x+5`B. `fog(x)=27x^(2)+18x-5`C. `gof(x)=9x^(2)-5`D. `gof(x)=9x^(2)+15`
Answer» Correct Answer - A

Discussion

156.	`"If "f:R to R, g:R and h:R to R` be three functions are given by `f(x)=x^(2)-1,g(x)=sqrt(x^(2)+1)and h(x)={{:(,0,x le0),(,x,xgt 0):}` Then the composite functions (ho fog) (x)) is given byA. `{{:(,-x^(2),x lt 0),(,0,x=0),(,x^(2),x gt0):}`B. `{{:(,x^(2),xne0),(,0,x=0):}`C. `{{:(,x^(2),xgt 0),(,0,xle0):}`D. None of these
Answer» Correct Answer - B We have, `("ho fog")(x)=h("fog (x)")=h(f(g(x))=h(f(sqrt(x^(2)+1))` `("ho fog")(x)=h(x^(2)+1-1)=h(x^(2))` `Rightarrow ("ho fog")(x)={{:(,x^(2),x^(2)ne0),(,0,x=0):}`

Discussion

157.	The function `f:R to R` given by `f(x)=x^(2)+x` isA. one-one nad ontoB. one-one and intoC. many-one and ontoD. many one and into
Answer» Correct Answer - D

Discussion

158.	Mark (√) against the correct answer in the following:Let f(x) = cos-1 2x. Then, dom (f) = ?A. [ - 1, 1]B. \([\frac{-1}{2},\frac{1}{2}]\)C. \([\frac{-\pi}{2},\frac{\pi}{2}]\)D. \([\frac{-\pi}{4},\frac{\pi}{4}]\)
Answer» Correct Answer is (B) \([\frac{-1}{2},\frac{1}{2}]\) f(x) = cos^-1 2x. domain of cos^-1x = [ - 1,1] on multiplying by an integer the domain decreases by same number ⇒ domain of cos^-1 2x = [-1/2, 1/2]

158.

Mark (√) against the correct answer in the following:Let f(x) = cos-1 2x. Then, dom (f) = ?A. [ - 1, 1]B. \([\frac{-1}{2},\frac{1}{2}]\)C. \([\frac{-\pi}{2},\frac{\pi}{2}]\)D. \([\frac{-\pi}{4},\frac{\pi}{4}]\)

Answer»

Correct Answer is (B) \([\frac{-1}{2},\frac{1}{2}]\)

f(x) = cos^-1 2x.

domain of cos^-1x = [ - 1,1]

on multiplying by an integer the domain decreases by same number

⇒ domain of cos^-1 2x = [-1/2, 1/2]

Discussion

159.	if `f:[1,oo)->[2,oo)` is given by `f(x)=x+1/x` then `f^-1(x)` equals to : a) `(x+sqrt(x^2-4))/2` b) `x/(1+x^2)` c) `(x-sqrt(x^2-4))/2` d) `1+sqrt(x^2-4)`A. `(x+sqrt(x^(2)-4))/(2)`B. `(x)/(1+x^(2))`C. `(x-sqrt(x^(2)-4))/(2)`D. `1+sqrt(x^(2)-4)`
Answer» Correct Answer - A Let `y=x+(1)/(x) rArr y=(x^(2)+1)/(x)` `rArr xy=x^(2)+1` `rArr x^(2)-xy+1=0 rArr x=(y pm sqrt(y^(2)-4))/(2)` `rArr f^(-1)(y)=(y pm sqrt(y^(2)-4))/(2)` `rArr f^(-1)(x)=(x pm sqrt(x^(2)-4))/(2)` Since, the range of the inverse function is `[1, oo),` then we take `f^(-1)(x)=(x+sqrt(x^(2)-4))/(2)` If we consider `f^(-1)(x)=(x-sqrt(x^(2)-4))/(2), " then " f^(-1)(x) gt 1` This possible only if `(x-2)^(2) gt x^(2)-4` `rArr x^(2)+4-4x gt x^(2)-4` `rArr 8 gt 4x` `rArr x lt 2, " where " x gt 2` Therefore, (a) is the answer.

Discussion

160.	If `f(x)=(2^x+2^(-x))/2`, then `f(x+y)f(x-y)`is equals to`1/2{f(2x)+f(2y)}`(b) `1/2{f(2x)-f(2y)}`(c) `1/4{f(2x)+f(2y)}``1/4{f(2x)-f(2y)}`A. `(1)/(2){f(2x)+f(2y)}`B. `(1)/(2){f(2x)-f(2y)}`C. `(1)/(4){f(2x)+f(2y)}`D. `(1)/(4){f(2x)-f(2y)}`
Answer» Correct Answer - A

Discussion

161.	Mark (√) against the correct answer in the following: Let f(x) = cos-1(3x – 1). Then, dom (f) = ?A. \((0, \frac{2}{3})\)B. \([0,\frac{2}{3}]\)C. \([\frac{-2}{3},\frac{2}{3}]\)D. None of these
Answer» Correct Answer is (B) \([0,\frac{2}{3}]\) f(x) = cos^-1(3x - 1). domain of cos^-1x = [ - 1,1] on multiplying by an integer the domain decreases by same number ⇒ domain of cos^-13x = [ - 1/3,1/3] ⇒ domain of cos^-1 (3x - 1) = [1/3 - 1/3,1/3 + 1/3] = [0,2/3]

161.

Mark (√) against the correct answer in the following: Let f(x) = cos-1(3x – 1). Then, dom (f) = ?A. \((0, \frac{2}{3})\)B. \([0,\frac{2}{3}]\)C. \([\frac{-2}{3},\frac{2}{3}]\)D. None of these

Answer»

Correct Answer is (B) \([0,\frac{2}{3}]\)

f(x) = cos^-1(3x - 1).

domain of cos^-1x = [ - 1,1]

on multiplying by an integer the domain decreases by same number

⇒ domain of cos^-13x = [ - 1/3,1/3]

⇒ domain of cos^-1 (3x - 1) = [1/3 - 1/3,1/3 + 1/3] = [0,2/3]

Discussion

162.	Mark (√) against the correct answer in the following: Let f(x) = √log (2x – x2). Then, dom (f) = ? A. (0, 2) B. [1, 2] C. ( - ∞, 1] D. None of these
Answer» Correct Answer is (C) ( - ∞, 1] f(x) = √log (2x – x²). ⇒ 2x - x² > 1 ⇒ x² - 2x + 1 < 0 ⇒ (x - 1)² < 0 ⇒ x - 1 < 0 ⇒ x < 1 log(2x - x²) > 0 ⇒ 2x - x²> e⁰ = 1 ⇒ x < 1 Dom (f) = ( - ∞, 1)

162.

Mark (√) against the correct answer in the following: Let f(x) = √log (2x – x2). Then, dom (f) = ? A. (0, 2) B. [1, 2] C. ( - ∞, 1] D. None of these

Answer»

Correct Answer is (C) ( - ∞, 1]

f(x) = √log (2x – x²).

⇒ 2x - x² > 1

⇒ x² - 2x + 1 < 0

⇒ (x - 1)² < 0

⇒ x - 1 < 0

⇒ x < 1

log(2x - x²) > 0

⇒ 2x - x²> e⁰ = 1

⇒ x < 1

Dom (f) = ( - ∞, 1)

Discussion

163.	Suppose `f(x)=(x+1)^2forxgeq-1.`If `g(x)`is the function whose graph is the reflection of the graph of `f(x)`with respect to the line `y=x ,`then `g(x)`equal.`a-sqrt(x)-1,xgeq0`(b) `1/((x+1)^2),x >-1``sqrt(x+1,)xgeq-1`(d) `sqrt(x)-1,xgeq0`A. `-sqrt(x)-1,x ge 0`B. `(1)/((x+1)^(2)), x gt -1`C. `sqrt(x+1), x ge -1`D. `sqrt(x) -1, x ge 0`
Answer» Correct Answer - D It is only to find the inverse. Let `y=f(x)=(x+1)^(2), " for " x ge -1` `pm sqrt(y)=x+1, x ge -1` `rArr sqrt(y)=x+1 rArr y ge 0, x+1 ge 0` `rArr x=sqrt(y) -1` `rArr f^(-1)(y)=sqrt(y)-1` `rArr f^(-1)(x)=sqrt(x)-1 rArr x ge 0`

Discussion

164.	Find the range of the function `f(x)=\|x\|`.
Answer» Correct Answer - `[0,oo]` `f(x)=\|x \|ge0."So,range "(f)={0,oo)`.

Discussion

165.	Let `f(x) =x^(2).` Then dom (f ) and range (f ) are respectivelyA. R and RB. `R^(+) " and " R^(+)`C. `R " and " R^(+)`D. `R " and " R-{0}`
Answer» Correct Answer - C f(x) `=x^(2) ` is defined for each ` x in R` .So dom (f) =R ` y =x^(2) rArr x = +- sqrt(y)` when `y lt 0` there is no real value of x .So `y ge 0` ` :. ` range `(f) = R^(+)`

Discussion

166.	Write the domain and the range of the function `f(x)=sqrt(x+[x])`.
Answer» Correct Answer - dom (f)=R, range `(f)=[0,1)` `x-[x]=0" for "x"inZandx-[x]gt0" for "x""inR-Z`. `:.x-{x}ge0" for "x""inR."So, dom "(f)=R` Also, `0lex-[x]lt1implies0lesqrt(x-[x])lt1."So, range "(f)=[0,1)`.

Discussion

167.	If `f(x)=(x-5)/(5-x)` then find dom (f) and range (f).
Answer» Correct Answer - dom `(f)=R-{5}," range "(f)={-1}` f(x) is not defined when x=5. So, dom `(f)=R-{5}`. `f(x)=((x-5))/((5-x))=-1."So,range "(f)={-1}`

Discussion

168.	Find the domain and the range of each of the following real function: f(x) = 1 – \|x – 2\|
Answer» Given: f(x) = 1 – \|x – 2\| Need to find: Where the functions are defined. Since \|x – 2\| gives real no. for all values of x, the domain set can possess any real numbers. So, the domain of the function, D_f(x) = (-∞, ∞). Now the given function is f(x) = 1 – \|x – 2\|, where \| x – 2 \| is always positive. So, the maximum value of the function is 1. Therefore, the range of the function, R_f(x) = (-∞, 1)

168.

Find the domain and the range of each of the following real function: f(x) = 1 – |x – 2|

Answer»

Given: f(x) = 1 – |x – 2|

Need to find: Where the functions are defined.

Since |x – 2| gives real no. for all values of x, the domain set can possess any real numbers.

So, the domain of the function, D_f(x) = (-∞, ∞).

Now the given function is f(x) = 1 – |x – 2|, where | x – 2 | is always positive. So, the maximum value of the function is 1.

Therefore, the range of the function, R_f(x) = (-∞, 1)

Discussion

169.	Find the domain and the range of each of the following real function: f(x) = \(\frac{\|x-4\|}{x-4}\)
Answer» Given: f(x) = \(\frac{\|x-4\|}{x-4}\) Need to find: Where the functions are defined. To find the domain of the function f(x) we need to equate the denominator of the function to 0. Therefore, x – 4 = 0 ⇒ x = 4 It means that the denominator is zero when x = 4 So, the domain of the function is the set of all the real numbers except 4. The domain of the function, D_f(x) = (- ∞, 4) ∪ (4, ∞). The numerator is an absolute function of the denominator. So, for any value of x from the domain set, we always get either +1 or -1 as the output. So, the range of the function is a set containing -1 and +1 Therefore, the range of the function, R_f(x) = { -1 , 1 }

169.

Find the domain and the range of each of the following real function: f(x) = \(\frac{|x-4|}{x-4}\)

Answer»

Given: f(x) = \(\frac{|x-4|}{x-4}\)

Need to find: Where the functions are defined.

To find the domain of the function f(x) we need to equate the denominator of the function to 0.

Therefore,

x – 4 = 0

⇒ x = 4

It means that the denominator is zero when x = 4

So, the domain of the function is the set of all the real numbers except 4.

The domain of the function, D_f(x) = (- ∞, 4) ∪ (4, ∞).

The numerator is an absolute function of the denominator. So, for any value of x from the domain set, we always get either +1 or -1 as the output. So, the range of the function is a set containing -1 and +1

Therefore, the range of the function, R_f(x) = { -1 , 1 }

Discussion

170.	Let `f={(1,6),(2,5),(4,3),(5,2),(8-1),(10,-3)}` and `g={(2,0),(3,2),(5,6),(7,10),(8,12),(10,16)}`. Find `(i) dom (f+g) (ii) dom ((f)/(g))`.
Answer» Correct Answer - (i) `{2,5,8,10}` (ii) `{5, 8,10}` dom `(f+g)="dom "(f)nn"dom "(g)={2,5,8,10}` dom `((f)/(g))="dom "(f)nn"dom "(g)-[x:g(x)=0}=(2,5,8,10}-{2}={5,8,10}`.

Discussion

171.	If `f(x)=1-1/x`then write the value of `f(f(1/x))dot`
Answer» Correct Answer - `(x)/(x-1)` `f((1)/(g))=((1)/(x)-1)/((1)/(x))=(1-x)impliesf{f((1)/())}=((1-x)-1)/(1-x)=(-x)/(1-x)=(x)/(x-1)`.

Discussion

172.	If `f`is an even function defined on the interval `(-5,5),`then four real values of `x`satisfying the equation `f(x)=f((x+1)/(x+2))`are ______, __________, _______ and______.
Answer» Correct Answer - `((pm3 pm sqrt(5))/(2))` Since, f is an even function, then `f(-x)=f(x), AA x in (-5,5)` Given, `f(x)=f((x+1)/(x+2)) " ...(i)" ` `rArr f(-x)=f((-x+1)/(-x+2))` `rArr f(x)=f((-x+1)/(-x+2))" "[ because f(-x)=f(x)]` Talking `f^(-1)` on both sides, we get `x=(-x+1)/(-x+2)` `rArr -x^(2)+2x= -x+1` `rArr x^(2)-3x+1=0` `rArr x=(3pm sqrt(9-4))/(2)=(3pm sqrt(5))/(2)` Again, `f(x)=f((x+1)/(x+2))` `rArr f(-x)=f((x+1)/(x+2)) " "[because f(-x)=f(x)]` Taking `f^(-1)` on both sides, we get `-x=(x+1)/(x+2)` `rArr x^(2)+3x+1=0` `rArr x=(-3 pm sqrt(9-4))/(2)=(-3pm sqrt(5))/(2)` Therefore, four values of x are `(pm 3pm sqrt(5))/(2)`

Discussion

173.	Write the range of the real function f(x) = \|x\|.
Answer» f(x) = \|x\| f(-x) = \|-x\| Therefore, f(x) will always be 0 or positive. Thus, Range of f(x) ϵ [0,∞).

173.

Write the range of the real function f(x) = |x|.

Answer»

f(x) = |x|

f(-x) = |-x|

Therefore,

f(x) will always be 0 or positive.

Thus,

Range of f(x) ϵ [0,∞).

Discussion

174.	If `f(x)=(kx)/(x+1)`, where `xne-1andf{f(x)}=x" for "xne-1` then find the value of k.
Answer» Correct Answer - `k=-1` `f(f(x))=f((kx)/(x+1))=((kxx(kx))/(x+1))/((kx)/(x+1)+1)=(k^(2)x)/(kx+x+1)=(k^(2)x)/(kx+x+1)=x`. `:.k^(2)=kx+x+1+impliesk^(2)-kx-(x+1)=0impliesk=(xpmsqrt(x^(2)+4(x+1)))/(2)=(xpm(x+2))/(2)`

Discussion

175.	Find the domain and range of the function given by `f(x)=1/(sqrt(x-[x]))`
Answer» We have, `f(x)=(1)/(sqrt(x-[x]))` We know that `0lex-[x]lt1"for all "x""inR` and `x-[x]=0"for all "x""inZ`. `:.0ltx-[x]lt1" for all "x""inR-Z` `impliesf(x)=(1)/(sqrt(x-[x]))` exists for all `x""inR-Z` `implies"dom "(f)=R-Z`. Also, `0ltx-[x]lt1 "for all "x""inR-Z` `implies0ltsqrt(x-[x])lt1 "for all "x""inR-Z` `implies1lt(1)/(sqrt(x-[x]))ltoo "for all "x""inR-Z` `implies1ltf(x)ltoo" for all "x""inR-Z` `implies"range "(f)=(1,oo)`. Hence, dom `(f)=R-Z" and range "(f)=(1,oo)`.

Discussion

176.	Let`f(x)=x^2`and`g(x)" "=" "2x" "+" "1`betwo real functions. Find `(f" "+" "g)``(x)`, `(f" "" "g)" "(x)`, `(fg)" "(x)`, `(f/g)(x)`.
Answer» Here, dom (f)=R and dom (g) =R. `:."dom "(f)nn"dom "(g)=(RnnR)=R`. (i) `(f+g):RtoR` is given by `(f+g)(x)=f(x)+g(x)=x^(2)+(2x+1)=(x+1)^(2)`. (ii) `(f-g):RtoR` is given by `(f-g)(x)=f(x)-g(x)=x^(2)-(2x+1)=(x^(2)-2x-1)`. (iii) `(fg):RtoR` is given by `(fg)(x)=f(x).g(x)=x^(2).(2x+1)=(2x^(3)+x^(2))` (iv) `{x:g(x)=0}={x:2x+1=0}={(-1)/(2)}`. `:."dom "((f)/(g))=RnnR-{(-1)/(2)}=R-{(-1)/(2)}`. The function `(f)/(g):R-{(-1)/(2)}toR` is given by `((f)/(g))(x)=(f(x))/(g(x))=(x^(2))/(2x+1),xne(-1)/(2)`.

Discussion

177.	Let `f:RtoR`, defined by `f(x)={{:(1",",ifx""inQ),(-1",",ifx""inQ):}` Find (i) `f(1)/(2)` (ii) `f(0.34)` (iii) `f(sqrt2)` (iv) `f(pi)` (v) range(f) (vi) `f^(-1)(1)` (vii) `f^(-1){1}`
Answer» Since each one of `(1)/(2)` and 0.34 is rational, we have (i) `f((1)/(2))=1and(ii)f(0.34)=1`. Since each one of `sqrt2andphi` is irrational, we have (iii) `f(sqrt2)=-1and(iv)f(pi)=-1`. (v) range `(f)={f(x):x""inR}` `={f(x):x""inQ}uu{f(x):x""inR-Q}={1,-1}`. (vi) `f^(1){1}={x:f(x)=1}=Q`. (vii) `f^(1){-1}={x:f(x)=-1}=(R-Q)`.

Discussion

178.	If `f(x)=(2x)/(1+x^2) ,`show that `f(tantheta)=sin2theta)`
Answer» `f(tantheta)=(2tantheta)/((1+tan^(2)theta))=(2tantheta)/(sec^(2)theta)=(2sintheta)/(costheta)xxcos^(2)theta=2sintheta.costheta=sin20`

Discussion

179.	If `f(x)=1/(1-x)`, show that`f[f{f(x)}]=xdot`
Answer» `f(x)=(1)/((1-x))impliesf{f(x)}=(1)/({1-(1)/((1-x))})=(1-x)/(-x) " "["replace x by "(1)/((1-x))]` `impliesf[f{f(x)}]=({(1)/((1-x))-1})/((1)/((1-x)))=(x)/(1-x)xx((1-x))/(1)=x`.

Discussion

180.	If the function `f:[1,oo)->[1,oo)` is defined by `f(x)=2^(x(x-1)),` then `f^-1(x)` is (A) `(1/2)^(x(x-1))` (B) `1/2 sqrt(1+4log_2x)` (C) `1/2(1-sqrt(1+4log_2x))` (D) not definedA. `((1)/(2))^(x(x-1))`B. `(1)/(2)(1+sqrt(1+4log_(2)x))`C. `(1)/(2)(1-sqrt(1+4log_(2)x))`D. not defined
Answer» Correct Answer - B Let `y=2^(x(x-1)), " where " y ge 1 " as " x ge 1` Taking `log_(2)` on both sides, we get `log_(2)y=log_(2)2^(x(x-1))` `rArr log_(2)y=x(x-1)` `rArr x^(2) -x-log_(2)y =0` `rArr x=(1pm sqrt(1+4log_(2)y))/(2)` For ` y ge 1, log_(2)y ge 0 rArr 4 log_(2) y ge 0 rArr 1+4 log_(2)y ge 1` `rArr sqrt(1+4log_(2)y) ge 1` `rArr -sqrt(1+4log_(2)y) le -1` `rArr 1-sqrt(1+4log_(2)y) le 0` But `x ge 1` so, `x=1- sqrt(1+4log_(2)y)` is not possible. Therefore, we take `x=(1)/(2)(1+sqrt(1+4log_(2)y))` `rArr f^(-1)(y) =(1)/(2)(sqrt(1+4log_(2)y))` `rArr f^(-1)(x)=(1)/(2)(sqrt(1+4log_(2)x))`

Discussion

181.	If `f(x)=1/(2x+1), x!=-1/2, `then show that `f(f(x))=(2x+1)/(2x+3)`, provided that `x!=-3/2dot`
Answer» `f(x)=(1)/((2x+1))impliesf{f(x)}=(1)/({2xx(1)/((2x+1))+1})=(2x+1)/(2x+_3)`.

Discussion

182.	If `f(x) = x^2 - 3x + 4`, then find the values of x satisfying the equation `f(x) = f(2x + 1)`.
Answer» Correct Answer - `x=-1orx=(2)/(3)` `f(x)=f(2x+1)impliesx^(2)-3x+4=(2x+1)^(2)-3(2x+1)+4` `impliesx^(2)-3x+4=4x^(2)-2x+2implies3x^(2)+x-2=0implies(x+1)(3x-2)=0`.

Discussion

183.	Which of the following functions is periodic?A. f(x) = x-[x], where [x] denotes the greatest integer less than or equal to the real number xB. `f(x)=sin(1/x) " for "x ne 0, f(0)=0`C. f(x) = x cos xD. None of the above
Answer» Correct Answer - A Clearly, `f(x)=x-[x]=[x]` which has period 1. And `(1)/(x),x cos x` are non-periodic functions.

Discussion

184.	The function `f:[0,oo] to R` given by f(x)=(x)/(x+1)` isA. one-one and intoB. one-one but not ontoC. onto but not one-oneD. neither one-one nor onto
Answer» Correct Answer - B We have, `f(x)=(x)/(x+1)=1-(1)/(1+x)` Let `x, y in R` be such that `f(x)=f(y)`. Then, f(x)=f(y) `Rightarrow 1-(1)/(1+x)=1-(1)/(1+y)Rightarrow 1+x=1+y Rightarrow x=y` `therefore ` f is one-one Clearly, f is not onto as f takes only values less than 1 i.e. Range `f=(-oo, 1) ne` co-domain of f.

Discussion

185.	Let `f:toRtoR:f(x)=x^(2)andg:CtoC:g(x)=x^(2)`, where C is the set of all complex numbers. Show that `fne""g`.
Answer» Dom (f) =R and dom `(g)=Cimplies"dom "(f)ne"dom "(g)`

Discussion

186.	Let `f: RvecR`be such that `f(x)=2^x`. Determine:Range of `f`(ii) `{x :f(x)=1}`(iii) Whether `f(x+y)=f(x)f(y)`holds.
Answer» (i) `f(x)=2^(x)gt0` for every `x""inR`. If `x""inR^(+)`, there exists `log_(2)x` such that `f(log_(2)x)=2^(log_(2)x)=x`. `:."range "(f)=R^(+)`. (ii) `f(x)=1implies2^(x)=1=2^(0)impliesx=0."So,"{x:f(x)=1}={0}`. (iii) `f(x+y)=2^(x+y)=2^(x)xx2^(y)=f(x).f(y)`.

Discussion

187.	If `f(x)=log_(e)(1-x)andg(x)=[x]` then find: (i) (f+g)(x) (ii) (fg)(x) (iii) `((f)/(g))(x)` (iv) `((g)/(f))(x)`. Also find `(f+g)(-1),(fg)(0),((f)/(g))(-1),((g)/(f))((1)/(2))`.
Answer» Clearly, `log_(e)(1-x)` is defined only when `1-xgt0,i.e.,xlt1`. `:."dom "(f)=(-oo,1)`. Also, dom (g)=R. `:."dom "(f)nn"dom "(g)=(-oo,1)nnR=(-oo,1)` (i) `(f+g)"(-oo,1)toR` is given by `(f+g)(x)=f(x)+g(x)=log_(e)(1-x)+[x]`. (ii) `(fg):(-oo,1)toR` is given by `(fg)(x)=f(x)xxg(x)={log_(e)(1-x)}xx[x]`. (iii) `{x:g(x)=0}={x:[x]=0}={0,1)`. `:."dom "((f)/(g))="dom "(f)nn"dom "(g)-{x:g(x)=0}` `=(-oo,1)nnR-[0,1)=(-oo,0)`. `:.(f)/(g):(-oo,0)toR` is given by `((f)/(g))(x)=(f(x))/(g(x))=(log_(e)(1-x))/([x])`. (iv) `{x:f(x)=0}={x:log_(e)(1-x)=0}={0}`. `:."dom "((g)/(f))="dom "(g)nn"dom "(f)-{x:f(x)=0}`. `=Rnn(-oo,1)-{0}=(-oo,0}uu(0,1)`. `:.(g)/(f):(-oo,0)uu(0,1)toR` `((g)/(f))(x)=(g(x))/(f(x))=([x])/(log_(e)(1-x))`. Now, we have: `(f+g)(-1)=f(-1)+g(-1)=[-1]+log_(e)(1+1)=(log_(e)2)-1`. `(fg)(0)=f(0)xxg(0)=log_(e)(1-0)xx[0]=(log_(e)1xx0)=(0xx0)=0`. `((f)/(g))(-1)=(f(-1))/(g(-1))=([-1])/(log_(e)(1+1))=(-1)/(log_(e)2)`. `((g)/(f))((1)/(2))=g((1)/(2))/(f((1)/(2)))=([(1)/(2)])/(log_(e)(1-(1)/(2)))=([05.])/(log_(e)((1)/(2)))=0`.

Discussion

188.	If `f:R->S` defined by `f(x)=sinx-sqrt(3)cosx+1` is onto , then the interval of S is :A. [0,1]B. [-1,1]C. [0,3]D. [-1,3]
Answer» Correct Answer - D We have `sqrt(1+(sqrt(3))^(2)) le sin x sqrt3 cos x le sqrt(1+(sqrt3)^(2))"for all "x in R` `Rightarrow -2 le sin x-sqrt3 cos x le 2" ""for all "x in R` `Rightarrow -1 le sin x-sqrt3 cos x+1 le 3" ""for all "x in R` `Rightarrow -1 le f(x)" "le sqrt3" ""for all "x in R` `Rightarrow "Range(f)"=[-1,3]` If `f:R to S` is onto, then S=Range(f)=[-1,3]

Discussion

189.	Let`f"":""NvecY`be a function defined as `f""(x)""=""4x""+""3`, where `Y""=""{y in N"":""y""=""4x""+""3`for some `x in N}`. Show that f is invertibleand its inverse is(1) `g(y)=(3y+4)/3`(2) `g(y)=4+(y+3)/4`(3) `g(y)=(y+3)/4`(4) `g(y)=(y-3)/4`A. `g(y)=(y+3)/(4)`B. `g(y)=(y-3)/(4)`C. `g(y)=(3y+4)/(3)`D. `g(y)=4+(y+3)/(4)`
Answer» Correct Answer - B For any `x, y in N` `f(x)=f(y) Rightarrow 4x+3=4y+3 Rightarrow x=y` `therefore` f is one-one Clearly, `Y=(y in N: y=4x+3"for some "x in N)="Range (f)"` `therefore` f:N-Y is onto. Thus, `f:N to Y` is a bijection and hence invertible. Let g be the inverse of f. Then. `fog (y)=y"for all "y inY` `Rightarrow f(g(y))=yRightarrow 4g(y)+3 =yRightarrow g(y)=(y-3)/(4)`

Discussion

190.	If `f(x)=3x^(3)-5x^(2)+10`, find `f(x-1)`.
Answer» We have, `f(x)=3x^(3)-5x^(2)+10" "....(i)` Replacing x by (x-1) in (i), we get `f(x-1)=3(x-1)^(3)-5(x-1)^(2)+10` `=3{x^(3)-1-3x(x-1)}-5(x^(2)-2x+1)+10` `=3(x^(3)-3x^(2)+3x(x-1)-5(x^(2)-2x+1)+10` `=3x^(3)-14x^(2)+19x+2`. `:.f(x-1)=3x^(3)-14x^(2)+19x+2`.

Discussion

191.	A function f from the set of natural number to integers defined by `f(n)={{:(,(n-1)/(2),"when n is odd"),(,-(n)/(2),"when n is even"):}`A. neither one-one nor ontoB. one-one but not ontoC. one but not one-oneD. one-one and onto both
Answer» Correct Answer - D Clearly, range (f)=Z (Set of integers) So,f is onto We observe that distinct odd natural numbers are mapped to distinct non-negative integers and distinct even natural numbers are mapped to distinct negative integers. So, f is one-one Hence, f is both one-one and onto.

Discussion

192.	The range of the function `f(x)=7-x p_(x-3)` , isA. {1,2,3,4,5}B. {1,2,3,4,5,6}C. {1,2,3,4}D. {1,2,3}
Answer» Correct Answer - D Clearly `f(x)=^(7-x)P_(x-3),is` defined for positive integer values of x satisfying `7-x gt 0, x-3 ge 0 and x-3 le 7-x` `i.e. x lt 7, x ge 3 and x-3 le 7 -x` i.e. x=3,4,5 `therefore "Domain(f)"=(3,4,5)` Hence, range(f)=`(f(3),f(4),F(5)={.^(4)P_(0),^(3)P_(1),^(2)P_(2)}` ={1,2,3}

Discussion

193.	Let `f:RtoR:f(x)=x^(2)+1`. Find (i) `f^(-1){-4}` (ii) `f^(-1){10}` (iii) `f^(-1){5,17}`.
Answer» It is given that `f(x)=x^(2)+1`. (i) Let `f^(1)(-4)=x`. Then, `f(x)=-4impliesx^(2)+1=-4impliesx^(2)=-5`. But, there is no real value of x whose square is -5. `:.f^(1){-4}=phi`. (ii) Let `:.f^(1)(10)=x`. `f(x)=10impliesx^(2)+1=-4impliesx^(2)=9impliesx=pm3`. `:.f^(1){10}={-3,3}`. (iii) Let `f^(1)(5)=x`. Then, `f(x)=5impliesx^(2)+1=5impliesx^(2)=4impliesx=pm2`. `:.f^(1){5}={-2,2}`. Let `f^(1)(17)=x`. Then, `f(x)=17impliesx^(2)+1=17impliesx^(2)=16impliesx=pm4`. `:.f^(1){17}={-4,4}`. Hence, `f^(1){5,17}=-{-2,2,-4,4}`.

Discussion

194.	Let f and g be real functions, defined by `f(x)=sqrt(x-1)andg(x)=sqrt(x+1)`. Find (i) `(f+g)(x)` (ii) `(f-g)(x)` (iii) `(fg)(x) `(iv) `((f)/(g))(x)`.
Answer» Clearly, `f(x)sqrt(x-1)` is defined for all real values of x for which `x-1ge0,i.e.,xge1."So, dom "(f)=[1,oo)`. Also, `g(x)=sqrt(x+1)` is defined for all real values of x for which `x+1ge0,i.e.,xge-1."So, dom "(g)=[-1,oo)`. `:."dom "(f)nn"dom "(g)={1,oo)nn[-1,oo)=[1,oo)`. (i) `(f+g):[1,oo)toR` is given by `(f+g)(x)=f(x)+g(x)=(sqrt(x-1)+sqrt(x+1))`. (ii) `(f-g):[1,oo)toR` is given by `(f-g)(x)=f(x)-g(x)=(sqrt(x-1)-sqrt(x+1))` (iii) `(fg):[1,oo)toR` is given by `(fg)(x)=f(x).g(x)=sqrt(x-1)xxsqrt(x+1)=sqrt(x^(2)-1)`. (iv) `{x:g(x)=0}={x:sqrt(x+1)=0}={x:x+1=0}={-1}`. `:."dom "(f)nn"dom "(g)-{x:g(x)=0}` `=[1,oo)nn[-1,oo)-{-1}=[1,oo)`. `:.(f)/(g)to[1,oo)toR` is given by `((f)/(g))(x)=(f(x))/(g(x))=sqrt(x-1)/sqrt(x+1)`.

Discussion

195.	Let `A={1,2}, B={3,6}`and `f: A->B`given by `f(x)=x^2+2 n d g: A->B`given by `g(x)=3xdot`Then we observe that `f a n d g`have the same domain and co-domain. Also we hve, `f(1)=3=g(1)a n d f(2)=6=g(2)dot`Hence `f=gdot`
Answer» Clearly, we have dom (f) =dom `(g)={1,2}`. Co-domain of f= co-domain of `g={3,6}`. Also, `f(1)=(3xx1)=3,f(2)=(3xx2)=6,` `g(1)=(1^(2)+2)=3,g(2)=(2^(2)+2)=6` `:.f(1)=g(1)andf(2)=g(2)`. Thus, `f(x)=g(x)" for all "x""inA`. Hence, f=g.

Discussion

196.	Show that `f(x)=(1)/sqrt(x-\|x\|)` is not defined for any `x""inR`. How will you define dom (f) and range (f)?
Answer» We have, `f(x)=(1)/sqrt(x-\|x\|)` Now, `implies\|x\|={{:(x",when "xge0),(-x",when "xlt0):}` `impliesx-\|x\|={{:(x-x",when "xge0),(x+x",when "xlt0):}` `impliesx-\|x\|={{:(0",when "xge0),(2x",when "xlt0):}` `implies(1)/(sqrt(x-\|x\|))` is not defined for any `x""inR`. `:."dom "(f)=phi" and range "(f)=phi`.

Discussion

197.	Find the domain and the range of the real function, `f(x)=(1)/(sqrt(x+\|x\|))`.
Answer» We have, `f(x)=(1)/(sqrt(x+\|x\|))` Now, `\|x\|={{:(x",when "xge0),(-x",when "xlt0):}` `impliesx+\|x\|={{:(x+x",when "xge0),(x-x",when "xlt0):}` `impliesx+\|x\|={{:(2x",when "xge0),(0",when "xlt0):}` `impliesx+\|x\|gt0,"when "xgt0` `impliesf(x)=(1)/(sqrt(x+\|x\|))` assumes real values only when `x+\|x\|gt0` and this happens only when `xgt0`. `:."dom "(f)=(0,oo)`.

Discussion

198.	If f : R → R and g : R → R are defined by f (x) = 2x + 3 and g(x) = x2 + 7, then the values of x such that g{f (x)} = 8 are(a) 1, 2 (b) – 1, 2 (c) – 1, – 2 (d) 1, – 2
Answer» Answer : (c) – 1, – 2 g{f (x)}= 8 ⇒ g{2x + 3} = 8 ⇒ (2x + 3)² + 7 = 8 ⇒ 4x² + 12x + 9 – 1 = 0 ⇒ 4x² + 12x + 8 = 0 ⇒ x² + 3x + 2 = 0 ⇒ (x + 1) (x + 2) = 0 ⇒ x = – 1, – 2.

198.

If f : R → R and g : R → R are defined by f (x) = 2x + 3 and g(x) = x2 + 7, then the values of x such that g{f (x)} = 8 are(a) 1, 2 (b) – 1, 2 (c) – 1, – 2 (d) 1, – 2

Answer»

Answer : (c) – 1, – 2

g{f (x)}= 8 ⇒ g{2x + 3} = 8

⇒ (2x + 3)² + 7 = 8

⇒ 4x² + 12x + 9 – 1 = 0

⇒ 4x² + 12x + 8 = 0

⇒ x² + 3x + 2 = 0

⇒ (x + 1) (x + 2) = 0

⇒ x = – 1, – 2.

Discussion

199.	If a function `g={(1,1),(2,3),(3,5),(4,7)}`is described by `g(x)=alphax+beta,`find the values of `alphaa n dbetadot`A. 2x-1B. 2x+1C. x+2D. x-2
Answer» Correct Answer - A Let f(x)=ax+b We have `f(1)=1,f(2)=3,f(3)=5 and f(4)=7` We have `Rightarrow a+b=1, 2a+b=3, 3a+b=5 and 4a+b=7` `Rightarrow a=2 and b=-1` `"Hence", f(x)=2x-1`

Discussion

200.	If `f(x)=(x-1)/(x+1),x!=-1,`. then show that `f(f(x))=-1/x`, prove that `x!=0`.
Answer» We have, `f(x)=(x-1)/(x+1)`, where `xne-1`. `:.f{f(x)}=f((x-1)/(x+1))={{(x-1)/(x+1)-1}/{(x-1)/(x+1)+1)}` `={{(x-1)-(x+1)}}/((x+1))xx(x+1)/{{(x-1)+(x+1)}}=(-2)/(2x)=(-1)/(x)`. Hence, `f{f(x)}=(-1)/(x)` where `xne0`.

Discussion

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If `f:R->S` defined by `f(x)=sinx-sqrt(3)cosx+1` is onto , then the interval of S is :A. [0,1]B. [-1,1]C. [0,3]D. [-1,3]

If `f(x)=3x^(3)-5x^(2)+10`, find `f(x-1)`.

A function f from the set of natural number to integers defined by `f(n)={{:(,(n-1)/(2),"when n is odd"),(,-(n)/(2),"when n is even"):}`A. neither one-one nor ontoB. one-one but not ontoC. one but not one-oneD. one-one and onto both

The range of the function `f(x)=7-x p_(x-3)` , isA. {1,2,3,4,5}B. {1,2,3,4,5,6}C. {1,2,3,4}D. {1,2,3}

Let `f:RtoR:f(x)=x^(2)+1`. Find (i) `f^(-1){-4}` (ii) `f^(-1){10}` (iii) `f^(-1){5,17}`.

Let f and g be real functions, defined by `f(x)=sqrt(x-1)andg(x)=sqrt(x+1)`. Find (i) `(f+g)(x)` (ii) `(f-g)(x)` (iii) `(fg)(x) `(iv) `((f)/(g))(x)`.

Let `A={1,2}, B={3,6}`and `f: A->B`given by `f(x)=x^2+2 n d g: A->B`given by `g(x)=3xdot`Then we observe that `f a n d g`have the same domain and co-domain. Also we hve, `f(1)=3=g(1)a n d f(2)=6=g(2)dot`Hence `f=gdot`

Show that `f(x)=(1)/sqrt(x-|x|)` is not defined for any `x""inR`. How will you define dom (f) and range (f)?

Find the domain and the range of the real function, `f(x)=(1)/(sqrt(x+|x|))`.

If f : R → R and g : R → R are defined by f (x) = 2x + 3 and g(x) = x2 + 7, then the values of x such that g{f (x)} = 8 are(a) 1, 2 (b) – 1, 2 (c) – 1, – 2 (d) 1, – 2

If a function `g={(1,1),(2,3),(3,5),(4,7)}`is described by `g(x)=alphax+beta,`find the values of `alphaa n dbetadot`A. 2x-1B. 2x+1C. x+2D. x-2

If `f(x)=(x-1)/(x+1),x!=-1,`. then show that `f(f(x))=-1/x`, prove that `x!=0`.

If `f`is an even function defined on the interval `(-5,5),`then four real values of `x`satisfying the equation `f(x)=f((x+1)/(x+2))`are , , _ and____.