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51.

The domain of the function` f(x)=(sec^(-1)x)/(sqrt(x-[x]))` , where [x] denotes the greatest integers less than or equal to x is defined for all x belonging toA. `R`B. `R-{(-1,1) cup {n:n in Z}}`C. `R^(+)-(0,1)`D. `R^(+)-[n:n in N]`

Answer» Correct Answer - B
Discussion
52.

Statement -1: Let f(x) be a function satisfying `f(x-1)+f(x+1)=sqrt(2)f(x)` for all ` x in R ` . Then f(x) is periodic with period 8. Statement-2: For every natural number n there exists a periodic functions with period n.A. Statement-1 is True, Statement-2 is True, statement-2 is a correct explanation for the statement-1 .B. Statement-1 is True, Statement-2 is True, statement-2 is not a correct explanation for the statement-1 .C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False , Statement-2 is True.

Answer» Correct Answer - B
Function f(x) satisfies the relation.
`f(X-1)+f(x+1)=sqrt(2)f(x)` for all x. ……..(i)
Replacing x by `x+1` in (i), we get .
`f(x)+f(x+2)=sqrt(f)(x+1)` ………(ii)
Replacing x by (x-1) in (i) ,we get
`f(x-2)+f(x) =sqrt(2)f(x-1)` …..(iii)
Adding (ii) and (iii) , we get
`2f(x)+f(x+2)+f(x-2)=sqrt(2)(f(x-1)+f(x+1))`
`implies 2f(x)+f(x+2)+f(x-2)=sqrt(2)xxsqrt(2)f(x)`
`implies f(x+2)+f(x-2)=0`
Replacing x by x+2 in (ii) we get
`f(x+4)+f(x)-0`
Replacing x by x+4 in (iv) we get
`f(x+8)+f(x+4)=0`
Subtracting (iv) from (v), we get
`f(x+8)-f(x)=0implies f(x)=f(x+8)`
So, f(x) is a periodic function with period 8.
Hence, statement-1 is true.
We know that g(x)=x-[x] is a periodic function with period
1. Therefore, for any ` n in N, phi (x)=(x)/(n)-[(x)/(n)]` is periodic with period n.
Discussion
53.

Which of the following functions is periodic ?A. `f(x)=x+sinx`B. `f(x)=cos sqrt(x)`C. `f(x)=cos x^(2)`D. `f(x)= cos^(2)x`

Answer» Correct Answer - D
Discussion
54.

The period of `f(x)=(1)/(2){(| sinx|)/(cos x)+(|cos|)/(sinx)}`, isA. `2pi`B. `pi`C. `(pi)/(2)`D. `(pi)/(4)`

Answer» Correct Answer - A
We observe that `(|sinx|)/(cosx) and (|cosx|)/(sinx)` are periodic functions each with period `2pi` . Also, there is no positive real number T between 0 and `2pi` such that `f(x+T)=f(x)` .
So, f(x) is periodic with period `2pi` .
Discussion
55.

Domain (D) and range (R) of `f(x)=sin^(-1)(cos^(-1)[x]),`where [.] denotes the greatest integer function, is`D-=x in [1,2],R in {0}`D`-=x in 90 ,1],R-={-1,0,1}``-=x in [-1,1],R-={0,sin^(-1)(pi/2),sin^(-1)(pi)}``-=x in [-1,1],R-={-pi/2,0,pi/2}`A. `[1, 2) and {0}`B. `[0, 1] and {-1, 0, 1}`C. `[-1, 1] and {0, sin ^(-1)((pi)/(2)),sin^(-1) (pi)}`D. `[-1 1] and {- (pi)/(2), 0, (pi)/(2)}`

Answer» Correct Answer - A
Let `f(x) = cos^(-1)[x] and g(x) = sin^(-1) x`. Then, `phi(x) = gof(x)`. Clearly
`D(f) = [-1,1), R(f) = [Cos^(-1)(-1), cos^(-1) (0), cos^(-1)(1)]= {pi,(pi)/(2),0}`
`D(g) = [-1,1] and R(g) =[-(pi)/(2),(pi)/(2)]`
`therefore" "D(phi) = D(gof) = {x : x in D(f) and f(x) in D(g)}`
`={x: x in [-1,2) and f(x)in [-1,1]}`
`{x:x in {-1,2) and cos^(-1) [x] in [-1,1]}`
`=[1,2)`
`R(phi) ={ phi (x) : x in |1,2)}= { sin^(-1) (cos^(-1)1)}={sin^(-1)(0)} = {0}`
Discussion
56.

The function `f(theta) = cos (pi sin^(3) theta)`, isA. not periodicB. periodic and its period is same as that of `cos theta `C. periodic and its period is same as that of ` cos 2 theta `D. periodic and its period is same as that of ` cos ( pi theta)`

Answer» Correct Answer - C
We have,
`f(theta) = cos(pi sin^(2) theta)`
Clearly, `-pi le sin^(3) theta le pi` for ` theta in [-pi//2,pi//2]` and cosine function is an even function.
Same as the period of cos `2 theta`
Discussion
57.

The function `f(x) = 3^(sin^(2)pi + sin^(4) pi x + x-[x])` where [x] denotes the greatest interger less than or equal to x, isA. a periodic function with period 1B. a periodic function with period 2C. a periodic function with period ` (1)/(2)`D. not a periodic function

Answer» Correct Answer - A
We have
`sin^(2) pi x + sin^(4) pi x`
`=(1)/(2) {1-cos 2 pi x} +(1)/(8) {3-4 cos 2 pi x + cos 4 pi x}`
`=(7)/(8) - cos 2 pi x +(1)/(8) cos 4 pi x`
Clearly, it is a peridic function equal to LCMof
`(2pi)/(2pi)` and `(2pi)/(4)` i.e., 1.
Also, `x - [x]` is periodic with period 1.
Hence, `f(x)` is periodic with period 1.
Discussion
58.

If `f(x) = log_([x-1])(|x|)/(x)`,where [.] denotes the greatest integer function,thenA. `D(f) = [3, oo), R(f) = {0, 1} `B. `D(f) = [ 3, oo), R(f) = [3, oo), R(f)= {0}`C. `D(f) = (2, oo), R(f) = {0, 1}`D. `D(f) = (3, oo), R(f) = {0}`

Answer» Correct Answer - B
f(x) is defined for all x satisfying.
`[x-1] gt 0, [x -1] ne 1 and (|x|)/(x) gt 0`
Now ,`[x-1] gt 0, [x-1] ne 1 rArr x in [3,oo)`
and, `(|x|)/(x) gt 0 rArr x gt 0`
`therefore " "f(x) = log_([x-1])(|x|)/(x) = log_([x-1])1=0`
So, `R(f) ={0}`
Discussion
59.

The domain of the function `f(x) = sqrt((4-x^(2))/([x]+2))` where [x] denotes the greatest integer less than or equal to x,isA. `[-1, 2]`B. `(-oo, -2 )`C. `(-oo, -2) uu [-1,2]`D. none of these

Answer» Correct Answer - C
For f(x) to be real,we must have
`(4-x^(2))/([x]+2) gt 0 and [x] + 2 ne 0`
`rArr (4-x^(2))/([x] +2)gt 0 and x in [-2,-1)`
CASE-I When `4-x^(2) gt 0 and [x]+ 2 ht 0`
In this case, we have
`x in [-2,2] and x in [-1,oo) rArr x in [-1,2]`
Also, `x in [-2,-1)`
`therefore " " x in [-1,2]`
CASE -II When `4-x^(2) lt 0 and [x] + lt0`
`
Discussion
60.

Which of the following functions is non-periodic? (1) `2^x/2^[x]=` (2) `sin^(-1)({x})`(3) `sin^(-1)(sqrt(cosx))` (4) `sin^(-1)(cos x^2)`A. f(x) = tan (3x+5)B. g(x)={x}, the fractional part of xC. `f(x)=1-(cos^(2)x)/(1+tanx)-(sin^(2))/(1+cotx)`D. `phi(x)=x+ cos x `

Answer» Correct Answer - D
We observe that
`f(x)=tan(3x+5)` is periodic with period `(pi)/(3)`
`g(x)={x}=x-[x]` has period 1
`h(x)=1-(cos^(3)x+ sin^(3)x)/(cosx+ sinx)`
`implies h(x)=1-(cos^(2)x+sin^(2)x-sinx cosx)`
`implies h(x)=(1)/(2) sin 2x` has period `pi`. and , `phi(x)=x+cosx` is non- periodic, because x is non- periodic.
Discussion
61.

Period of the function `f(x) =(1)/(3){sin 3x + |sin 3x | + [sin 3x]}` is (where [.] denotes the greatest integer function )A. `(pi)/(3)`B. `( 2pi )/(3)`C. `( 4pi )/(3)`D. `pi`

Answer» Correct Answer - B
We obseve that the periods of sin3x, |sin 3x| and
`[sin3x]` are `(2pi)/(3),(pi)/(3)` and `(2pi)/(3)` respectivel.
`therefore ` Period of `f(x)` is `(2pi)/(3)`
Discussion
62.

If `f: R to R ` is defined by `f(x)=x-[x]-(1)/(2)` for all ` x in R `, where [x] denotes the greatest integer function, then `{x in R: f(x)=(1)/(2)}` is equal toA. ZB. NC. `phi`D. R

Answer» Correct Answer - C
Discussion
63.

Let `f(x)=sin sqrt([a]) x` (where [ ] denotes the greatest integer function). If f is periodic with fundamental period `pi`, then a belongs toA. `lambda in [4,5)`B. `lambda in [4,5]`C. `lambda=4,5`D. none of these

Answer» Correct Answer - A
We have , `f(x)=sin(sqrt[lambda])x)`
Clearly , period of `f(x)` is ` (2pi)/(sqrt([lambda]))`
But , period of f() is given as `pi`.
`:. (2pi)/(sqrt([lambda]))=pi=2implies[lambda]=4implieslambda in [4,5)`.
Discussion
64.

If `T_(1)` is the period of the function `f(x)=e^(3(x-[x])) and T_(2)` is the period of the function `g(x)=e^(3x-[3x])``([*]` denotes the greatest integer function ), thenA. `T_(1)=T_(2)`B. `T_(1)=(T_(2))/(3)`C. `T_(1)=3T_(2)`D. none of these

Answer» Correct Answer - C
Discussion
65.

If f(x) and g(x) are two real functions such that `f(x)+g(x)=e^(x) and f(x)-g(x)=e^(-x)` , thenA. f(x) is an odd functionB. g(x) is an even functionC. f(x) and g(x) are periodic functions.D. none of these

Answer» Correct Answer - D
Discussion
66.

The range of the function `f(x)=sin{log_(10)((sqrt(4-x^(2)))/(1-x))}` ,isA. [0,1]B. (-1,0)C. [-1,1]D. (-1,1)

Answer» Correct Answer - C
Discussion
67.

Let `f(x)=(1)/(x) and g(x)=(1)/(sqrt(x))`. Then,A. f(g(x)) and g(f(x)) have different domainsB. f(g(x)) and g(f(x)) have same domainC. g(f(x)) is a bijective mappingD. f(g(x)) is neither odd or even.

Answer» Correct Answer - B
Discussion
68.

Domain of `(sqrt(s^(2)-4x+3)+1) log_(5)""((x)/(5))+(1)/(x)(sqrt(8x-2x^(2)-6)+1) le 1` isA. `(-oo,1] cup [3,oo)`B. `[1,3]`C. {1,3}D. {1}

Answer» Correct Answer - D
Discussion
69.

The domain of the function `f(x)=log_(3+x)(x^2-1)` isA. `(-3,-1) cup (1,oo)`B. `[-3,-1) cup [1,oo)`C. `(-3,-2) cup (-2,-1) cup (1,oo)`D. `[-3,-2) cup (-2,-1) cup [1,oo)`

Answer» Correct Answer - C
Discussion
70.

If `f(x)=cos^(-1)((2-|x|)/(4))+[log_(10)(3-x)]^(-1)`, then its domain isA. [-2,6]B. `[-6,2)cup(2,3)`C. `[-6,2]`D. `[-2,2) cup (2,3]`

Answer» Correct Answer - B
Discussion
71.

If D is the set of all real x such that `1-e^((1)/(x)-1)` is positive , then D is equal toA. `(-oo,1]`B. `(-oo,0)`C. `(1,oo)`D. `(-oo,0)cup(1,oo)`

Answer» Correct Answer - D
Discussion
72.

The function `f(x)=log_(2x-5)(x^(2)-3x-10)` is defined for all belonging toA. `[5,oo)`B. `(5,oo)`C. `(-oo,+5)`D. none of these

Answer» Correct Answer - B
Discussion
73.

If `f : R -> R` is defined by `f(x) = [2x] - 2[x]` for `x in R`, where [x] is the greatest integer not exceeding x, then the range of f isA. [0,1]B. {0,1}C. `(0,oo)`D. `(-oo,0]`

Answer» Correct Answer - B
Discussion
74.

If ` x in R`, then `f(x)=sin^(-1)((2x)/(1+x^(2)))` is equal toA. `2 tan^(-1)x`B. `{{:( -pi-2tan^(-1)x, -oo lt x lt -1),(2 tan^(-1)x , -1 le x le 1),(pi -2 tan^(-1)x, 1 lt x lt oo):}`C. `{{:( -pi-2tan^(-1)x, -oo lt x lt -1),(2 tan^(-1)x , -1 le x le 1),(pi -2 tan^(-1)x, 1 lt x lt oo):}`D. `{{:( -pi+2tan^(-1)x, -oo lt x le -1),(2 tan^(-1)x , -1 lt x lt 1),(pi -2 tan^(-1)x, 1 le x lt oo):}`

Answer» Correct Answer - B
Discussion
75.

If f(x) is defined on [0,1], then the domain of `f(3x^(2))` , isA. `[0,1//sqrt(3)]`B. `[-1//sqrt(3),1//sqrt(3)]`C. `[-sqrt(3),sqrt(3)]`D. none of these

Answer» Correct Answer - B
Discussion
76.

If `f(x)=sinx+cosx `and `g(x)=x^2-1`, then `g(f (x)) `is invertible in the domain .A. `[0, pi//2]`B. `[-pi//4, pi//4]`C. `[-pi//4, pi//2]`D. `[0, pi]`

Answer» Correct Answer - B
Clearly, range `(f) = [-1,1] sub` domina (g)= R
`therefore" " gof: R to R` is given by
`" "gof(x) = g(f(x))`
`" "gof(x) = g(sin x + cosx) = (sin x + cos x)^(2) -1= sin x`
For `gof` to be bijective, we must have
`-(pi)/(2) le 2x le (pi)/(2)rArr x in [-pi//4,pi//4]`
Hence, `gof:[-pi//4,pi//4] uu [-1,1]`
Discussion
77.

The domain of the function `f(x) =(1)/(sqrt(|x|-x))`, isA. `(0,oo)`B. `(-oo,0)`C. `R-{0}`D. none of these

Answer» Correct Answer - B
Let `f(x)=(1)/(sqrt(|x|-x))`.
Clearly , f(x) is defined for all x satisfying
`|x| -x gt0 0implies |x| gt x implies x lt 0`
Hence , domain (f) `=(-oo,0)` .
Discussion
78.

Prove that for `n=1, 2, 3...``[(n+1)/2]+[(n+2)/4]+[(n+4)/8]+[(n+8)/16]+...=n`where `[x]` represents Greatest Integer FunctionA. nB. n-1C. n+1D. n+2

Answer» Correct Answer - A
For any ` x in R `, we have
`[x]=[(x)/(2)]+[(x+1)/(2)]" "`….(i)
`implies [n]=[(n+1)/(2)]+[(n)/(2)]`
`implies n=[(n+1)/(2)]+[(n)/(2)]`
`implies n=[(n+1)/(2)]+[(n)/(2)]+[((n)/(2)+1)/(2)]" " ` [ using (i)]
`implies n=[(n+1)/(2)]+[(n+2)/(4)+[(n)/(4)]`
`implies n=[(n+1)/(2)]+[(n+2)/(4)]+[(n)/(8)]+[((n)/(4)+1)/(2)]" "`[using (i)]
`implies n[(n+1)/(2)]+[(n+2)/(4)]+[(n+4)/(8)]+[(n)/(8)]`
Continuing in this manner , we have
`[(n+1)/(2)]+[(n+2)/(4)]+[(n+4)/(8)]+[(n+8)/(16)]+....=n`.
Discussion
79.

The number of integral solutions of the equation {x+1}+2x=4[x+1]-6 , is

Answer» Correct Answer - B
We known that {x}=x-[x]
`:. {x+1}+2x=4[x+1]-6`
`implies x+1-[x+1]+2x=4[x+1]-6`
`implies 3x+1 =4[x+1]-6`
`implies 3x+1=5[x+1]-6`
`implies 3x+1=5([x]+1)-6`
` implies 3x=5[x]-2 " "` (i)
`implies 3([x]+{x})=5[x]-2`
` 3{x}=2[x]-2 " "`(ii)
Now,
`0 le {x} lt 1`
`implies 0 le 3 {x} lt 3 `
`implies 0 le 2 [x]-2 lt 3 " "`[Using (ii)]
`implies 2 le 2[x] lt 5 implies 1 [x] lt (5)/(2) implies 1 le [x] lt (5)/(2) implies [x]=1,2`
From (i) , we have
`[x]=-1implies x=1 and , [x]=2implies x=(8)/(3)`
Hence, is x=1 the only integral solution of the given equation.
Discussion
80.

The domain of the function `f(x)=(cos^(-1)x)/( [x])` , isA. `[-1,1]`B. `[-1,1]-{0}`C. `[-1,00) cup {1}`D. `[-1, 0)`

Answer» Correct Answer - C
Let `g(x)= cos ^(-1) x and h(x)=[x]`,
Then , `f(x)=(g(x))/(h(x))`
`:.` Domain (f)= Domain (g) `cap` Domain (h)- {x: h(x)=0}
Clearly , Domain (g) =[-1,1], Domain h=R,
and , `{x: h(x)=0}={x:[x]=0}=[0,1). `
`:. ` Domain (f)=[-1,1] `cap` R -[0,1)=[-1,0) cup {1}` .
Discussion
81.

The domain of the function `f(x)=sqrt(x^2-[x]^2)`, where `[x]`is the greatest integer less than or equal to `x ,`is`R`(b) `[0,+oo]``(-oo,0)`(d) none of theseA. `R`B. `[0, oo)`C. `(-oo, 0]`D. none of these

Answer» Correct Answer - D
`f(x) = sqrt(x^(2) -[x]^(2))` is defined, if
`x^(2) - [x]^(2) le 0`
`rArr x^(2) ge [x]^(2)`
`rArr x ge 0` or, x is a negative integer.
`rArr x in [0,oo) uu Z^(-),Z^(-)` denotes the set of all negative intergers.
Discussion
82.

The domain of definition of `f(x)=cos^(- 1)(x+[x])` isA. `[0, 1) `B. `R-Z`C. `(0, oo)`D. none of these

Answer» Correct Answer - A
`f(x)= cos^(-1) (x + [x])` is defined, if
`-1 le x+[x] le1`
`rArr -1 le [x] +{x} + [x] le 1 " "[because x = [x= [x] + {x}]`
`rArr - 1 le 2[x] + {x} le 1 rArr [x] = 0 rArr x in [0,1)`.
Discussion
83.

The domain of definition of the functions `f(x-[x])`, isA. `R`B. `R-Z`C. `(0, oo)`D. none of these

Answer» Correct Answer - B
`f (x) = log_(e)(x-[x])` is defined
`0 lt x -[x] lt 1`
But, `0 le x -[x] lt 1 and x - [x] = 0 "for" x in Z`
`therefore 0 lt x 0- [x] lt 1 rArr x notin Z rArr x in R-Z`
Discussion
84.

If `f(x) = [x] and g(x) = {x}=` fraction part of x, then for any two real numbers x and y.A. `f(x+ y) = f(x) + f(y)`B. ` g (x + y ) = g (x) + g (y)`C. `f (x + y ) = f (x) + f ( y + g(x))`D. none of these

Answer» Correct Answer - C
We have,
`[x+y]=[x] +[y +{x}]` for all `x,y in R`
`rArr f(x+y)=f(x) +f(y+ g(x)) ` for all `x,y in R`.
Discussion
85.

The domain of definition of `f(x) = log_(2) (log_(3) (log_(4) x))`, isA. `[4, oo)`B. `(4, oo)`C. `(-oo, 4)`D. none of these

Answer» Correct Answer - B
f(x) is denfined, if
`(log_(3) (log_(4)x)) gt 0 rArr log_(4) x gt 3^(0) rArr x gt 4 rArr x in (4,oo)`
Hence, domain `(f)=(4,oo)`
Discussion
86.

If `e^x + e^(f(x)) = e` then domain of f(x) isA. `(-oo, 1)`B. `(-oo, 0)`C. `(1, oo)`D. none of these

Answer» Correct Answer - A
We have,
`e^(x) + e^(f(x)) = e`
`rArr e^(f(x)) = e-e^(x)`
`rArr f(x) = log_(e) (e-e^(x))`
`rArr f(x) = log_(e) e+ log_(e) (1-e^(x-1))`
`rArr f(x) = 1+ log_(e) (1-e^(x-1))`
Clearly, for (f) to be real, we must have
`1-e^(x-1) gt 0 rArr e^(x) lt e rArr x lt 1 rArr x in (-oo,1)`
Discussion
87.

The domain of `f(x)i s(0,1)dot`Then the domain of `(f(e^x)+f(1n|x|)`is`(-1, e)`(b) `(1, e)``(-e ,-1)`(d)`(-e ,1)`A. `(-1, e)`B. `(1, e)`C. `(e, 1)`D. `(-e, 1)`

Answer» Correct Answer - C
Since f(x) is defined on (0,1) and
`g(x) = f(e^(x)) + f(log_(e)|x|)`. Therefore, g(x) exists, if
` 0 lt e^(x) lt and 0 log_(e)|x| lt 1`
`rArr -oo lt x lt 0 and 1 lt |x| lt e`
`rArr - oo, lt x lt 0 and 1 le |x| lt e`
`rArr x in (-oo, 0) and x in (-e,-1) uu (1,e)`
`rArr x in (-e,-1)`.
Discussion
88.

The domain of the function `f(x)=log_2[log_3(log_4(x^2-3x+6)}]i s` .A. `(1, 2)`B. `[1, 2 ] `C. `(-oo, 1] uu (2, oo)`D. `(-oo, 1]uu[2, oo)`

Answer» Correct Answer - C
`f(x) - log_(2) [log_(3) {log_(4) (x^(2) - 3x+6)}]` is defined,
if
`log{log_(4)(x^(2) - 3x+ 6)}le0`
`rArr log_(4) (x^(2) - 3x+6) gt 3^(0)`
`rArr x^(2) - 3x + 6 gt 4`
`rArr x^(2)-3x+ 2gt 0`
`rArr (x-1) (x-2) gt = 0 rArr x in (-oo,1)uu(2,oo)`
Discussion
89.

The domain of definition of the function `f(x)=sqrt(log_(10) ((2-x)/(x))) ` isA. `(0, 1)`B. `[0, 1 ]`C. `(0, 1]`D. `(0, 2 )`

Answer» Correct Answer - C
We have,
`f(x) = sqrt(log_(10)((2-x)/(x)))`
Clearly, f(x) is defined, if
`rArr (2-x)/(x) ge 10^(0) and (x-2)/(x) lt0`
`rArr (2-x)/(x) ge 1 and (x-2)/(x) le 0`
`rArr 2((x-1)/(x)) le 0 and (x-2)/(x) lt 0 rArr 0 x le 1 rArr x in (0, 1]`
Discussion
90.

The domain of definition of `f(x) = sqrt(e^((cos^-1(log_(4) x^(2)))`isA. `[1//2, 2 ]`B. `[-2, - 1//2] uu [1//2, 2]`C. `[-2, - 1//2]`D. none of these

Answer» Correct Answer - B
For f(x) to be real, we must have
`-1 le log_(4) x^(2) le 1 and x ne 0`
`rArr 4^(-1) le x^(2) le 4^(1) and x ne 0`
`rArr (1)/(4) x^(2) le 4 and x ne 0`
`rArr x in [-2,-1//2] uu [1//2,2] and ne 0`
`rArr x in [-2,-1//2] uu [1//2,2]`
Discussion
91.

The domain of definition of the function `f(x) = sqrt(log_(x^(2)-1)) x` isA. `( sqrt2, oo)`B. `(0, oo)`C. `(1, oo)`D. none of these

Answer» Correct Answer - A
We have,
`f(x) = sqrt(log_((x^(2)-1))x)`
Clearly,f(x) is define,if
`log_((x^(2)-1)) x ge 0, x ge 0 ,x^(2)ge -1 ge 0 and x^(2) - 1 ne1`
Now,
`log_((x^(2)-1)) x gt 0`
`rArr {{:(,x gt 1, "if" x^(2) - 1 gt 1),(,0 lt x lt 1, "if" x^(2) - 1 lt 1):}`
`rArr {{:(,x ge1, "if" x^(2) - 2 lt 0 ),(,0 lt x lt 1,"if" x^(2)-2 lt 0 ):}`
`rArr { {:(, x ge 1,"if" x lt - sqrt(2) or "," xgt sqrt(2)),(,0lt x lt 1 , "if"-sqrt(2) lt x ltsqrt(2)):}`
`rArr x gt sqrt(2) or , 0 lt x lt 1`
`rArr x in (0,1) uu (sqrt(2) , ooo)`
And ,
`x gt 0, x^(2) -1 gt 0 and x^(2) - 1 ne 1`
`rArr x in (1,sqrt(2))`
From (i) and (ii), we get `x in (sqrt(2),oo)`
Hence, domain of `f(x)` is `(sqrt(2),oo)` .
Discussion
92.

The domain of definition of the function `f(x) = log_(3) {-log_(1//2)(1+(1)/(x^(1//5)))-1}`A. `(-oo, 1)`B. `(0, 1 )`C. `(1, oo)`D. none of these

Answer» Correct Answer - C
Clearly, f(x) is defined if
`- log_(1//2)(1+(1)/(x^(1//5)) - 1 gt 0, 1+(1)/(x^(1//5)) gt 0 and x ne 0`
`log_(1//2) (1+(1)/(x^(1//5))) lt -1,1+x^(-1//5) gt 0 and x ne 0`
`rArr 1+(1)/(x^(1//5)) gt ((1)/(2))^(-1),x^(-1//5) gt - and x in 0`
`rArr (1)/(x^(1//5)) gt 1, (1)/(x^(1//5)) gt - and x ne 0`
`rArr lt x lt 1,x gt - 1 and x ne 0 rArr 0 lt x lt1`
Discussion
93.

find the value of theA. `pm sqrt(npi), n in {0, 1, 2, …. } `B. `pm sqrt(npi ), n in {1, 2, …}`C. `(pi)/(2)+ 2npi, n in {…, - 2, -1, 0, 1 , 2, …}`D. `2n pi, n in {… -2, -1, 0, 1, 2,… }`

Answer» Correct Answer - A
We have, `f(x) = x^(2) and g (x) = sin x`
`therefore (fogogof) (x) = (fog) (gof(x)) =(fog)(g(f(x))) = (fog) = (sin^(2))`
`= f(g(sinx^(2)) = f(sin(sin x^(2)) = sin^(2) (sin x^(2))`
and, `(gogof) (x) = (gog) (f(x)) = (gog) (x^(2)) = g(g(x^(2))`
`therefore (fogogof) (x)= (gofog) (x)`
`rArr sin^(2) (sin x^(2)) = sin (sin x^(2))`
`rArr sin (sin^(2)) = sin (sins^(2))`
`rArr sin (sin x^(2)){ sin (sin x^(2))-1)} = 0`
`rArr sin(xin x^(2)) = 0 or, sin (sin x^(2)) = 1`
`sin x^(2) = npi, n in Z or, sin x^(2) = 2mpi +(pi)/(2), m in Z`
`rArr sin x^(2) = 0`
`rArr x^(2) = n pi, n in Z`
`rArr = pm sqrt(npi, n in {0,1,2,3,......}`
Discussion
94.

The domain of the function `f(x)=sin^(-1)((8(3)^(x-2))/(1-3^(2(x-1))))`isA. ` (-oo, 0] `B. `[2, oo)`C. `(-oo, 0) uu [2, oo)`D. `(-oo, -1] uu [1, oo)`

Answer» Correct Answer - C
f(x) is defined for
`1 le (8.3^(x-2))/( 1-3^(2(x-1))) le 1`
`hArr - 1 le(3^(x) - 3^(x-2))/(1-3^(2x-2)) le 1`
`hArr (3^(x) - 3^(x-2))/(1-3^(2x-2))+ 1 ge 0 and (3^(x)-3^(x-2))/(1-3^(2x - 2)) -1 le 0 `
`hArr (1+3^(x) - 3^(x-2) - 3^(2x-2))/(1-3^(2x-2))ge 0 and (3^(x) - 3^(x-2)-1+3^(2x-2))/(1-3^(2x-2))`
`hArr ((1+3^(x))(1-3^(x-2)))/(3^(x).3^(x-2)) ge and ((3^(x) - 1)(3^(x-2)-1+))/((3^(2x-2) - 1))`
`hArr((3^(x)-3^(2)))/((3^(2x)-3^(2))) ge 0 and ((3^(x) -1))/((3^(2x) - 1)) ge 0`
`hArr ((3^(x) - 3^(2)))/((3^(x) - 3)) ge 0 and ((3^(x) -1))/((3^(x)- 3)) ge 0`
`hArr x in (-oo,1] uu[2,oo) and x in (-oo,0]uu(1,oo)`
`hArr x in (-oo,0] uu [2,oo)`
Discussion
95.

A function whose graph is symmetrical about y-axis isA. `f(x)=x((3^(x)-1)/(3^(x)-1))`B. `f(x)=log_(2)(x+sqrt(x^(2)+1))`C. `f(x+y)=f(x)+f(y)`D. `f(x)=sinx+cosx`

Answer» Correct Answer - A
We know that the graph of an even function is symmetrical about y-axis .
Clearly , `f(x)=x((3^(x)-1)/(3^(x)+1))` is an even function.
So, option (a) is correct.
Discussion
96.

If the graph of the function y = f(x) is symmetrical about the line x = 2, thenA. ` f (x) =- f (-x)`B. `f ( 2+x ) = f ( 2 - x )`C. `f (x) = f (-x)`D. ` f (x +2 ) = f (x - 2 )`

Answer» Correct Answer - B
It the graphof f(x) is symmetrical about the line x = 2, then
`f(2+x) = f(2-x)` for all `x in R`.
Discussion
97.

If a real valued function `f(x)` satisfies the equation `f(x +y)=f(x)+f (y)` for all `x,y in R` then `f(x)` isA. a periodic functionB. an even functionC. an odd functionD. none of these

Answer» Correct Answer - C
We know that the function f(x) satisfying the property
`f(x+y)=f(x)+f(y)` for all ` x, y in R` has the formula
`f(x)=xf(1)` for all ` x in R`.
Clearly , it is an odd function.
Discussion
98.

A function whose graph is symmetrical in opposite quadrants isA. `f(x)=e^(x)+e^(-x)`B. `f(x)=log_(e)x`C. `f(x+y)=f(x)+f(y)`D. ` f(x)=cos(x)+sinx`

Answer» Correct Answer - C
We know that the graph of an odd function is symmetrical in opposite quadrants and the function satisfying the equation .
`f(x+y)=f(x)+f(y)` for all ` x , y in R ` has the forumla
`f(x)=xf(1)` for all ` x in R`.
Clearly , it is an odd function . Hence, its graph is symmetrical in opposite quadrants.
Discussion
99.

The domain of definition of `f(x) = sqrt(sec^(-1){(1-|x|)/(2)})` isA. `(-oo, -3)`B. `[3, oo)`C. `phi`D. `(-oo, -3] uu [3, oo)`

Answer» Correct Answer - D
For (x) to real, must have
`sec^(-1){(1-|x|)/(2)} gt 0 and (1-|x|)/(2) le -1 or,(1-|x|)/(2) ge1`
`rArr 1-|x| le -2 or, 1- ge2[because 0 le sec^(-1){(1-|x|/(2)} ge pi]`
`|3| ge 3 " "[because |x| le- 1 "absurd"]`
`rArr x in (-oo, -3]uu [3,oo)`
Discussion
100.

Fin the domain of defintion of the function `f(x)=(1)/(sqrt([x]^(2)-[x]-6)`, isA. `(-oo,-2) cup [4,oo)`B. `(-oo,-2]cup[4,oo)`C. `(-oo,-2) cup( 4,oo)`D. none of these

Answer» Correct Answer - A
f(x) is defined for
`[x]^(2)-[x]-6 gt 0`
`implies {[x]-3}{[x]+2} gt 0`
`implies [x] lt -2 or, [x] gt 3 implies x in (-oo,-2] cup [4,oo)`
Discussion