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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
If one the roots fo the equation `x^(2) +x f(a) + a=0` is the cube of the othere for all `x in R`, then f(x)=A. `x ^(1//4) + x ^(3//4)`B. ` - (x^(1//4) + x ^(3//4))`C. ` x + x ^(3)`D. none of these |
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Answer» Correct Answer - B Let `alpha ` and `alpha^(3)` be the roots of the equation `x^(2) + x f(a) + a=0`. Then, `alpha + alpha^(3) = - f (a) and alpha^(4) = a` `rArr f (a) = -alpha a alpha^(3) = - (a_^(1//4) - a^(3//4)` `rArr f(a) = - (a^(1//4) + a^(3//4))`, where `a = alpha^(4) gt 0` `rArr f(x) = - (x^(1//4) + x^(3//4)), x gt 0` |
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| 152. |
If the graph of a function f(x is symmetrical about the line x = a, thenA. ` f (a + x ) = f (a- x )`B. `f (a + x) = f (x - a )`C. `f(x) = f (-x)`D. none of these |
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Answer» Correct Answer - A Trivial |
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| 153. |
If f(x) is an even function, then the curve y=f(x) is symmetric aboutA. x-axisB. y-axisC. both the axesD. none of these |
| Answer» Correct Answer - B | |
| 154. |
Find the domain of the following functions : `f(x)=sqrt(log_10((log_10x)/(2(3-log_10x)))`A. `(10^(2), 10^(3)) `B. `[10^(2), 10^(3)]`C. ` [ 10^(2), 10^(3))`D. `(10, 10^(3))` |
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Answer» Correct Answer - A We, have `f(x) = sqrt(log_(10){(log_(10)x)/(2(3-log_(10)x))})` Clearly, f(x) is defined, if `rArr { (log_(10)x)/(2-log _(10)x)} gt 0 (log_(10)x)/(2(3-log_(10)x)) gt 0 ` and `x gt 0` `rArr (log_(10)x)/(2(3-log_(10)x))gt 1, (log_(10)x)/(3-log_(10)x) gt 0 and x gt 0` `rArr (3(log_(10)x-2))/(2(log_(10) x-3)) le 0 ,(log_(10))/(log_(10)x-3) lt 0 and x gt 0` `2le log_(10) x lt3, 0 ltlog_(10) x lt 3 and x gt 0` `rArr 10^(2) lex le10^(3), 10^(0) lt x lt 10^(3) and x gt 0` `rArr 10^(2) le x lt 10^(3) rArr x in [ 10,10^(3))` |
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| 155. |
Find the range of `f(x)=sqrt(sin(cos x))+sqrt(cos(sin x))`.A. `[sqrt(cos1),sqrt(sin1)]`B. `[sqrt(cos1),1+sqrt(sin1)]`C. `[1-sqrt(cos1),sqrt(sin 1)]`D. none of these |
| Answer» Correct Answer - B | |
| 156. |
The range of the function `f(x) = sqrt(2-x)+sqrt( 1+x)`A. `[sqrt3, sqrt6]`B. `[0, sqrt6]`C. `(sqrt3, sqrt6)`D. none of these |
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Answer» Correct Answer - A We have, `f(x) = sqrt(2-x) + sqrt(1+x)` is Clearly, `f(x)` is defined for `2-x le 0 and 1+ x ge 0 rArr x le 2 and x ge - 1 rArr x in [-1,2]` So, domain `(f) = [-1,2]` Let `y = sqrt(2-x) + sqrt(1+x)" "….(i)` ` rArr y^(2) = 3+2 sqrt(2+x-x^(2))" "......(ii)` `rArr ((y^(2) -3)/(2))^(3) = 2+ x-x^(2)` `rArr x^(2) - x = 2 -((y^(2)-3)/(2))^(2)` `rArr (x-(1)/(2))^(2) = (9)/(4) -((y^(2)-3)/(2))^(2)` `rArr x -(1)/(2) = pm sqrt((9)/(4)-((y^(2)-3)/(2))^(2))` `rArr x = (1)/(2) pm sqrt((9)/(4) - ((y^(2)-3)/(2))^(2))` For x to be real, we must have `(9)/(4) -((y^(2)-3)/(2))^(2) ge 0` `rArr ((y^(2)-3)/(2))^(2) -((3)/(2))^(2) le 0` `-(3)/(2) le (y^(2) -3)/(2) le (3)/(2)` `rArr 0 ley^(2) le 6 rAra - sqrt(6) le y le sqrt(6) rArr y in [-sqrt(6),sqrt(6)]` Alos, from (i) and (ii), we have `y^(2) ge 3 and y ge 0 y ge sqrt(3)` Form (iii) and (iv) ,we have `y in [sqrt(3),sqrt(6)]` Hence range `(f) = [ sqrt(3), sqrt(6)]` |
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| 157. |
Let `f(x) = sec^(-1)[1 + cos^(2) x]`, wheb denotes the greatest integer function. Then the range of (x) isA. ` [ 1, 2 ]`B. `[ 0, 2] `C. `{sec ^(-1) 1, sec ^(-1) 2} `D. none of these |
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Answer» Correct Answer - C Clearly, `f(x)` is defined for all `x in R` for any ` x in R`we have `0 le cos^(2) x le 1` `rArr 1 le 1 + cos ^(2) x le 2` `rArr [1 + cos^(2) x] = {{:(,1,"for" x in R -{ pi : n in Z}),(,2,"for"x = pi"," n in Z):}` `rArr f(x) =sec^(-1) [ 1+ cos^(2)]={{:(sec^(-1)1,"for" x in R - {n pi : n in Z}),(sec^(-1)2,"for" x = pi n in Z):}` Hence, range `(f) = {sec^(-1), sec^(-1)}` |
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| 158. |
The range of the function is `f(x)=log_5(25-x^2)` isA. `[0, 5]`B. ` [0, 2)`C. ` (0, 2)`D. none of these |
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Answer» Correct Answer - D Cleary, f(x) is defined, if `25-x^(2) gt 0 rArr - 5 lt x lt 5` Now, let `y = log_(5) (25-x^(2))`. Then `5^(y) = 25- x^(2) rArr - 5 lt x lt 5` For x to real, we must have `25-5^(y) ge 0 rArr 5^(y) le 25 rArr y le 2` Also, `y =f(x) to - oo` as `x to pm5`. Hence, range `(f) = (-oo,2]` |
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| 159. |
If `f(x) = log_(e^2x) ((2lnx+2)/(-x))` and `g(x) = {x}` then range of `g(x)` for existance of `f(g(x))` isA. `(0, 2//e)`B. `(0, 1//e) - {1//e^(2)}`C. `(0, 3//e)`D. none of these |
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Answer» Correct Answer - B We have `f(x) log_(e^(2)x)((2 In x +2)/(-x)) and g(x) = {x}` For f(x) to real, we must have `x lt 0, x ne (1)/(e^(2)) and ((2 In (x) +2)/(-x)) gt 0` Now ,`((2 In x +2))/(-x) lt 0 and x gt 0` `rArr Inx + 2 lt 0 rArr In x lt-1 rArr x lte^(-k) rArr x lt(1)/(2)` Thus, f(x) is defined if `x in (0,1//e) - [1//e^(2)]` |
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| 160. |
Let `f(x)=4cossqrt(x^2-pi^2/9` Then the range of `f(x)` isA. `[-1, 1] `B. `[-4, 4]`C. `[0, 1 ]`D. none of these |
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Answer» Correct Answer - B Clearly, f(x) is defined for `x^(2) - (pi^(2))/(9) ge 0` `rArr x le -(pi)/(3) or x ge (pi)/(3)` `rArr x in (-oo, -pi//3]uu[pi//3, oo)` Let `y = 4 cos sqrtx^(2) - (pi)/(2))` `rArr cos^(-1) (y)/(4) = sqrt(x^(2) - (pi^(2))/(9))` `rArr x^(2) - (pi)/(9) =(cos^(-1) .(y)/(4))^(2) rArr x = pm sqrt((pi^(2))/(9) + (cos^(-1).(y)/(4))^(2))` clearly, x is real all y for which `cos^(-1).(y)/(4)`is meaningful. Now, `cos^(-1)((y)/(4))` is defined, if `-1 le (y)/(4) lt 1 rArr - 4 lt y lt 4rArr y in [-4,4]` Hence, range of `f =[-4,4]` |
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| 161. |
`f(x) = 9^x/ ( 1 + 9^x)` then value of `f(1/2015) + f(2/2015) .......+ f(4029/2015)`A. 1007B. `(4029)/(2)`C. 2014D. 2015 |
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Answer» Correct Answer - D We have `f(x) = (9^(x))/(9^(x)+9)` `therefore " " f(2-x) = (9^(x-2))/(9^(2-x)+9)` So,`f(x) +f(2-x)=(9^(x))/(9^(x)+9) +(9^(x-2))/(9^(2-x) +9) = (9^(x))/(9^(x)+9)+(9)/(9+9^(x)) =1` `rArr f(x) + f(2-x) = 1` `therefore " "f{(1)/(2015)}+f((2)/(2015)) +((3)/(2015))+......+f((4029)/(2015))` `= {f((1)/(2015)) +f((4029)/(2015))}+{f((2)/(2015))+f((4028)/(2015))}+.....+{f((20+f((2014)/(2015))+f((2016)/(2015))}+f((2015)/(2015))` `={1 +1+....+1(2014 "times")} +f(1) = 2014 +(1)/(2) = (4029)/(2)` |
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| 162. |
The range of the function `f(x) = tan sqrt((pi^(2))/(9)-x^(2))`, isA. `[0, sqrt3]`B. `(0, sqrt3)`C. `[0, sqrt3)`D. `(0, sqrt3]` |
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Answer» Correct Answer - A Clearly, = , f(x) is defined for`x in [-pi//3,pi//3]` Since tan x is an incresing function in `[0,pi//2)` and `0 le (pi^(2))/(9) - x^(2)(pi)/(9) "for" x in [(-pi)/(3),(pi)/(3)]` ` rArr 0 le sqrt((pi^(2))/(9)-x^(2)) le (pi)/(3) "for" x in [(-pi)/(3),(pi)/(3)]` Also, `f(-x) = f(x) ` for all `x in [-pi//3, pi//3]`. Thererfore , Range `(f) = [f((-pi)/(3)),f(0)]=[tan0, tan,(pi)/(3)]=[0,sqrt(3)]` |
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| 163. |
The range of the functions `f : [0,1] to R`, given by `f(x) = x^(3) - x^(2) + 4x + 2 sin^(-1)x` , isA. ` [-pi-2, 0] `B. `[2, 3 ] `C. `[ 0, 4 + pi]`D. `[0, 2+ pi]` |
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Answer» Correct Answer - C We have, `f(x) = x^(3) - x^(20 + 4x + 2 sin ^(1) x` `rArr f(x) = 3x^(2) - 2x + 4 +(2)/(sqrt(1=x^(2)))` `rArr f(x) lt 0 ` for all `x in [0,1]" "[because 3x^(2) -2x+ 4 gt 0"for al" x in R]` `rArr f(x)` is increasing function on `[0,1]` Hence,range`(f) =[(f (0),f(1)]=[0,4 + pi]` |
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| 164. |
The range of the function `f(x) = log_(3) (5+4x - x^(2))`, isA. `(0, 2 ]`B. `(-oo, 2]`C. `(0, 9]`D. none of these |
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Answer» Correct Answer - B For f(x) to be real, we must have `5 + 4x - x^(2) gt 0` `rArr x^(2) - 4x - 5 lt 0 rArr - 1 lt x lt 5` Now, let `y = f(x)`. Then , `y = log_(3) (5+4x -x^(2))` `rArr 3^(y) = 5+ 4x -x^(2)` `x^(2) -4x + 3^(y) - 5=0` `rArr x= (4pmsqrt(16-4.3^(y) + 20))/(2) rArr = 2 pm sqrt(9-3y)` For x to be real, we have must have `9-3^(y) ge 0 rArr 3^(y) ge 9 rArr le rArr y le 2` Also , `y to - oo` as `x to -1 or 5` Hence, range of `(f) = (-oo,2]` |
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| 165. |
Range of the function `f(x)=sqrt(cos^(- 1)(sqrt(log_4x))-pi/2)+sin^(- 1)((1+x^2)/(4x))` is equal to (A) `(0,pi/2+sqrt(pi/2)]` (B) `[pi/2,pi/2+sqrt(pi/2)]` (C) `[pi/6,pi/4)` (D) `{pi/6}`A. ` (0, (pi)/(2)+ sqrt (( pi)/(2))]`B. ` [ (pi)/(2), (pi)/(2)+ sqrt ((pi)/(2))] `C. ` [ (pi)/(6), ( pi)/(2)]`D. `{ (pi)/(2)}` |
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Answer» Correct Answer - D The domain of f(x) is {1} . `therefore ` Range `(f) = {f(1)} = {(pi)/(2)}` |
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| 166. |
The function `f(x)=((1)/(2))^(sinx)`, isA. periodic with period `2pi`B. an odd functionC. not expressible as the sum of an even function and an odd functionD. none of these |
| Answer» Correct Answer - A | |
| 167. |
If[.] and {.} denote greatest integer and fractional part functions respectively, then the period of `f(x) = e^(sin 3pi{x} + tan pi [x])` isA. `2//3`B. `1`C. `3`D. none of these |
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Answer» Correct Answer - B We have, `f(x) = e^(sin 3 pi {x} + tan pi{x} = e^(sin 3 pi {x}))" "[because tan pi[x] = 0]` Since {x} is a periodic with period 1. Therefore f(x) is peridic with period 1. |
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| 168. |
The set of value of a for which the function `f(x)=sinx+[(x^(2))/(a)]` defined on [-2,2] lies an odd function , isA. `(4,oo)`B. `[-4,4]`C. `(-oo,4)`D. none of these |
| Answer» Correct Answer - C | |
| 169. |
Let `f:[pi,3pi//2] to R ` be a function given by `f(x)=[sinx]+[1+sinx]+[2+ sinx]` Then , the range of f(x) isA. `{0,3}`B. `{1}`C. `{0,2}`D. `{3}` |
| Answer» Correct Answer - A | |
| 170. |
The range of the function `f(x)=1+sinx+sin^(3)x+sin^(5)x+……` when `x in (-pi//2,pi//2)`, isA. (0,1)B. RC. (-2,2)D. none of these |
| Answer» Correct Answer - B | |
| 171. |
Find the domain of the function `f(x)=sqrt(1-2x)+3sin^(- 1)((3x-1)/2)`A. `[-1//3,1]`B. `(-oo,1//2]`C. `[-1//3,1//2]`D. `[-1//3,1//2)` |
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Answer» Correct Answer - C Let f(x)=g(x)+h(x), where `g(x)=sqrt(1-2x) and h(x)=3 sin^(-1)((3x-1)/(2))` . Clearly , Domain (g)`=(-oo,1//2]` Also , h(x) is defined when `-1 le (3x-1)/(2) le 1 implies -2 le 3x-1 le 2 implies -(1)/(3) le x le 1` So, domain (h) `=[-1//3,1]` Hence , domain `(f)=(-oo,1//2] cap [-1//3,1]=[-1//3,1//2]` |
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| 172. |
The domain of definition of the function `f(x)=tan((pi)/([x+2]))`, isA. `[-2,1]`B. `(-2,-1)`C. `R-[-2,-1]`D. none of these |
| Answer» Correct Answer - D | |
| 173. |
If `f(x)=sqrt(|3^(x)-3^(1)|-2) and `g(x)=tan pi x,` then domain of fog(x) , isA. `[n+(1)/(3),n+(1)/(2)] cup [n+(1)/(2),n+1], n in Z `B. `(nx+(1)/(4),n+(1)/(2)) cup (n+(1)/(2),n+1), n cup Z `C. `(n+(1)/(4), n+(1)/(2)) cup [n-(1)/(2),n+1],n in Z `D. `[n+(1)/(4),x +(1)/(2)) cup (n+(1)/(2), n+2), n in Z ` |
| Answer» Correct Answer - B | |
| 174. |
The value of `f(x)=3sin((pi^2)/(16)-x^2)`lie in the interval____A. `[-pi//4,pi//4]`B. `[0,3//sqrt(2)]`C. `(-3,3)`D. none of these |
| Answer» Correct Answer - B | |
| 175. |
the period of the `f(x)=sin^4x+cos^4x` isA. `pi`B. `pi//2`C. `2pi`D. none of these |
| Answer» Correct Answer - B | |
| 176. |
if for nonzero x, `af(x) + bf(1/x) =1/x-5`, where `a!=b` then f(2) =A. `(3(2b+3a))/(2(a^(2)-b^(2))`B. `(3(2b-3a))/(2(a^(2)-b^(2))`C. `(3(3a-2b))/(2(a^(2)-b^(2))`D. `(6)/(a+b)` |
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Answer» Correct Answer - B We have, `a f(x)+bf((1)/(x))=(1)/(x)-5" "`…(i) On replacing x by `(1)/(x)`, we get `af((1)/(x))+bf(x)=x-5` `implies bf(x)+af((1)/(x))=x-5" "`……(ii) Multiplying (i) by a (ii) by b and then subtracting, we get `(a^(2)-b^(2))f(x)=((a)/(x)-bx)-5(a-b)` `implies f(x)=(1)/(a^(2)-b^(2))((1)/(x)-bx)-(5)/(a+b)` `implies f(2)=(3(2b-3a))/(2(a^(2)-b^(2))` |
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| 177. |
If `f(x)=sinx+cosa x`is a periodic function, show that `a`is a rational numberA. `a in Z `B. `a in N `C. `a in Q`D. `a in R ` |
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Answer» Correct Answer - C Let T be the period of f(x). Then, `f(x+T)=f(x)` for all x. `implies sin(T+x) + cos a (T+x) = sin x + cos ax ` for all `x in R ` Putting x=0 and x=-T respectively , we get ` sinT+ cos a T =1` and ` -sin T + cos a T=1` Solving these two equations, we get `sin T =0 and cos a T =1` `implies T n pi and a T =2m pi `, where ` m, n in Z`. `implies (aT)/(T)=(2mpi)/(npi)` `implies a=(2m)/(n)`, which is a rational number `=a in Q` |
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| 178. |
The interval in which the function `y = f(x) = (x-1)/(x^2-3x+3)` transforms the real line isA. `(0,oo)`B. `(-oo,oo)`C. `[0,1]`D. `[-1//3,1]-{0}` |
| Answer» Correct Answer - D | |
| 179. |
Let `f(x)=|x-1|`. Then,A. `f(x^(2))=[f(x)]^(2)`B. `f(|x|)=|f(x)|`C. `f(x+y)=f(x)+f(y)`D. none of these |
| Answer» Correct Answer - D | |
| 180. |
The function `f: C -> C` defined by `f(x) = (ax+b)/(cx+d)` for `x in C` where `bd != 0` reduces to a constant function ifA. `a=c`B. `b=c`C. `ad=bc`D. `ab=cd` |
| Answer» Correct Answer - C | |
| 181. |
If f(x)=ax+b and g(x)=cx+d, then f(g(x))=g(f(x)) is equivalent toA. f(a)=g(c )B. `f(b)=g(b)`C. `f(d)=g(b)`D. `f(c )=g(a)` |
| Answer» Correct Answer - C | |
| 182. |
`x(1+2(x+4)^(- 0. 5))/(2-(x+4)^(0. 5))+5(x+4)^(0. 5)` Find the domain of the following functionA. RB. (-4,4)C. `R^(+)`D. `(-4,0) cup (0,oo)` |
| Answer» Correct Answer - D | |
| 183. |
The maximum possible domain and thecorresponding range of `f(x)=(-1)^x`A. `D=R,E=[-1,1]`B. D=I( the set of integers ), E=[-1,1]C. D=R,E=[-1,1]D. `D=I,E={{:(+1, " when " x=0 or even ),(-1," when x is odd "):}` |
| Answer» Correct Answer - D | |
| 184. |
Which of the following functions is not an are not an insjective map(s) ?A. `f(x)=|x+1|,x in [-1,oo)`B. `g(x)=x+(1)/(x), x in (0,oo)`C. `h(x)=x^(2)+4x-5, x in (0,oo)`D. `k(x)=e^(-x), x in [0,oo)` |
| Answer» Correct Answer - B | |
| 185. |
The function f(x) given by `f(x)={{:(x^(4)tan""(pix)/(2),|x| lt1),(x|x|,|x| ge1):} is `A. an odd functionB. an even functionC. a periodic function with period `(2pi)/(sqrt(5))`D. none of these |
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Answer» Correct Answer - A We have , `f(x)={{:(-x^(2),if x le-1),(x^(4)tan""(pix)/(2), if -1 lt x lt 1),(x^(2), if x gt1):}` If is evident from this definition that `f(-x)=f(x)` for all ` x in R `. So, f(x) is an odd functions. |
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