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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.
| 101. |
The domain of the function `f(x)=sqrt(cos^(- 1)((1-|x|)/2))` isA. `[-3, 3]`B. `(-oo, -3) uu( 3, oo)`C. `(-oo, -3] uu [3, oo)`D. `1` |
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Answer» Correct Answer - A For f (x) to be real , we must have `-1 ge(1-|x|)/(2) ge 1` `rArr -2 ge1 - |x| ge 2` `rArr -3 -|x| ge 2` `rArr - 3le - |x| le 1` `rArr -1 le|x| le 3 rArr |x| le 3 x in [-3,3]` Hence, domain (f) = [-3,3] |
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| 102. |
The domain of definiton of the function `f(x)=(1)/(sqrt(x^(12)-x^(9)+x^(4)-x+1))` , isA. `(-oo,-1)`B. `(1,oo)`C. `(-1,1)`D. R |
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Answer» Correct Answer - D f(x) assumes real values , if `x^(12)-x^(9)+x^(4)-x+1 gt0` `implies (x^(12)+x^(4))-(x^(9)+x)+1) gt 0` `implies x^(4)(x^(8)+1)-x(x^(8)+1)+1 gt 0` `implies x(x^(8)+1) (x^(3)-1)+1 gt 0` Clearly , it is true for all ` x ge 1 or , x le 0`. For ` 0 lt x lt 1`, we have `x^(4) gt x^(8)` `implies x^(4)+1 gt x^(8)+1` `implies x^(4)+1 gt x(x^(8)+1)` `implies -x(x^(8)+1)+x^(4)+1 gt 0` `implies x^(12)-x(x^(8)+1)+x^(4)1 gt 0` Thus , `x^(12)-x^(9)+x^(4)-x+1 gt 0` for all ` x in R. ` Hence domain of f(x) is R. |
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| 103. |
The domain of definiton of the function ` f(x) = cot^(-1) {(x)/(sqrt(x^(2)-[x^(2)]))}` isA. `R - {pm sqrtn : n in N }`B. `R - { pm sqrtn : n ge 0, ne Z)`C. `R`D. `R-{0}` |
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Answer» Correct Answer - B We know that `cos^(-1) x` is defined for all `x in R` . Therefore, `cot^(-1){(x)/(sqrt(x^(2) -[x]))}` is defined, if `x^(2) -[x^(2)] gt0` `rArr x^(2) gt [x^(2)]` `rArr x^(2) in R and x^(2) ne 0,1,2,3,....` `rArr x ne R and x ne p, sqrt(n),n=0,1,2,.....` `rArr x in R- {pm sqrt(n) : ngt 0, n in z}` Hence, domina `f = R -{pm sqrt(n) :n gt 0, n in Z}` |
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| 104. |
The function `f(x) = cot^(-1) sqrt(x(x+3)) + cos^(-1) sqrt(x^(2) + 3x +1)` is defined on the set S, where S is equal toA. `{-3,0}`B. `[-3, 0]`C. `[0, 3]`D. `phi` |
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Answer» Correct Answer - A We know that `cot^(-10 x` is defined for all `x in R` and `cos^(-1) x ` is defined for all `in [-1,1]` , Therefore,f(x) is defined if `x(x + 3) ge 0` and `0 le x^(2)+ 3x + 1 le1` `rArr x^(2) + 3x ge 0 and -1le x^(2) + 3x le 0 rArr x = - 3,0` |
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| 105. |
Let g(x) be a function defined on [-1,1]. If the area of the equilateral triangle with two of its vertices at (0,0) and `(x,g(x))` is `(sqrt(3))/(4)` , then the function g(x) , isA. `pm sqrt(1-x^(2))`B. `-sqrt(1-x^(2)) or sqrt(1-x^(2))`C. `sqrt(1-x^(2))` onlyD. `sqrt(1+x^(2))` |
| Answer» Correct Answer - B | |
| 106. |
The domain of definition of the function `f(x)=sin^(-1)((x-3)/(2))-log_(10)(4-x)` , isA. `1 le x le 5`B. `1 lt x lt 4`C. `1 le x le 4`D. `1 le x le 4` |
| Answer» Correct Answer - C | |
| 107. |
The domain of definition of `f(x)=sin^(-1)(|x-1|-2)` isA. `[-2,0] cup [2,4]`B. `(-2,0),cup(2,4)`C. `[-2,0] cup [1,3]`D. `[-2,0] cup [1,3]` |
| Answer» Correct Answer - A | |
| 108. |
The period of the function `f(x)=|sinx|-|cosx|` , isA. `pi//2`B. `pi`C. `2pi`D. none of these |
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Answer» Correct Answer - B We have, `f(x)=|sinx)-|cos x|` `:. f(pi+x)=|sin(pi+x)|-|cos(pi+x)|` `implies f(pi+x)=| -sinx|-|-cos x|` `implies f(pi+x)=|sin x|-|cosx|=f(x)` for all ` x in R ` ` implies f(pi+x)=f(x)` for all ` x in R ` So, f(x) is periodic with period `pi`. |
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| 109. |
Let f be a real valued periodic function defined for all real umbers x such that for some fixed ` a gt 0` , `f(x+a)=(1)/(2)+sqrt(f(x)-{f(x)}^(2))` for all x . Then , the period of f(x) isA. aB. 2aC. 3aD. 4a |
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Answer» Correct Answer - B We have , `f(x+a)=(1)/(2)+sqrt(f(x)-{f(x)}^(2))` `:. f(x+a+a)=(1)/(2)+sqrt(f(x+a)-{f(x+a)}^(2))` `implies f(x+2a)=(1)/(2)+sqrt(f(x+a){1-f(x+a)})` `impliesf(x+2a)=(1)/(2)+ sqrt({(1)/(2)+sqrt(f(x)-{f(x)}^(2))}{(1)/(2)-sqrt(f(x)-{f(x)}^(2))})` `implies f(x+2a)+(1)/(2)+sqrt((1)/(4)-{f(x)-{f(x)}^(2)})` `implies f(x+2a)=(1)/(2)+sqrt((1)/(4)-f(x)+{f(x)}^(2))` `implies f(x+2a)=(1)/(2)+sqrt({f(x)-(1)/(2)}^(2))` `implies f(x+2a)=(1)/(2)+|f(x)-(1)/(2)|" "[ :. sqrt(x^(2))=|x|]` `implies f(x+2a)=(1)/(2)+f(x)-(1)/(2)" " [ :. f(x) ge (1)/(2)]` `implies f(x+2a)=f(x)` for all x . Hence, f(x) is a periodic function with period 2a. |
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| 110. |
Let f(x) be a real valued periodic function with domain R such that `f(x+p)=1+[2-3f(x)+3(f(x))^(2)-(f(x))^(3)]^(1//3)` hold good for all ` x in R ` and some positive constant p, then the periodic of f(x) isA. pB. 3pC. 2pD. `p^(2)` |
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Answer» Correct Answer - C We have , `f(x+p)=1+[2-3f(x)+3(f(x))^(2)-(f(x))^(3)]^(1//3)` `implies f(x+p)=1+[1+{1-f(x)}^(3)]^(1//3)` `implies f(x+p)-1=[1-{f(X)-1}^(3)]^(1//3)` ` implies g(x+p)=[1-{g(x)}^(3)]^(1/3)" "`…..(i) where g(x) =f(x)-1. `implies g(x+2p)=[1-{g(x+p)}^(3)]^(1//3)" "`[ On replacing x by x+p] `implies g(x+2p)=[1-{1-g(x)}^(3)]^(1//3)" "`[Using (i)] `implies g(x+2p)=g(x)` for all ` x in R ` `implies g(x+2p)-1=f(x)-1` `implies f(x+2p)=f(x)` Hence , f(x) is a period function with period 2p. |
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| 111. |
If the function `f: RvecA`given by `f(x)=(x^2)/(x^2+1)`is surjection, then find `Adot`A. RB. [0,1]C. (0,1]D. [0,1) |
| Answer» Correct Answer - D | |
| 112. |
The range of the function `f(x)=(x)/(1+x^(2))` isA. `[0,1//2]`B. `[-1//2,1//2]`C. `[-1//2,0]`D. `[-1//2,0) cup (0,1//2]` |
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Answer» Correct Answer - B Clearly, f(x) is defined for all ` x in R. ` So, domain (f)=R In order to find the range of f(x) , let y=f(x) `implies y=(x)/(x+x^(2)` `implies x^(2)y-x+y=0` `implies x=(1 pm sqrt(1-4y^(2)))/(2y)` For x to be real, we must have `1-4y^(2) ge 0 and y ne 0 implies -(1)/(2) le y le (1)/(2) and y ne 0` Also , y=-0 for x=0 Hence, range of f(x)=[-1/2,1/2] |
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| 113. |
Range of the function `f(x)=(1+x^2)/(x^2)` is equal toA. (0,1)B. [0,1]C. `(1,oo)`D. `[1,oo)` |
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Answer» Correct Answer - C We observe that f(x) is defined for all ` x in R -{-0}` . So, domain (f)=R-{0} Let ` y=(1+x^(2))/(x^(2))`. Then, `y=(1+x^(2))/(x^(2))implies pm sqrt((1)/(y-1))` for x to be real, we must have `y-1 gt 0 implies ygt 1implies y in (1,oo)` Hence, range `(f)=(1,oo)` |
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| 114. |
Range of the function `f(x)=(x^2-3x+2)/(x^2+x-6)` isA. `R-[1//5,1] `B. `R`C. `R-{1}`D. none of these |
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Answer» Correct Answer - C We observe that f(x) is defined for ` x^(2)+x-6 ne 0`, i.e., ` x ne -3,2` `:. ` Domain (f)=R-{-3,2} Let `y=(x^(2)-3x+2)/(x^(2)+x-6) =(x-1)/(x+3)implies s=(3y+1)/(y-1)` Clearly , x is real for ` y-1 ne 0, i.e., y ne 1`. Hence, range (f) =R-{1}` |
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| 115. |
Range of the function `f(x)=log_esqrt(4-x^2)` isA. `(0,oo)`B. `(-oo,oo)`C. `(-oo, log_(e) 2]`D. `(log_(e)2,oo)` |
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Answer» Correct Answer - C Clearly , `f(x)=log_(e)sqrt(4-x^(2))` is defined for all ` x in (-2,2)` Let y =f(x). Then, `y=logesqrt(4-x^(2)) implies e^(2y)=4-x^(2)implies x=sqrt(4-e^(2y))` Clearly , x is real , if `4-e^(2y) ge 0 implies 0 lt e^(y) le 2 " "[ :. e^(y) gt 0]` `implies -oo lt 6 le log_(e)2implies y in (-oo, log_(e)2]` |
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| 116. |
If `f(x)=cos[pi^2]x ,`where `[x]`stands for the greatest integer function, then`f(pi/2)=-1`(b) `f(pi)=1``f(-pi)=0`(d) `f(pi/4)=1`A. `f((pi)/(2))=-1`B. `f(pi)=1`C. `f(-pi)=-1`D. `f((pi)/(4))=2` |
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Answer» Correct Answer - A We have , `f(x) =cos[pi^(2)]x+cos[-pi^(2)]x` `implies f(x)=cos9x + cos (-10x)` `implies f(x)=cos9x+ cos 10 x` `:. f((pi)/(2)) =cos""(9x)/(2)+ cos 5 pi =0+(-1)^(5)=-1` `f(pi) = cos 9 pi + cos 10 pi=(-1)^(9) +(-1)^(10)=-1+1=0` `f(-pi)= cos (-9pi)+ cos (-10pi) = cos 9 pi + cos 10 pi=0` ` f((pi)/(4))= cos (9pi)/(4)+ cos (10 pi)/(4)=(1)/(sqrt(2))+0=(1)/(sqrt(2))` Hence , option (a) is correct. |
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| 117. |
If `f : R -> R` is defined by `f(x) = 1 /(2-cos3x)` for each `x in R` then the range of `f` isA. `[-1//3,0]`B. RC. [1/3,1]D. none of these |
| Answer» Correct Answer - C | |
| 118. |
If `f: R to R and g: R to R `are defined by `f(x)=2x+3 and g(x)=x^(2)+7` , then the values of x such that `g(f(x))=8` areA. 1,2B. `-1,2`C. `-1,-2`D. `1,-2` |
| Answer» Correct Answer - C | |
| 119. |
The domain of definition of the function `f(x)=(1)/(sqrt(|x|-x))` isA. RB. `(0,oo)`C. `(-oo,0)`D. none of these |
| Answer» Correct Answer - C | |
| 120. |
The set of values of x for which the function `f(x)=(1)/(x)+2^(sin^(-1)x)+(1)/(sqrt(x-2))` exists isA. RB. R-{0}C. `phi`D. none of these |
| Answer» Correct Answer - C | |
| 121. |
`f(x)=sqrt(sin^(- 1)(log_2x))`A. `x in (1,2)`B. `s in [1,2]`C. `x in [2,oo)`D. `x in (0,oo)` |
| Answer» Correct Answer - B | |
| 122. |
The domain of the definition of the function `f(x)=sqrt(1-sqrt(1-sqrt(1-x^(2))))` , isA. `(-oo,1)`B. `(-1,oo)`C. `[0,1]`D. `[-1,1] |
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Answer» Correct Answer - D Clearly, f(x) is defined , if `1-sqrt(1-sqrt(1-x^(2))) ge 0, 1-sqrt(1-x^(2)) ge 0 and 1-x^(2), ge 0` `implies 1 ge 1-sqrt(1-x^(2)), 1 ge 1-s^(2) and x^(2)-1 le 0` `implies sqrt(1-x^(2)) ge 0, x^(2) ge 0 and x^(2) -1 le 0` `implies x^(2)-1 le 0` `implies -1 le x le 1` `implies x in [-1,1]` |
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| 123. |
Let `f: R to R ` be given by `f(x+(5)/(6))+f(x)=f(x+(1)/(2))+f(x+(1)/(3))` for all ` x in R ` . Then ,A. f(x) is periodicB. f(x) is evenC. `f(x+2)-f(x+1)=f(x+1)-f(x)`D. none of these |
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Answer» Correct Answer - C we have, `f(x+(5)/(6))+f(x)=f(x+(1)/(2))+f(x+(1)/(3))` for all ` x in R ` Clearly , `(x+(5)/(6))+x=(x+(1)/(2))+(x+(1)/(3))` Thus, f(x) satisfies the property f(u)+f(v)=f(a)+f(b) if u+v=a+b We observe that option (c ) satisfied this property . |
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| 124. |
Let f be a real valued function satisfying `f(x+y)=f(x)f(y) ` for all `x, y in R ` such that f(1)=2 . If ` sum_(k=1)^(n)f(a+k)=16(2^(n)-1) `, then a=A. 3B. 4C. 2D. none of these |
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Answer» Correct Answer - A From illustration 11 , we have `f(x)=[f(1)]^(x)=2^(x) ` for all ` x in R ` `: . sum_(k=1)^(n)f(a+k)=16(2^(n)-1)` `implies sum_(k=1)^(n)2^(a+k)=16(2^(n)-1)` `implies 2^(a) sum_(k=1)^(n)2^(k)=16(2^(n)-1)` `implies 2^(a)xx2 ((2^(n)-1)/(2-1))=16(2^(n)-1)implies 2^(a+1)=2^(4)implies a=3` |
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| 125. |
Let f be a real valued function satisfying ` f(x+y)=f(x)f(y)` for all ` x, y in R ` such that `f(1)=2 `. Then , `sum_(k=1)^(n) f(k)=`A. `2^(n+1)-2`B. `2^(n+1)-1`C. `2^(n)-1)`D. `2^(n)-2` |
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Answer» Correct Answer - A We have , `f(x)=[f(1)]^(x)=2^(x)` for all ` x, y in R . ` `:. sum_(k=1)^(n)f(k)=sum_(k=1)^(n)2^(k)=2((2^(n)-1)/(2-1))=2x^(n+1)-2` |
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| 126. |
Let f be a real valued function satisfying `f(x+y)=f(x)+f(y)` for all ` x, y in R and f(1)=2`. Then `sum_(k=1)^(n)f(k)=`A. `(n(n+1))/(2)`B. `n(n+1))`C. `(n+1)`D. `n` |
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Answer» Correct Answer - B We have, `f(x)=xf(1)=2x ` for all `x in R `. `:. Sum_(k=1)^(n)f(k)=sum_(k=1)^(n) 2k=2 sum_(k=1)^(n)k=n(n+1)` |
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| 127. |
For ` x in R , x ne0, 1, ` let `f_(0)(x)=(1)/(1-x) and f_(n+1)(x)=f_(0)(f_(n)(x)),n=0,1,2…..` Then the value of `f_(100)(3)+f_(1)((2)/(3))+f_(2)((3)/(2))` is equal toA. `(4)/(3)`B. `(1)/(3)`C. `(5)/(3)`D. `(8)/(3)` |
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Answer» Correct Answer - C We have , `f_(0)(x)=(1)/(1-x) and f_(n+1)(x)=f_(0)(f_(n)(x))`. `:. f_(1)(x)=f_(0)(f_(0)(x))=f_(0)((1)/(1-x))=(1)/(1-(1)/(1-x))=(x-1)/(x)` `and , f_(2)(x)=f_(0)(f_(1)(x))=f_(0)((x-1)/(x))=(1)/(1-(x-1)/(x))=x` Thus, ` f_(3)(x)=f_(0)(f_(2)(x))=f_(0)(x)` `f_(4)(x)=f_(0)(f_(3)(x))=f_(0)(f_(0)(x))=f_(1)(x)` `f_(5)(x)=f_(0)(f_(4)(x))=f_(0)(f_(1)(x))=f_(2)(x)=x` `f_(6)(x)=f_(0)(f_(5)(x))=f_(0)(f_(2)(x))=f_(0)(x)` and so on. In general `f_(3n)(x)=f_(0)(x)` ` f_(3n_+1)(x)=f_(1)(x) and f_(3n+2)(x)=f_(2)(x)=x ` for ` n=1,2,` `:. f_(100)(x)=f_(1)(x)=(x-1)/(x)`. Hence, `f_(100)(x)+f_(1)((2)/(3))+f_(2)((3)/(2))=f_(1)(3)+f_(1)((2)/(3))+f_(2)((3)/(2))` `=(30-1)/(3)+((2)/(3)-1))/((2)/(3))+(3)/(2)=(2)/(3)-(1)/(2)+(3)/(2)=(5)/(3)` |
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| 128. |
If `f(x)=(x)/(x-1)`,then`underset( 19 "times")( (fofofo…of))` (x) is equal toA. `(x)/(x-1)`B. `((x)/(x-1))^(19)`C. `(19x)/(x-1)`D. `x` |
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Answer» Correct Answer - A We have, `f(x)=(x)/(x-1)` `:. Fof(x)=f(f(x))=f((x)/(x-1))=((x)/(x-1))/((x)/(x-1)-1)=x` `implies fofof(x)=f(x)=(x)/(x-1)` ` :. underset(19" times ")((fofofo ...of))(x)=f(x)=(x)/(x-1)` |
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| 129. |
If `f(x)+2f((1)/(x))=3x,x ne 0`, and `S={x in R : f (-x)},` then S:A. is an empty setB. contains exactly one elementC. Contains exactly two elements .D. contains more than two elements |
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Answer» Correct Answer - C We have, `f(x)+2f((1)/(x))=3x" " `….(i) Replacing x by `(1)/(x)`, we get `f((1)/(x))+2f(x)=(3)/(x)" " `….(ii) Solving (i) and (ii) , we obtain `f(x)=3((2)/(x)-x)` `:. f(x)=f(-x)` `implies 3((2)/(x)-x)=3(-(2)/(x)+x)` `implies (4)/(x)-2x=0implies 2x^(2)-4=0implies x=pm sqrt(2)` Hence , `S={x in R: f(x)=f(-x)}={-sqrt(2), sqrt(2)}` |
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| 130. |
If `3f(x)-f((1)/(x))= log_(e) x^(4)` for `x gt 0` ,then `f(e^(x))=`A. xB. `log_(e)x `C. `e^(x)`D. none of these |
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Answer» Correct Answer - A We have , `3f(x)-f((1)/(x))=log_(e) x^(4)` `implies 3f(x)-f((1)/(x))=4 log_(e)x " " ` ….(i) `implies 3f((1)/(x))-f(x)=4 log_(e)((1)/(x))" "`[Replacing x by 1/x] `implies 3f((1)/(x))-f(x)=-4 log_(e)x" "`…..(ii) Solving (i) and (ii) , we get `8f(x)=8log_(e)x implies f(x)=log_(e)ximplies f(e^(x))=x` |
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| 131. |
If [x] denotes the greatest integer less than or equal to ` x and n in N , ` then ` f(X)= nx+n-[nx+n]+tan""(pix)/(2)` , isA. a periodic function with period 1B. a periodic function with period 4 .C. not periodicD. a periodic function with period 2. |
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Answer» Correct Answer - D For ` n in N`, we find that `nx+n-[nx+n]=nx-[nx]` is a periodic with period `(1)/(n)`, because x-[x] is periodic with period 1. Also , ` tan""(pix)/(2)` is periodic with period `(pi)/(pi//2)=2` Hence , f(x) is periodic with period 2, i.e., the LCM of `(1)/(n)` and 2 . |
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| 132. |
Let `f(x)=(sin 2n x)/(1+cos^(2)nx), n in N ` has `(pi)/(6)` as its fundamental period , then n=A. 2B. 4C. 6D. none of these |
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Answer» Correct Answer - C We have, `f(x)=( sin 2n x)/(1+ cos^(2)nx)implies f(x)=(2 sin 2nx)/(3+cos 2nx)` Clearly , period of `f(x)` is ` (2pi)/(2n)=(pi)/(n)` `:. (pi)/(n)=(pi)/(6)implies n=6` |
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| 133. |
Which of the following functions is an odd functions ?A. `f(x)`= constB. f(x)=sinx+ cosxC. `f(x)=sin{log_(10)(x+sqrt(x^(2)+1))}`D. `f(x)=1+x+2x^(2)` |
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Answer» Correct Answer - C Clearly , g(x)= sin x and `h(x)=log(x+sqrt(x^(2)+1))` both the odd functions. Therefore , f(x)=goh(x) is also an odd functions. |
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| 134. |
Which of the following functions is an odd functions ?A. `f(x)=sqrt(1+x+x^(2))-sqrt(1-x+x^(2))`B. `f(x)=x((a^(x)+1)/(a^(x)-1))`C. `f(x)=log_(10)((1-x^(2))/(1+x^(2)))`D. f(x)=k (constant ) |
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Answer» Correct Answer - A If ` f(x)=sqrt(1+x+x^(2))-sqrt(1-x+x^(2))`, then `f(-x)=sqrt(1-x+x^(2))-sqrt(1+x+x^(2))` `implies f(-x)=-f(x)` So, f (x) is an odd functions Thus, option (a) is correct . If `f(x)=x((a^(x)+1)/(a^(x)-1))`, then ` f(-x)=x((a^(-x)+1)/(a^(-x)-1))=-x((1+^(x))/(1-a^(x)))=x((a^(x)+1)/(a^(x)-1))-f(x)` So, f(x) is an even function. Thus , option (b) is not correct. If `f(x)=log_(10)((1-x^(2))/(1+x^(2)))`, then `f(-x)=log_(10)((1-x^(2))/(1+x^(2)))=f(x)` So, f(x) is an even function. Thus, option (c ) is not correct If f(x)=k for all `x ` then `f(-x)=f(x) ` for all `ximplies f(x)` is an even functions. |
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| 135. |
Statement-1: The function f(x) given by `f(x)=sin^(-1){log(x+sqrt(x^(2)+1))}` is an odd function. Statement:2 The composition of two odd functions is an odd function.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. |
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Answer» Correct Answer - A Let `phi(x) and Psi(x)` be two odd functions. Then, `(phi 0 Psi)(-x)=phi(Psi(-x))=phi(-Psi(x))=-phi(Psi(x))=-phi0Psi(x)` `implies phi 0 Psi` is an odd function. So,statement-2 is true. If `phi(x)=sin^(-1)x and Psi(x)=log(x+sqrt(x^(2)+1))`. Then, `phi` and `Psi` are odd functions such that `phi 0 Psi=f`. Hence, f is an odd function. so, statement-1 is true. Also, statement-2 is a correct explanation for statement-1 . |
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| 136. |
Let `f(x)= cos^(-1)((x^(2))/(x^(2)+1))`. Then , the range of the f , isA. `(0, pi//2]`B. `[-pi//2, pi//2]`C. `[-pi//2, 0]`D. none of these |
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Answer» Correct Answer - A Since `0 le (x^(2))/(x^(2) + 1) lt 1` for all `x in R`. Therefore `0 lt cos^(-1) ((x^(2))/(x^(2)+1)) lt (pi)/(2)` ` 0 lt f(x) le (pi)/(2) rArr` Range of `f(x) = (0,pi//2)]` |
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| 137. |
Let X be the set of all positive such that `f(x+y) = f(xy)` for all `x ge 4, y ge 4`. If `f(8)= 9`, then f(9) is equal to.A. 8B. 9C. 81D. 64 |
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Answer» Correct Answer - B We have, `f(x+y)=f(xy)` for all x,y` ge 4` `therefore f(9) = f(5+4) = f(5xx4) = f(20) = f(16+4) = f(16 xx 4)` `= f(8xx8) = f(8+8) = f((16) = f(4xx4) = f(4+4) = f(8) = 9` |
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| 138. |
The range of the function ` f(x) = (1)/(2-sin 3x)` isA. `(1//3, 1 )`B. `[1//3, 1 )`C. `[1//3, 1]`D. none of these |
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Answer» Correct Answer - C Cleary, domain of `f(x)` is R and `f(x) lt 0` for all `x in R` Let `y = f(x)`. Them `y = (1)/(2-sin 3x)` `rArr sin 3x = (2y-1)/(y) rArr x = (1)/(3) sin^(-1)((2y-1)/(y))` For x to be real, we must have `rArr -1 le (2y-1)/(y) le 1` `rArr - y le 2y -1 le y" "[beacuse y = f(x) gt 0]` ` rArr -y le 2y - 1 and 2y - 1 le y` `rArr y gt (1)/(3) and y le 1 rArr y in [1//3,1]` Hence, range `(f) = [1//3,1]` |
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| 139. |
The domain of the function `f(x)=(sin^(-1)(x-3))/(sqrt(9-x^(2)))`, isA. `[1,2)`B. `[2,3)`C. `[1,2]`D. `[2,3]` |
| Answer» Correct Answer - B | |
| 140. |
Which of the following functions has period `pi` ?A. `|-tanx|+cos 2x `B. `2 sin""(pix)/(3)+ 3 cos ""(2pix)/(3)`C. `6 cos (2pix+(pi)/(4))+5 sin (pi x+(3pi)/(4))`D. `| tan 2x|+| sin 4x|` |
| Answer» Correct Answer - A | |
| 141. |
If `f : R -> R` are defined by `f(x) = x - [x]` and `g(x) = [x]` for `x in R`, where [x] is the greatest integer not ex-ceeding x, then for every` x in R, f(g(x))=`A. xB. 0C. f(x)D. g(x) |
| Answer» Correct Answer - B | |
| 142. |
Let `f(x)=min{x,x^2}`, for every `x in R`. ThenA. `f(x)={{:(x"," x ge1),(x^(2)","0 lex lt1),(x" ," x lt 0):}`B. `f(x)={{:(x^(2)","x ge1),(x"," x lt1):}`C. `f(x)={{:(x","x ge1),(x^(2)"," x lt1):}`D. `f(x)={{:(x^(2)","x ge1),(x"," 0 lex lt1),(x^(2)"," x lt0):}` |
| Answer» Correct Answer - A | |
| 143. |
A polynomial function f(x) satisfies the condition `f(x)f(1/x)=f(x)+f(1/x)` for all `x inR`,`x!=0`. If f(3)=-26, then f(4)=A. `-35`B. `-63`C. 65D. none of these |
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Answer» Correct Answer - B We know that a polynomial function f(x) of degree n satisfying. `f(x)f((1)/(x))= f((1)/(x)) ` for all `x(ne0) in R`is of the form `f(x) = 1 pm x^(n) ` for all `x (ne 0) in R`. We are given that f(3) =- 26. `therefore f(x) = 1 - x^(n)" ".....(i)` `rArr f(3) = 1- 3^(n)` `rArr -26= 1-3^(n)" "[becausef (3) = - 26]` `rArr 3^(n) = 27 rArr 3^(n) = 3^(3) rArr n = 3` Substituting n = 3 in (i), we get `f(x) = 1-x^(3) rArr f(4) = 1-4^(3) = - 63` |
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| 144. |
If `f: R to R, g , R to R` be two funcitons, and `h(x) = 2 "min" {f(x) - g(x),0}` then `h(x)=`A. `f(x) + g(x) -|g(x)-f(x)|`B. `f(x) + g(x) +|g(x) - f(x)|`C. `f(x)-g(x)+|g(x) +|g(x) - f(x)|`D. `f(x) -g(x) -|g(x) - f(x)|` |
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Answer» Correct Answer - D Following cases arise. CASE-1 When `f(x) ge g(x)` In this case, we have `f(x)-g(x) ge 0` `therefore " ""min" {f(x) - g(x), 0 } = 0` `rArr h(x) = 0` `rArr h(x) = {f(x) - g(x)}-{f(x) -g(x)}` `rArr h(x) = {f (x) - g(x)} - |f (x) - g(x)|` `[{:(,because f(x) - g(x) ge 0),(,therefore|f(x) - g(x)|=f(x) - g(x)):}]` CASE II When `f(x) lt g(x)` In this case, we have `f(x) - g(x) lt 0` `therefore "min" {f(x) - g(x),0}=f(x) - g(x)` `rArr h(x) = 2 min {f(x) - g(x),0}=2 {f(x) - g(x)}` `rArr h (x) = {f(x) - g(x)} + {f(x) - g(x)}` `rArr h(x) = {f(x) - g(x)}-| f(x) - g(x)|`[{:(, therefore f(x) -g(x) lt 0),(,therefore |f(x) - g(x)|),(,=-{f(x) -g(x)}):}]` Hence, option (d) is correct . |
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| 145. |
If `f(x) = cos [pi]x + cos [pi x]`, where `[y]` is the greatest integer function of y then `f(pi/2)` is equal toA. `cos3`B. 0C. `cos 4`D. none of these |
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Answer» Correct Answer - C We have, `f (x) = cos [pi] x + cos[pix]` `f((pi)/(2)) = cos([pi].(pi)/(2)] + cos[(pi^(2))/(2)]= cos.(3pi)/(2)+ cos 4 = cos4` |
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| 146. |
Let ` f : R to R` be a function given by `f(x+y) = f(x) + f(y) ` for all x,y `in`R such that `f(1)= a` Then, `f (x)=`A. `a^(x)`B. `ax`C. `a^(x)`D. none of these |
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Answer» Correct Answer - A We know that the general expression for the function satsifying `f(x+y) = f(x) f(y) ` for all x,y `in R," "[because f(1) =a]` |
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| 147. |
If `f(x) = (x-1)/(x+1),` then `f(alpha, x)=`A. `(f(x)+alpha)/(1+alpha f(x))`B. `((alpha-1)f(x) + alpha +1)/((alpha+1)f(x)+(alpha-1))`C. `((alpha+1)f(x) +alpha-1)/((alpha-1) f(x) + (alpha+1))`D. none of these |
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Answer» Correct Answer - c We have `f(x) =(x-1)/(x+1)rArr (f(x) +1)/(f(x) -1) = (2x)/(-2) rArr x = (1+f(x))/(1-f(x))` `therefore f (alphax) = (alphax -1)/(ax + 1) =(alpha {(1+f(x))/(1-f(x))}-1)/(alpha{(1+f(x))/(1-f(x))} + 1)` `rArr f(alpha,x) =((alpha +1)f (x) + alpha -1)/((alpha -1)f(x) + alpha f(x) + alpha +1)` |
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| 148. |
Let ` f : R to R` be a function given by `f(x+y) = f(x) + f(y) ` for all x,y `in`R such that `f(1)= a` Then, `f (x)=`A. `a^(x)`B. `ax`C. `a^(x)`D. `a+x` |
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Answer» Correct Answer - B We know that the general formula for the function `f: R to R ` given by `f (x)f(y)` for all x,y `in`R `f(x) = xf(1) rArr f (x) =ax" "[because f(1) =a]` |
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| 149. |
If `f(x)= cos [(pi^(2))/(2)] x + sin[(-pi^(2))/(2)]x,[x]` denoting the greatest integer function,thenA. `f(0) = 0`B. `f((pi)/(3)) = sqrt((3)-1)/(2)`C. `f((pi)/(2)) = -1`D. `f(pi) = 0` |
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Answer» Correct Answer - B We have , `pi ^(2) ~= 9.8696 `therefore [(pi^(2))/(2)] = 4 and [(-pi^(2))/(2)] = -5` `therefore f(x) = cos 4x sin (-5x) = cos 4 x -sin 5x ` Thus, we have, `f(0) =1` `f((pi)/(3)) = cos.(4pi)/(3)-sin .(-5x)/(3) = - (1)/(2) +(sqrt(3))/(2)= (sqrt(3)-1)/(2)` `f((pi)/(2)) = cos 2 pi - sin.(5pi)/(2) = 1-1=0` and ,`f(pi)= cos 4 pi - sin5 pi = 1` |
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| 150. |
If `f (x) = 27x^(3) -(1)/(x^(3))` and `alpha, beta` are roots of `3x - (1)/(x) = 2` thenA. `f (alpha) = f(beta)`B. `f(alpha) =10`C. `f (beta) = -10`D. none of these |
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Answer» Correct Answer - A `f(x) =27 x^(3) -(1)/(x^(3)) = (3x-(1)/(x))^(3) + 9(3x-(1)/(x))` Since `alpha ` and `beta` are the roots of `3x -(1)/(x) =2` `therefore 3 alpha - (1)/(alpha) = 2` and `3beta -(1)/(beta) =2` Now, `f(alpha) = (3alpha -(1)/(3))^(3) + 9 (3alpha -(1)/(alpha))=2^(3) + 9xx 2 = 26` Similarly, we have `f (beta) = 26` |
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