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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. |
Which of the following function is (are) even, odd, or neither?`f(x)=x^2sinx``f(x)=sqrt(1+x+x^2)-sqrt(1-x+x^2)``f(x)=log((1-x)/(1+x))``f(x)=log(x+sqrt(1+x^2))``f(x)=sinx-cosx``f(x)=(e^x+e^(-x))/2` |
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Answer» (i) `f(-x)=(-x)^(2) sin(-x)= -x^(2)sinx= -f(x)` Hence, f(x) is odd. (ii) `f(-x)=sqrt(1+(-x)+(-x)^(2))-sqrt(1-(-x)+(-x)^(2)) ` `=sqrt(1-x+x^(2))-sqrt(1+x+x^(2))` `= -f(x)` Hence, f(x) is odd. (iii) `f(-x)=log{(1-(-x))/(1+(-x))}` ` log((1+x)/(1-x))` `= -f(x)` Hence, f(x) is odd. (iv) `f(-x)=log(-x+sqrt(1+(-x)^(2)))` `=log{((-x+sqrt(1+x^(2)))(x+sqrt(1+x^(2))))/((x+sqrt(1+x^(2))))}` `log((1)/(x+sqrt(1+x^(2))))= -f(x)` Hence, f(x) is odd. (v) `f(-x)=sin(-x)-cos(-x)= -sinx-cosx` Clearly, `f(-x) ne f(x) " and " f(-x) ne -f(x).` Hence, f(x) is neither even nor odd. (vi) `f(-x)=(e^(-x)+e^(-(-x)))/(2)=(e^(-x)+e^(x))/(2)=f(x)` Hence, f(x) is even. |
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| 102. |
Let `**` be a binary operation on the set Q of rational numbers as follows: (i) `a**b=a-b`(ii) `a**b=a^2+b^2`(iii) `a**b=a+a b`(iv) `a**b=(a-b)^2`(v) `a*b=(a b)/4`(vi) `a**b=a b^2`. Find which of the binary operations are commutative and which are associative. |
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Answer» Correct Answer - (ii),(iv),(v) are commutative (v) is associative (v) has identify element |
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| 103. |
If `f: R to R ` is defined as f(x)=2x+5 and it is invertible , then `f^(-1) (x)` isA. `(x-5)/(2)`B. `(x-2)/(5)`C. `(x+5)/(2)`D. none of these |
| Answer» Correct Answer - A | |
| 104. |
{ . } denotes the fractional part function and [.] denotes the greatest integer function. Now, match the following lists: |
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Answer» Correct Answer - `a to s; b to r; c to s; d to p` a. `f(x)=e^(cos^(4)pix+x-[x]+cos^(2)px)` `cos^(2)pi x " and " cos^(4)pi x` have period 1. `x-[x]={x}` has period 1. Then the period of `f(x)` is 1 b. `f(x)=cos 2pi {2x}+sin 2pi{2x}` The period of {2x} is 1/2. Then the period of `f(x)` is 1/2. c. Clearly, `tan pi[x]=0 AA x in R` and the period of `sin 3pi{x}` is equal 1. d. `f(x)=3x-[3x+a]-b=3x+a-[3x+a]-(a+b)` `={3x+a}-(a+b)` Thus, the period of `f(x)` is 1/3. |
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| 105. |
State whether the following statements are true or false. Justify.(i) For an arbitrary binary operation `∗` on a set N, `a ∗ a = a ∀a ∈ N`.(ii) If `∗` is a commutative binary operation on `N`, then `a ∗ (b ∗ c) = (c ∗ b) ∗ a` |
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Answer» (i) false (ii) true |
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| 106. |
Is `*`defined on the set `{1, 2, 3, 4, 5} b y a * b = LdotCdotMdot`of a and b a binary operation? Justify your answer.A. 6B. 24C. 36D. none of these |
| Answer» Correct Answer - C | |
| 107. |
If `**` defined on the set `{1, 2, 3, 4, 5}` by `a**b = LCM`of `a` and `b` a binary operation? Justify your answer. |
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Answer» Given set A = { 1,2,3,4,5} Now `2,3 in A` and 2*3 = L.C.M of 2 and 3 = `6 notinA` `therefore` In the give set, the operation *is not binary. |
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| 108. |
Find the domain of function`f(x)=(log)_4[(log)_5{(log)_3(18 x-x^2-77}]` |
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Answer» f(x) is defined if `log_(5){log_(3)(18-x^(2)-77)} gt 0 " and " 18x-x^(2)-77 gt 0` or `log_(3)(18x-x^(2)-77) gt 5^(0) " and " x^(2)-18x+77 lt 0` or `log_(3)(18x-x^(2)-77) gt 1 " and " (x-11)(x-7) lt 0` or `18x-x^(2) -77 gt 3^(1) " and " 7 lt x lt 11` or `18x -x^(2)-80 gt 0 " and " 7 lt x lt 11` or `x^(2) -18x +80 lt 0 " and " 7 lt x lt 11` or `(x-10)(x-8) lt 0 " and " 7 lt x lt 11` or `8 lt x lt 10 " and " 7 lt x lt 11` or `8 lt x lt 10` or `x in (8,10)` Hence, the domain of f(x) is `(8,10)`. |
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| 109. |
Write the function `f(x) ={sinx}` where {.} denotes the fractional part function) in piecewise definition. |
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Answer» We have `f(x)={sinx}` Clearly, `{sinx}=0` if sin x is integer. ` :. {sinx}=0 " when " x=(n pi)/(2)` Now when ` sin x in (0,1)` ` {sin x} =sin x-[sin x]=sin x -0=sinx` When ` sinx in (-1,0),` ` {sin x} =sin x-[sin x]=sin x -(-1)=sinx+1` Thus, `f(x)={(0","sinx in{-1,0,1}),(sinx","sinx in (0,1)),(sinx+1","sinx in (-1,0)):}` `f(x)={(0",",x=(n pi)/(2)","n in Z),(sinx",",x in underset(n in Z)(cup)(2n pi","(2n+1)pi)),(sinx+1",",x in underset(n in Z)(cup)"((2n+1)"pi","(2n+2)pi")"):}` |
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| 110. |
State whether the following statements are true orfalse. Justify.(i) For an arbitrary binary operation `*`on a set `N ," "a" "*" "a" "=" "a""""""AA""""""""a in N`.(ii) If `*`is a commutative binary operation on N, then`a" "*" "(b" "*" "c)" |
| Answer» Correct Answer - (i) False (ii) False | |
| 111. |
Find the domain of`f(x)(sqrt((1-sinx)))/((log)_5(1-4)^2)+cos^(-1)(1-{x})dot` |
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Answer» Correct Answer - `(-1//2,1//2)-{0}` (a) `1-sinx ge 0 " or " sin x le 1 " or " x in R` (b) ` 1-4x^(2) gt 0 " or " x in (-1//2,1//2)` (c ) `log_(5)(1-4x^(2)) ne 0 " or " 1-4x^(2) ne 1 " or " x ne 0` (d) `-1 le 1-{x} le 1 " or " 0 le {x} le 2 " or " x in R` Hence, domain is common values of (a),(b),(c) and (d), i.e., `x in (-(1)/(2),(1)/(2))-{0}` |
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| 112. |
If`f(x)={xcosx+(log)_e((1-x)/(1+x))a; x=0; x!=0`is odd, then `a_______,` |
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Answer» For `x ne 0, f(x)=x cosx + log_(e)((1-x)/(1+x))` ` :. f(-x)=(-x)cos(-x)+log_(e)((1+x)/(1-x))` `= -x cosx-log_(e)((1-x)/(1+x))` `= -f(x)` For `x=0`, we must have `f(-x)= -f(x)` ` :. a= -a` ` :. a=0` |
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| 113. |
Find the domain and range of `f(x)=sin^(-1)[x]w h e r[]`represents the greatest function). |
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Answer» `f(x)=sin^(-1) [x]` is defined if `-1 le [x] le1` `implies [x]= -1,0,1` `implies x in [-1,2)` Also, range is `{sin^(-1)(-1),sin^(-1)0,sin^(-1)1} -= {-pi//2,0,pi//2}` |
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| 114. |
The domain of the function `f(x)=1/(sqrt({sinx}+{sin(pi+x)}))`where `{dot}`denotes the fractional part, is`[0,pi]`(b) `(2n+1)pi/2, n in Z``(0,pi)`(d) none of these |
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Answer» `f(x)=(1)/(sqrt({sinx}+{sin(pi+x)}))=(1)/(sqrt({sinx}+{-sinx}))` Now, `{sinx} +{-sinx}={(0",",sinx " is an integer"),(1",", sinx " is not an integer"):}` For f(x) to get defined, `{sinx} +{-sin} ne 0` or ` sinx ne ` integer or `sinx ne +-1,0` or ` x ne (n pi)/(2), n in I` |
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| 115. |
The identity element for the binary operation `**` defined on Q - {0} as `a ** b=(ab)/(2), AA a, b in Q - {0}` isA. 1B. 0C. 2D. None of these |
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Answer» Correct Answer - C Given that, `a ** b = (ab)/(2), AA a,b in Q - {0}.` Let e be the identity element for `**`. ` :. A ** e = (ae)/(2)` `implies a=(ae)/(2)implies e=2` |
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| 116. |
Find the domain of `f(x)=sqrt((log)_(0. 4)((x-1)/(x+5)))` |
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Answer» `f(x)=sqrt(log_(0.4)((x-1)/(x+5)))` exists if `log_(0.4)((x-1)/(x+5)) ge 0 " and " ((x-1)/(x+5)) gt 0` or `(x-1)/(x+5) le (0.4)^(0) " and " (x-1)/(x+5) gt 0` or `(x-1)/(x+5) le1 " and " (x-1)/(x+5) gt 0` or `(x-1)/(x+5) -1 le 0 " and " (x-1)/(x+5) gt 0` or `(-6)/(x+5) le 0 " and " (x-1)/(x+5) gt 0` or `x+5 gt 0 " and " (x-1)/(x+5) gt 0` or `x gt -5 " and " x-1 gt 0 " " ("As " x+5 gt 0)` or `x gt -5 " and " x gt 1 ` Thus, the domain f(x) is `(1,oo).` |
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| 117. |
If `f(x)=sin(log)_e{(sqrt(4-x^2))/(1-x)}`, then the domain of `f(x)`is _____ and its range is __________. |
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Answer» We have `f(x)=sin(log_(e)((sqrt(4-x^(2)))/(1-x)))` We must have `4-x^(2) gt0 " and " 1-x-0` `implies x^(2) lt 4 " and " x lt 1` `implies -2 lt x lt 2 " and " x lt 1` `implies -2 lt x lt 1` Thus, domain of f(x) is `(-2,1)` When x approaches to 1 from its left-hand side `(sqrt(4-x^(2)))/(1-x)` approaches to infinity. When x approaches to -2 from its right-hand side `(sqrt(4-x^(2)))/(1-x)` approaches to zero. Also, `(sqrt(4-x^(2)))/(1-x)` exists continuously for ` x in (-2,1)`. Thus `0 lt (sqrt(4-x^(2)))/(1-x) lt oo` `implies -oo lt "log"_(e) (sqrt(4-x^(2)))/(1-x) lt oo` `implies sin("log"_(e)(sqrt(4-x^(2)))/(1-x)) in [-1 ,1]` |
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| 118. |
Find the domain of the function :`f(x)=1/(sqrt((log)_(1/2)(x^2-7x+13)))` |
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Answer» Correct Answer - `(3,4)` `f(x)=(1)/(sqrt(log_(1//2)(x^(2)-7x+13)))` exists if `log_(1//2)(x^(2)-7x+13) gt 0` or `x^(2)-7x+13 lt 1 " (1) " ` and `x^(2)-7x+13 gt 0 " (2)" ` or `x^(2)-7x+12 lt 0 " and " (x-(7)/(2))^(2)+(3)/(4) gt 0` or `3 lt x lt 4 " and " x in R` or ` 3 lt x lt 4` |
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| 119. |
Find the domain and range of `f(f)=log{x},w h e r e{}`represents the fractional part function). |
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Answer» We know that `o le {x} lt 1 AA x in R.` Now, when `{x} =0, log{x}` is not defined. So, x cannot be an integer. Hence, the domain is `R-I`. Now, for `0 lt {x} lt 1, -oo lt log {x} lt 0. ` So, range is `(-oo,0).` |
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| 120. |
find the value of: `intsqrt(1-sin2x)`dx |
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Answer» `I = int sqrt(1-sin2x) dx` `=>I = int sqrt(cos^2x+sin^2x-2sinxcosx) dx` `=>I = int sqrt((cosx-sinx)^2) dx` `=>I = int (cosx-sinx) dx` `=> I = sinx + cosx +c` |
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| 121. |
Give examples of two functions `f: N->Z" and "g: Z->Z`such that gof is injective but g is not injective. (Hint: Consider `f(x) = x" and "g(x) = |x|`) |
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Answer» A function `f(x)` is injective when, `f(x_1) = f(x_2)` only if `x_1 = x_2`. Let `f(x) = x and g(x) = |x|`. Here, `g(1) =g(-1) = 1`. So, `g(x)` is not injective. Now, `gof = gof(x) = g(f(x)) = g(x) = |x|` But, `f: N->Z`. It means `x` can not be negative, so, `gof(x_1) = gof(x_2)` only when `x_1 = x_2`. `:. gof` is injective. |
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| 122. |
The domain of definition of the function `f(x)`given by the equation `2^y=2`is`0A. `0 lt x le 1`B. ` 0 le x le 1`C. `-oo lt x le 0`D. `-oo lt x lt 1` |
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Answer» Correct Answer - D It is given that `2^(x)+2^(y)=2 AA x,y in R` or `2^(y)=2-2^(x)` or `y=log_(2)(2-2^(x))` Therefore, function id defined only when `2-2^(x) gt 0 " or " 2^(x) lt 2 " or " x lt 1`. |
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| 123. |
Find the domain and range of `f(x)=cos^(-1)sqrt((log)_([x])((|x|)/x))` |
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Answer» Correct Answer - Domain: `[2,oo),` Range: `{pi//2}` ` "log"_([x])(|x|)/(x)` is defined if `(|x|)/(x) gt 0, [x] gt 0, " and " [x] ne 1` or `x gt 0,x in [1,oo), " and " x notin [1,2)` or `x in [2,oo)` For `x in [2,oo)` we have `log_([x])(|x|)/(x)=log_([x])1=0.` Therefore, `f(x)=cos^(-1)0=(pi)/(2)` for all ` x in [2,oo).` Hence, domain `(f)` is `[2,oo)` and range `(f)` is `{pi//2}.` |
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| 124. |
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. RB. `[0,+oo)`C. `(-oo,0]`D. none of these |
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Answer» Correct Answer - D `x^(2)-[x]^(2) ge 0 " or " x^(2) ge [x]^(2)` This is true for all non-negative values of `x` and all negative integers `x`. |
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| 125. |
Is the pair of the functions `e^(sqrt(log_(e)x))` and `sqrt(x)` identical ? |
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Answer» `e^(sqrt(log_(e)x))=(e^(logx))^((1)/(2))=sqrt(x)=sqrt(x), " where " x ge 1` For `sqrt(x),x ge0,` Thus, functions are not identical as domain is not same. |
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| 126. |
The domain of the function `f(x)=(1)/(sqrt(|cosx|+cosx))` isA. `[-2n pi, 2 n pi], n in Z`B. `(2n pi,bar(2n+1) pi), n in Z`C. `(((4n+1)pi)/(2),((4n+3)pi)/(2)), n in Z`D. `(((4n-1)pi)/(2),((4n+1)pi)/(2)), n in Z` |
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Answer» Correct Answer - D `|cosx|+cosx={(0",",cosx le 0),(2 cosx",",cosx gt 0):}` For `f(x)` to be defined, `cos x gt 0` or ` x in (((4n-1)pi)/(2),((4n+1)pi)/(2)), n in Z` (1st and 4th quadrant). |
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| 127. |
Find the domain of`f(x)=(log)_(10)(log)_2(log)_(2/pi)(t a n^(-1)x)^(-1)` |
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Answer» We must have `log_(2)log_(2//pi)(tan^(-1)x)^(-1) gt 0` or `log_(2//pi)(tan^(-1)x)^(-1) gt 1` or `0 lt (tan^(-1)x)^(-1) lt (2)/(pi)` or `(pi)/(2) lt tan^(-1)x lt oo`, which is not possible . Hence, the domain is `phi`. |
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| 128. |
Which of the following functions is an injective (one-one) function in its respective domain? (A) `f(x) = 2x + sin 3x ` (B) `x. [x]` , (where `[.]` denotes the G.I.F) (C) `f(x)=(2^x-1)/(4^x+1)` (D) `f(x)=(2^x+1)/(4^x-1)`A. `f(x)=2x+sin 3x`B. `f(x)=x*[x],` (where [.] denotes the G.I.F)C. `f(x)=(2^(x)-1)/(4^(x)+1)`D. `f(x)=(2^(x)+1)/(4^(x)-1)` |
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Answer» Correct Answer - D `f(x)=(2^(x)+1)/(4^(x)-1)=(1)/(2^(x)-1)` is one-one function. |
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| 129. |
The domain of the function `f(x)=(sin^(-1)(3-x))/("In"(|x|-2))` isA. `[2,4]`B. `(2,3) cup (3,4)`C. `[2,oo)`D. `(-oo,-3) cup [2,oo)` |
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Answer» Correct Answer - B `f(x)=(sin^(-1)(3-x))/(log(|x|-2))` Let `g(x)=sin^(-1)(3-x)` or `-1 le 3 -x le 1` The domain of g(x) is [2, 4]. Let `h(x)=log(|x|-2)` I.e., `|x|-2 gt 0 " or " |x| gt 2` i.e., `x lt -2 " or " x gt 2` ` :. " Domain " (-oo, -2) cup (2 ,oo)` We know that `(f//g)(x)=(f(x))/(g(x)) AA x in D_(1) cap D_(2)-{x in R:g(x)=0}` Therefore, the domain of `f(x) " is " (2,4]-{3}=(2,3) cup (3,4].` |
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| 130. |
Let `f(x) =1/2-tan((pix)/2) `, -1 |
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Answer» `P:x in (-1,1)` `g(x)=sqrt(3+4x-4x^2)` `3+4x-4x^2>=0` `4x^2-4x-3<=0` `(2x+1)(2x-3)<=0` `x in [-1/2,3/2]` `x in [-1/2,1)` Option 3 is correct. |
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| 131. |
The domain of the function `f(x)=1/(sqrt(4x-|x^2-10 x+9|))`is `(7-sqrt(40),7+sqrt(40))`(b) `(0,7+sqrt(40))`(c)`(7-sqrt(40),oo)` (d) none of theseA. `(7-sqrt(40), 7+sqrt(40))`B. `(0,7+sqrt(40))`C. `(7-sqrt(40),oo)`D. none of these |
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Answer» Correct Answer - D `f(x)=(1)/(sqrt(4x-|x^(2)-10x+9|))` For `f(x)` to be defined , `|x^(2)-10x+9| lt 4x` or ` x^(2)-10x +9 lt 4x " and " x^(2)-10x+9 gt -4x` or ` x^(2)-14x +9 lt 0 " and " x^(2) -6x+9 gt 0` or ` x in (7-sqrt(40), 7 + sqrt(40)) " and " x in R -{3}` or ` x in (7-sqrt(40), 3) cup (3,7+sqrt(40))` |
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| 132. |
Find the domain and range of following functions (i) `f(x)=log_(e)(sinx)` (ii) `f(x)=log_(3)(5-4x-x^(2))` |
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Answer» (i) `f(x)=log_(e) sin x` is defined if `sin x in (0,1]` ` :. x in … (-4pi,-3pi)cup(-2pi,-pi) cup (0,pi) cup (2pi,3pi) cup (4pi,5pi)…` Also, `0 lt sinx le 1` ` :. -oo lt log_(e)(sinx)le 0` Thus, range is `(- oo,0]` (ii) `f(x)=log_(3)(5-4x-x^(2))` `=log_(3)(9-(x+2)^(2))` f(x) is defined if `5-4x-x^(2) gt 0` or `x^(2)+4x-5 lt 0` `implies (x-1)(x+5) lt 0` `implies -5 lt x lt 1` So, domain is `(-5,1)`. Also `9-(x+2)^(2) le 9` Thus, `0 lt 9-(x+2)^(2) le 9` `implies -oo lt log_(3)(9-(x+2)^(2)) le log_(3)9` Hence, range of f(x) is `(-oo,2]` |
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| 133. |
Find the domain and range of the following (i) ` f(x)=sqrt(x^(2)-3x+2) " (ii) " f(x)=sqrt(x^(2)-4x+6)` |
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Answer» (i) Clearly f(x) is defined if `x^(2)-3x+2 ge 0` `implies (x-1)(x-2) ge 0` `implies x in (-oo,1] cup [2,oo)` Now, `f(x)=sqrt(x^(2)-3x+2)` `=sqrt((x-(3)/(2))^(2)+2-(9)/(4))` `=sqrt((x-(3)/(2))^(2)-(1)/(4))` Clearly, `(x-(3)/(2))^(2)-(1)/(4) ge 0` ` :. sqrt((x-(3)/(2))^(2)-(1)/(4)) ge 0` Therefore, range is `[0,oo).` Here least value of f(x) occurs when `x-(3)/(2)= +-(1)/(2), ` i.e., (ii) `f(x) =sqrt(x^(2)-4x+6)` `=sqrt((x-2)^(2)+2)` Clearly `(x-2)^(2) +2 ge 2, AA x in R.` So, domain of f(x) is R. Also, `sqrt((x-2)^(2)+2) ge sqrt(2)` Hence, range is `[ sqrt(2),oo).` |
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| 134. |
Find the domain and range function `f(x) =(x^(2)-3x+2)/(x^(2)-4x+3)`. |
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Answer» `y=f(x)=((x-1)(x-2))/((x-3)(x-1))=(x-2)/(x-3), x ne 1,3` Domain of function `=R-{1,3}` Now `yx-3y=x-2` ` :. x= (3y-2)/(y-1)` Clearly `y ne 1`. Also, when ` x=1, y=1//2` Therefore, range of function is ` R-{(1)/(2),1}` |
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| 135. |
The domain of the function `f(x)=(1)/(sqrt(""^(10)C_(x-1)-3xx""^(10)C_(x)))` isA. `{9,10,11}`B. `{9,10,12}`C. all natural numbersD. `{9,10}` |
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Answer» Correct Answer - D Given function is defined if ` "'^(10)C_(x-1) gt 3""^(10)C_(x)` or `(1)/(11-x) gt (3)/(x) " or " 4x gt 33` or `x ge 9` But `x le 10` ` :. x=9,10` |
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| 136. |
Given the function `f(x)=(a^x+a^(-x))/2(w h e r ea >2)dotT h e nf(x+y)+f(x-y)=``2f(x)dotf(y)`(b) `f(x)dotf(y)``(f(x))/(f(y))`(d) none of these |
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Answer» Correct Answer - A We have `f(x+y)=f(x-y)=(1)/(2)[a^(x+y)+a^(-x-y)+a^(x-y)+a^(-x+y)]` `=(1)/(2)[a^(x)(a^(y)+a^(-y))+a^(-x)(a^(y)+a^(-y))]` `=(1)/(2)(a^(x)+a^(-x))(a^(y)+a^(-y))=2f(x)f(y)` |
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| 137. |
Which of the following functions has inverse function?`f: Zvecd efin e db yf(x)=x+2``f: Zvecd efin e db yf(x)=2x``f: Zvecd efin e db yf(x)=x``f: Zvecd efin e db yf(x)=|x|` |
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Answer» `f:Z to Z ,f(x) = x+2` is both one-one and onto. Hence invertible. `f:Z to Z ,f(x) = 2x` is one-one but not onto as range is set `{…,-4,-2,0,2,4,6, …}`. Hence not invertible. `f:Z to Z ,f(x) = x` is one-one and onto, Hence invertible. `f:Z to Z ,f(x) = |x|` is many-one, Also, range is `{0,1,2,3,...}`, So, into, Hence not invertible. |
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| 138. |
The domain of the function `f(x)=sqrt("In"_((|x|-1))(x^(2)+4x+4))` isA. `[-3,-1] cup [1,2]`B. `(-2, -1) cup [2,oo)`C. `(-oo,-3] cup (-2,-1) cup (2,oo)`D. None of these |
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Answer» Correct Answer - C Case I: ` 0 lt |x|-1 lt 1 " or " 1 lt |x| lt 2` Then `x^(2)+4x+4 le 1` or `x^(2)+4x+3 le 0` or `-3 le x le -1` So, `x in (-2,-1) " (1)" ` Case II: ` |x|-1 gt 1 " or " |x| gt 2` Then `x^(2)+4x+4 ge 1` or `x^(2)+4x+3 ge 0` or `x ge -1 " or " x le -3` So, ` x in (-oo, -3] cup (2,oo)` So, ` x in (-oo, -3] cup (2,oo) " (2) " ` From (1) and (2), ` x in (-oo, -3] cup (-2, -1) cup (2,oo)`. |
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| 139. |
If `f(x)=cos (logx)`, then `f(x)f(y)-1/2[f(x/y)+f(xy)]=`A. -1B. `1//2`C. -2D. 0 |
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Answer» Correct Answer - D `f(x)=cos(logx)` or `f(x)f(y)-(1)/(2)[f((x)/(y))+f(xy)]` `=cos(logx)cos(logy)-(1)/(2)[cos(logx-logy)]+[cos(logx+logy)]` `=cos(logx)cos (logy)-(1)/(2)[2cos(logx)cos(logy)]` `=0` |
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| 140. |
`f(x)=x^2-2a x+a(a+1),f:[a ,oo)vec[a ,oo)dot`If one of the solution of the equation `f(x)=f^(-1)(x)i s5049 ,`then the other may be5051 (b) 5048 (c)5052 (d) 5050A. 5051B. 5048C. 5052D. 5050 |
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Answer» Correct Answer - B::D `f(x)=x^(2)-2ax+a(a+1)` ` f(x)=(x-a)^(2)+a, x in [a,oo)` Let `y=(x-a)^(2)+a. " Clearly " y ge a.` Thus, `(x-a)^(2)=y-a` `or x=a+sqrt(y-a)` ` :. f^(-1)(x)=a+sqrt(x-a)` Now, ` f(x) =f^(-1)(x)` `or (x-a)^(2)+a=a+sqrt(x-a)` `(x-a)^(2) =sqrt(x-a)` ` or (x-a)^(4)=(x-a)` i.e., `x= a or (x-a)^(3) =1` i.e., ` x=a or a+1 ` If ` a=5049, " then " a+1=5050.` If `a+1=5049, " then " a=5048.` |
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| 141. |
Find the inverse of the function:`f:[-1,]vec[-1,1]d efin e db yf(x)=x|x|` |
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Answer» Correct Answer - `f^(-1)(x)={(-sqrt(-x)",",x lt 0),(sqrt(x)",",x ge 0):}` `f:[-1,1] to [-1,1]` is defined by `f(x)=x|x|={(x^(2)",",x ge 0),(-x^(2)",",x lt 0):}` or `f^(-1)(x)={(sqrt(x)",",x ge 0),(-sqrt(-x)",",x lt 0):}` |
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| 142. |
An even polynomial function `f(x)`satisfies a relation `f(2x)(1-f(1/(2x)))+f*16^2y)=f(-2)-f(4x y)AAx ,y in R-{0}a n df(4)=-255 ,f(0)=1.`Then the value of `|(f(2)+1)//2|`is_________ |
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Answer» Correct Answer - -15 We have `f(2x)-f(2x) f((1)/(2x))+f(16x^(2)y)=f(-2)-f(4xy)` Replacing y by `(1)/(8x^(2)),` we get `f(2x)-f(2x)((1)/(2x))+f(2)=f(-2)-f((1)/(2x))` ` :. f(2x)+f((1)/(2x))=f(2x)f((1)/(2x)) ` [As `f(x)` is even] ` :. f(2x)=1+-(2x)^(n)` `or f(x)=1+-x^(n)` Now, `f(4)=1+-4^(n)= -255 " "` (Given) Taking negative sign, we get `256=4^(n) or n=4.` Hence, `f(x)=1-x^(4)`, which is an even function. Therefore, `f(2)= -15.` |
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| 143. |
Find the inverse of the function:`f: Rvec(-oo,1)gi v e nb yf(x)=1-2^(-x)` |
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Answer» Correct Answer - `f^(-1)(x)=-log_(2)(1-x)` Let `y=1-2^(-x)` or `2^(-x)=1-y` or `-x=log_(2)(1-y)` or `f^(-1)(x)=g(x)= -log_(2) (1-x)` |
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| 144. |
Let `f: Rvec[0,pi/2)`be defined by `f(x)=tan^(-1)(x^2+x+a)dot`Then the set of values of `a`for which `f`is onto is`(0,oo)`(b) `[2,1]`(c) `[1/4,oo]`(d) none of theseA. `[0,oo)`B. `[2,1]`C. `[(1)/(4),oo)`D. none of these |
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Answer» Correct Answer - C Since co-domain`=[0,(pi)/(2))` This is possible only when `x^(2) +x+a` is perfect square. ` :. 1-4a=0 " or " a=(1)/(4)` |
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| 145. |
The function `f`satisfies the functional equation `3f(x)+2f((x+59)/(x1))=10 x+30`for all real `x!=1.`The value of `f(7)i s`8 (b)4 (c) `-8`(d) 11A. 8B. 4C. -8D. 11 |
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Answer» Correct Answer - B `3f(x)+2f((x+59)/(x-1))=10x+30` For `x=7, 3f(7)+2f(11)=70+30=100.` For `x=11, 3f(11)+2f(7)=140.` `(f(7))/(-20)=(f(11))/(-220)=(-1)/(9-4) or f(7)=4` |
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| 146. |
If `a , b`are two fixed positive integers such that `f(a+x)=b+[b^3+1-3b^2f(x)+3b{f(x)}^2-{f(x)}^3]^(1/3)`for all real `x ,`then prove that `f(x)`is periodic and find its period. |
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Answer» `f(a+x)=b+[b^(3)+1-3b^(2)f(x)+3b{f(x)}^(2)-{f(x)}^(3)]^(1//3)` `=b+[1+{b-f(x)}^(3)]^(1//3)` `or f(a+x)-b=[1-{b-f(x)}^(3)]^(1//3)` ` or phi(a+x)=[1-{phi(x)}^(3)]^(1//3) " (1)" ` where `phi (x)=f(x)-b` `or phi (2a+x)=[1-{phi(x+a)}^(3)]^(1//3)=phi(x) " From (1)]" ` `or f(x+2a)-b=f(x)-b` `or f(x+2a)=f(x)` Therefore, `f(x)` is periodic with period 2a. |
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| 147. |
Find the inverse of the function:`f:(-oo,1]vec[1/2,oo],w h e r ef(x)=2^(x(x-2))` |
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Answer» Correct Answer - `f^(-1)(x)=1-sqrt(1+log_(2)x)` `y=2^(x(x-2))` or ` x^(2)-2x=log_(2)y` or ` x^(2)-2x-log_(2)y=0` or `x=1+-sqrt(1+log_(2)y)` or `f^(-1)(x)=1-sqrt(1+log_(2)x)` as `f^(-1):[(1)/(2),oo) to (-oo,1]` |
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| 148. |
Let `f : R to R` be a function such that `f(0)=1` and for any `x,y in R, f(xy+1)=f(x)f(y)-f(y)-x+2.` Then `f` isA. one-one and ontoB. one-one but not ontoC. many one but ontoD. many one and into |
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Answer» Correct Answer - A `f(xy+1)=f(yx+1)` `f(x)f(y)-f(y)-x+2=f(y)f(x)-f(x)-y+2` `f(x) -f(y)=x-y` Putting `y=0` `f(x)-1=x-0` `f(x) = x+1` |
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| 149. |
If p, q are positive integers, f is a function defined for positive numbers and attains only positive values such that `f(xf(y))=x^p y^q`, then prove that `p^2=q`. |
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Answer» Given `f(xf(y))=x^(p)y^(q)` `or x=({f(xf(y))}^(1//p))/(y^(q//p)) " (1)" ` Let `xf(y)=1 or x=(1)/(f(y))`. Then from (1), `f(y)=(y^(q//p))/({f(1)}^(1//p))` ` or f(1)=(1)/({f(1)}^(1//p))` ` :. f(1)=1` ` :. f(y)=y^(q//p)" (2)" ` Now, `f(xy^(q//p))=x^(p)y^(q). " Put "y^(q//P)=z.` Then `f(xz)=(xz)^(p)` `or f(x)=x^(p) " (3)" ` From (2) and (3), `x^(P)=x^(q//p) or p^(2)=q.` |
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| 150. |
If `f: RvecS ,`defined by `f(x)=sinx-sqrt(3)cosx+1,i son to,`then find the set `Sdot` |
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Answer» Correct Answer - [-1, 3] `f(x)=sinx-sqrt(3)cosx+1` `=2("sin"x(1)/(2)-"cos"(sqrt(3))/(2))+1` `=2(sinx"cos"(pi)/(3)-cosx"sin"(pi)/(3))+1` `=2"sin"(x-(pi)/(3))+1` Clearly, `f` is onto, when the interval of S is `[-1,3].` |
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