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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.
| 51. |
Find the value of `4 tan^-1 (1/5) - tan^-1 (1/239) ` |
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Answer» Here, we will use the following properties, `tan^(-1)(x)+tan^(-1)(y) = tan^(-1)((x+y)/(1-xy))` `tan^(-1)(x)-tan^(-1)(y) = tan^(-1)((x-y)/(1+xy))` Now, `4tan^(-1)(1/5) = 2(tan^(-1)(1/5)+tan^(-1)(1/5))` `=2(tan^(-1)((2/5)/(1-1/25)))` `=2(tan^(-1)(5/12))` `=tan^(-1)(5/12)+tan^(-1)(5/12)` `= tan^(-1) (((5/12)+(5/12))/(1- (5/12)(5/12)))` `=tan^(-1)((10/12)/(119/144))` `4tan^(-1)(1/5)= tan^(-1) (120/119)` Now, `4tan^(-1)(1/5) - tan^(-1)(1/239) = tan^(-1) (120/119)-tan^(-1)(1/239)` `= tan^(-1)((120/119-1/239)/(1+(120/119)(1/239)))` `=tan^(-1)((120*239-119)/(119*239+120))` `=tan^(-1)(28561/28561) = tan^(-1)(1) = pi/4` |
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| 52. |
If `vec(a)` and `vec(b)` are unit vectors inclined at an angle `theta` then prove that(i)`sin(theta/2)=1/2|vec(a-vec(b)|` (ii) `tan(theta/2)=(|vec(a)-vec(b)|)/(|vec(a)-vec(b)|)` |
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Answer» Here, `veca` and `vecb` are unit vectors. `:. |hata| = |hatb| = 1` (i)`|hata-hatb|^2 = (hata)^2+(hatb)^2-2hata*hatb` `=|hata|^2+|hatb|^2-2|hata||hatb|cos theta` `=1^2+1^2-2(1)(1)costheta` `=>|hata-hatb|^2=2(1-costheta)` `=>|hata-hatb|^2/2 = 1-costheta` `=>|hata-hatb|^2/2 = 2sin^2(theta/2)` `=>|hata-hatb|^2/4 = sin^2(theta/2)` `=>sin (theta/2) =1/2 |hata-hatb| ->(1)` (ii)`|hata+hatb|^2 = (hata)^2+(hatb)^2+2hata*hatb` `=|hata|^2+|hatb|^2+2|hata||hatb|cos theta` `=1^2+1^2+2(1)(1)costheta` `=>|hata+hatb|^2=2(1+costheta)` `=>|hata+hatb|^2/2 = 1+costheta` `=>|hata+hatb|^2/2 = 2cos^2(theta/2)` `=>|hata+hatb|^2/4 = cos^2(theta/2)` `=>cos (theta/2) =1/2 |hata+hatb| ->(2)` Now, dividing (1) by (2), `sin (theta/2)/cos (theta/2) = |hata-hatb|/|hata+hatb|` `=> tan (theta/2) = |hata-hatb|/|hata+hatb|` |
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| 53. |
The value of `sec 20+sec 40 + sec 160` is |
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Answer» `sec40+sec80=1/cos40+1/cos80` `=(cos40+cos80)/(cos40cos80)` `=(2cos60*cos20)/(1/2*cos120+cos40)` `=(4*1/2*cos20)/(cos40-1/2)` `(4cos(20))/((2cos40)-1)` `(4cos20)/(2cos40-1)+1/cos160` `(4cos20)/(2cos40-1)-1/cos20` `=(4cos^2(20)-2cos(40)+1)/((2cos40-1)cos20)` `(2(1+cos40)-2cos40+1)/(cos60+cos20-cos20)` `(2+1)/cos60=3/0.5=6`. option c is correct. |
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| 54. |
For what value of k is the followinf function continuous at x = `-pi/6`. Where `f(x) = (sqrt3 sin x + cos x)/( x + pi/6) , x != -pi/6 and f(x) = k, x = - pi/6` |
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Answer» `lim_(x->a^+)f(x)=f(a)=lim_(x->0^-)f(x)` `lim_(x->-pi/6^+)(sqrt3sinx+cosx)/(x+pi/6)` `lim_(x->-pi/6^+)(sqrt3/2sinx+cosx*1/2)^2/(x+pi/6)` `lim_(h->0)(sinh/h)*2=2` `f(-pi/6)=2`. |
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| 55. |
Given: `vec a + vec b + vec c = 0`. Out off three vectors `vec a,vec b and vec c` two are equal in magnitude. The magnitude of third vector is `sqrt2` times that of either of the two having equal magnitude. The angle between vectors is |
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Answer» SD=11Rs VR=7Rs VC=8Rs Price of one each=11+7+8=26Rs Now we can spend(19-24)Rs `3VR,3VC,2VR and 1VC` `1VR and 2VC` there are total 6 cases. |
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| 56. |
If the vectors `3 vec p + vec q`; `5 vec p - 3 vecq` and `2 vec p + vec q`; `4 vec p - 2 vec q` are pairs of mutually perpendicular then sin( `vec p` ^ `vec q)` is : |
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Answer» We know dot product of two perpendicular vectors is `0`. `:. (3vecp+vecq).(5vecp-3vecq) = 0` `=>15|vecp|^2-3|vecq|^2-4vecp*vecq = 0->(1)` Now, `(2vecp+vecq).(4vecp-2vecq) = 0` `=>8|vecp|^2-2|vecq|^2 = 0` `=>|vecq|^2 = 4|vecp|^2` `=>|vecq| = 2|vecp|` Putting value of `|vecq|` in (1), `15|vecp|^2-12|vecp|^2-4|vecp|*2|vecp|*costheta = 0` `3|vecp|^2-8|vecp|^2costheta = 0` `cos theta = 3/8` `:. sin theta = sqrt(1-(3/8)^2) = sqrt55/8` |
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| 57. |
The maximum number of equivalence relations on the set A = {1, 2, 3} areA. 1B. 2C. 3D. 5 |
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Answer» Correct Answer - D Given that, A = {1, 2, 3} Now, number of equivalence relations as follows `R_(1)={(1,1),(2,2),(3,3)}` `R_(2)={(1,1),(2,2),(3,3),(1,2),(2,1)}` `R_(3)={(1,1),(2,2),(3,3),(1,3),(3,1)}` `R_(4)={(1,1),(2,2),(3,3),(2,3),(3,2)}` `R_(5)={(1,2,3) hArr A xx A= A^(2)}` ` :. ` Maximum number of equivalence relation on the set A = {1, 2, 3} = 5 |
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| 58. |
If ` vec a, vec b and vec c` be any three vectors then show that ` vec a + ( vec b + vec c) = ( vec a + vec b) + vec c` |
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Answer» We can show it with the help of the diagram. Please refer to video for the diagram. `veca +(vecb+vecc) = veca+vecd` Here, `vecd` is a vector joining tail of `vecb` and head of `vecc`. ` =>veca +(vecb+vecc)=veca+vecd = vece->(1)` Here, `vece` is a vector joining tail of `veca` and head of `vecd`. Now, `(veca+vecb)+vecc = vecf+vecc` Here, `vecf` is a vector joining tail of `veca` and head of `vecb`. If we join tail of `vecf` and head of `vecc`, then it will come `vece`. `=>(veca+vecb)+vecc = vecf+vecc = vece->(2)` From (1) and (2), `veca +(vecb+vecc) = (veca+vecb)+vecc` |
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| 59. |
`int_0^pi(xtanx)/(tanx+secx).dx=[pi(pi-2)]/2` |
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Answer» `I=int_0^pi (xtanx)/(tanx+secx)dx` `=int_0^pi (xsinx/cosx)/(sinx/cosx+1/cosx)dx` `I=int_0^pi (xsinx)/(1+sinx)-(1)` `I=int_0^pi((pi-x)sin(pi-x))/(1+sin(pi-x)` `I=int_0^pi((pi-x)sinx)/(1+sinx)-(2)` `I=int_0^pi[(xsinx)/(1+sinx)+((pi-x)sinx)/(1+sinx)]dx` `2I=int_0^pi[(xsinx+pisinx-xsinx)/(1+sinx)]dx` `2I=pi int_0^pi sinx/(1+sinx)dx` `I=pi/2 int_0^pi sinx/(1+sinx)dx` `[secpi-sec0]-int_0^pi sec^2x dx+int_0^pi 1dx` `-1-1-[tanx]_0^pi+(x)_0^pi` `-2+pi` `pi-2` `I=pi/2(pi-2)`. |
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| 60. |
The number of integral values of n such that the equation 2n{x} = 3x + 2[x]has exactly 5 solutions. |
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Answer» x=I+F 2nf=3(I+F)+2I `f=(5I)/(2n-3)` `0<=(5I)/(2n-3)<1` `0<=I/(2n-3)<1/5` `f=(5I)/(2n-3)=(5I)/(17)` `f=(5I)/14` `f=(5I)/21` `f=(5I)/23` `f=(5I)/25` `f=(5I)/27` option b is correct. |
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| 61. |
lf a relation R on the set `{1, 2, 3}` be defined by `R ={(1,2)}`, then R is:A. reflexiveB. transitiveC. symmetricD. None of these |
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Answer» Correct Answer - B R on the set {1, 2, 3} be defined by R = {(1, 2)} It is clear that R is transitive. |
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| 62. |
If ` vec a and vec b` are two non-zero vectors such that `| vec a xx vec b| = vec a . vec b`, then angle between `vec a` and `vec b` |
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Answer» `|veca*vecb|=|veca|*|vecb|*sintheta` `veca*vecb=|veca|*|vecb|costheta` `|veca*vecb|=veca*vecb` `|veca|*|vecb|sintheta=|veca|*|vecb|costheta` `sintheta=costheta` `tantheta=1` `theta=pi/4` anglebetween`veca and vecb=pi/4`. |
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| 63. |
Let `f(x)=x^[135]+x^[125] -x^[115]+x^5+1`.lf f(x) is divided by `x^3-x` then the remainder is some function of x say g(x). Find the value of g(10) |
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Answer» after dividing g(x)=2x+1 g(10)=2*10+1=21 |
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| 64. |
`f(x)=sin^(-1)[e^(x)]+sin^(-1)[e^(-x)]` where [.] greatest integer function thenA. domain of `f(x) " is " (-log_(e)2,log_(e)2)`B. range of `f(x) ={pi}`C. Range of `f(x) " is " {(pi)/(2),pi}`D. `f(x) =cos^(-1)x` has only one solution |
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Answer» Correct Answer - A::C We have `f(x)=sin^(-1)[e^(x)]+sin^(-1)[e^(-x)]` We must have `0 lt e^(x) lt 2 and 0 lt e^(-x) lt 2` `implies x in (-log_(e) 2, log_(e)2)` `implies f(x) ={(pi",",x=0),((pi)/(2)"," , x in (-log_(e)2","0)cup (0","log_(e)2)):}` |
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| 65. |
Prove that: ` cos ((2pi)/15) . cos ((4pi)/15) . cos ((8pi)/15) . cos ((16pi)/15) = 1/16` |
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Answer» `cos2pi/15*cos4pi/15*cos8pi/15*cos(pi+pi/15)` `(2sinpi/15*cospi/15*cos2pi/15*cos4pi/15*cos8pi/15)/(2sinpi/15` `(-2sin2pi/15cos2pi/15*cos4pi/15*cos8pi/15)/(2*2sinpi/15)` `(-sin4pi/15cos4pi/15*cos8pi/15)/(24sinpi/15)` `(sinpi/15)/(16sinpi/15)=1/16` proved. |
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| 66. |
lf the fundamental period of function `f(x)=sinx + cos(sqrt(4-a^2))x` is `4pi`, then the value of a is/areA. `(sqrt(15))/(2)`B. `-(sqrt(15))/(2)`C. `(sqrt(7))/(2)`D. `-(sqrt(7))/(2)` |
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Answer» Correct Answer - A::B::C::D Period of `sinx " is " 2pi` and period of `cos (sqrt(4-a^(2))x) " is " (2pi)/(sqrt(4-a^(2))).` `implies LCM (2pi,(2pi)/(sqrt(4-a^(2))))=4pi` (given) i.e., `sqrt(4-a^(2))=(p)/(2) " where " p=1,3.` Hence ` a^(2) =(15)/(4),(7)/(4),a=+-(sqrt(15))/(2),+-(sqrt(17))/(2)` |
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| 67. |
`cos^2(3pi/5)*cos^2(4pi/5)` |
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Answer» Here, we will use , `cos 36^@ = (sqrt5+1)/4 and cos72^@ = (sqrt5-1)/4` Now, `cos^2((3pi)/5) = (cos(108^@))^2 = (cos(pi - 72^@))^2 = (-cos72^@)^2` `=((sqrt5-1)/4)^2 = (5+1-2sqrt5)/16 = (3-sqrt5)/8` Now, `cos^2((4pi)/5) = (cos(144^@))^2 = (cos(pi - 36^@))^2 = (-cos36^@)^2` `=((sqrt5+1)/4)^2 = (5+1+2sqrt5)/16 = (3+sqrt5)/8` `:. cos^2((3pi)/5)*cos^2((4pi)/5) = (3-sqrt5)/8 ** (3+sqrt5)/8 = (9-5)/64 = 4/64 = 1/16` `:. cos^2((3pi)/5)*cos^2((4pi)/5) = 1/16` |
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| 68. |
If `theta` is the fundamental period of the function `f(x) = sin^99 x+sin^99(x+(2pi)/3)+sin^99(x+(4pi)/3)` , then the complex number `z=|z|(cos theta+i sintheta)` lies in the quadrant number. |
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Answer» Correct Answer - 3 Clearly, fundamental period is `(4pi)/(3)`. Then z lies in the third quadrant. |
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| 69. |
If `f(x)=10x-7 and (fog)(x)=x`, then `g(x)` is equal to:A. `(x+7)/(10)`B. `(x-7)/(10)`C. `(1)/(10x-7)`D. None of these |
| Answer» Correct Answer - a | |
| 70. |
If the function `f:A to B` is one-one onto and `g:B to A `, is the inverse of f, then fog =?A. fB. gC. `I_(A)`D. `I_(B)` |
| Answer» Correct Answer - D | |
| 71. |
Solve the following : (a) `1 le |x-2| le 3 " (b) "0 lt |x-3|le 5` (c ) `|x-2|+|2x-3|=|x-1| " (d) " |(x-3)/(x+1)| le1` |
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Answer» Correct Answer - (a) `[-1,1]cup [3,5] " (b) " [-2,8]-{3} " (c ) " [3//2,2] " (d) "[1,oo)` (a) ` 1 le |x-2| le 3` `implies-3 le x-2 le -1 " or " 1 le x-2 le 3` `implies -1 le x le 1 " or "3 le x le 5` (b) ` 0 lt |x-3| le 5` `implies -5 le x-3 le 5, x-3 ne 0` `implies -2 le x le 8, x ne 3` (c ) `|x-2| +|2x-3|=|x-1|` `implies |x-2|+|2x-3|=|(2x-3)-(x-2)|` `implies (x-2) (2x-3) le 0` `implies 3//2 le x le 2` (d) `|(x-3)/(x+1)| le 1` `implies -1 le (x-3)/(x+1) le 1` `implies 0 le (x-3)/(x+1)+1 " and " (x-3)/(x+1)-1 le 0` `implies0 le (2x-2)/(x+1) " and " (-4)/(x+1) le 0` `implies x le -1 " or " x ge 1 " and " x gt -1` `implies x ge 1` |
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| 72. |
If `f: R to R` be defined by `f(x)={(2x:xgt3),(x^(2):1lt x le 3),(3x:x le 1):}` Then,`f(-1)+f(2)+f(4)` isA. 9B. 14C. 5D. None of these |
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Answer» Correct Answer - A Given that, `f(x)={(2x:xgt3),(x^(2):1lt x le 3),(3x:x le 1):}` `f(-1)+f(2) +f(4)=3(-1)+(2)^(2)+2xx4` ` " " = -3 +4 +8=9` |
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| 73. |
Consider the function `f(x)` satisfying the identity `f(x) +f((x-1)/(x))=1+x AA x in R -{0,1}, and g(x)=2f(x)-x+1.` The domain of `y=sqrt(g(x))` isA. `(-oo,(1-sqrt(5))/(2)] cup [1,(1+sqrt(5))/(2)]`B. `(-oo,(1-sqrt(5))/(2)] cup (0,1)cup [(1+sqrt(5))/(2),oo)`C. `[(-1-sqrt(5))/(2),0] cup [(-1+sqrt(5))/(2),1)`D. None of these |
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Answer» Correct Answer - B `f(x)+f((x-1)/(x))=1+x " (1)" ` In (1), replace x by `(x-1)/(x)`. Then `f((x-1)/(x))+f(((x-1)/(x)-1)/((x-1)/(x)))=1+(x-1)/(x)` `or f((x-1)/(x))+f((1)/(1-x))=1+(x-1)/(x) " (2)" ` Now, from `(1) -(2)`, we have `f(x)-f((1)/(1-x))=x-(x-1)/(x) " (3)" ` In (3), replace x by `(1)/(x-1)`. Then `f((1)/(1-x))-f((x-1)/(x))=(1)/(1-x)-((1)/(1-x)-1)/((1)/(1-x))` ` or f((1)/(1-x))-f((x-1)/(x))=(1)/(1-x)-x " (4)" ` Now, from `(1)+(3)+(4)`, we have `2f(x)=1+x+x-(x-1)/(x)+(1)/(1-x)-x` ` or f(x)=(x^(3)-x^(2)-1)/(2x(x-1))` `g(x)=(x^(3)-x^(2)-1)/(x(x-1))-x+1=(x^(2)-x-1)/(x(x-1))` Now, for `y=sqrt(g(x)),` we must have `(x^(2)-x-1)/(x(x-1)) ge 0` `or ((x-(1-sqrt(5))/(2))(x-(1+sqrt(5))/(2)))/(x(x-1)) ge 0` `or x in (-oo,(1-sqrt(5))/(2)] cup (0,1)cup [(1+sqrt(5))/(2),oo)` |
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| 74. |
If f = {(2,3) ,(3,4) , (4,5}, then its inverse is :A. {(3,4) , (4,5) , (3,2)}B. {(3,2) , (4,3) , (5,4)}C. {(2,3) , (4, 3) , (5,4)}D. None of these |
| Answer» Correct Answer - b | |
| 75. |
If `f(x)=(a^x)/(a^x+sqrt(a ,)),(a >0),`then find the value of`sum_(r=1)^(2n1)2f(r/(2n))` |
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Answer» `f(x)=(a^(x))/(a^(x)+sqrt(a))` or `f(1-x)=(a^(1-x))/(a^(1-x)+sqrt(a))` `=(a^(1))/(a^(1)+sqrt(a)a^(x))` `=(sqrt(a))/(sqrt(a)+a^(x))` ` or f(x)+f(1-x)=1` ` :. f((1)/(2))=(1)/(2)` so, `sum_(r=1)^(2n-1)2f((r)/(2n))=2[f((1)/(2n))+f((2)/(2n))+ ... +f((n-1)/(2n))+f((n)/(2n))+f((n+1)/(2n))+ ... +f((2n-1)/(2n))]` `=2{[f((1)/(2n))+f((2n-1)/(2n))]+[f((2)/(2n))+f((2n-2)/(2n))]+ ... +[f((n-1)/(2n))+f((n+1)/(2n))]+f((1)/(2))}` `=2{[f((1)/(2n))+f(1-(1)/(2n))]+[f((2)/(2n))+f(1-(2)/(2n))]+ ... +[f((n-1)/(2n))+f(1-(n-1)/(2n))]+(1)/(2)}` `=2[1+1+1+...+(n-1)" times "]+1` `=2n-1` |
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| 76. |
Let `f(x)` be defined on `[-2,2]` and be given by `f(x)={(-1",",-2 le x le 0),(x-1",",0 lt x le 2):} and g(x)=f(|x|) +|f(x)|`. Then find `g(x)`. |
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Answer» We have `f(x)={(-1",",-2 le x le 0),(x-1",",0 lt x le 2):}` or `f(|x|)={(-1",",-2 le |x| le 0),(|x|-1",",0 le |x| le 2):}` `=|x|-1,0 le |x| le 2` ` " " `(As `-2 le |x| lt 0` is not possible) `={(-x-1",",-2 le x le 0),(x-1",",0 lt x le 2):} " (1)" ` Again, `f(x)={(-1",",-2 le x le 0),(x-1",",0 lt x le 2):}` or `|f(x)|={(|-1|",",-2 le x le 0),(|x-1|",",0 lt x le 2):}` or `|f(x)|={(1",",-2 le x le 0),(-(x-1)",",0 lt x le 1),(+(x-1)",",1 lt x le 2):} " (2)" ` Therefore, `g(x)=f(|x|)+|f(x)|` can be expressed as `g(x)={((-x-1)+1",",-2 le x le 0),((x-1)+(1-x)",",0 lt x le 1),((x-1)+(x+1)",",1 lt x le 2):}` [Using (1) and (2) ] `={(-x",",-2 le x le 0),(0",",0 lt x le 1),(2(x-1)",",1 lt x le 2):}` |
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| 77. |
`f(x)={(log_(e)x",",0lt x lt 1),(x^(2)-1 ",",x ge 1):}` and `g(x)={(x+1",",x lt 2),(x^(2)-1 ",",x ge 2):}`. Then find `g(f(x)).` |
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Answer» Correct Answer - `g(f(x))={(1+"In "x",",0lt x lt 1),(x^(2)",",1 le x lt sqrt(3)),((x^(2)-1)^(2)-1 ",",x ge sqrt(3)):}` `f(x)={(log_(e)x",",0lt x lt 1),(x^(2)-1 ",",x ge 1):}` and `g(x)={(x+1",",x lt 2),(x^(2)-1 ",",x ge 2):}` `g(f(x))={(f(x)+1",",f(x) lt 2),((f(x))^(2)-1 ",",f(x) ge 2):}` `={(log_(e)x+1",",log_(e)x lt 2","0lt x lt 1),(x^(2)-1+1",",x^(2)-1 lt 2","x ge 1),((log_(e)x)^(2)-1",",log_(e)x ge 2","0lt x lt 1),((x^(2)-1)^(2)-1 ",",x^(2)-1 ge 2"," x ge 1):}` `={(log_(e)x+1",", x lt e^(2)","0lt x lt 1),(x^(2)",",-sqrt(3)lt x lt sqrt(3)","x ge 1),((log_(e)x)^(2)-1",",x ge e^(2)","0lt x lt 1),((x^(2)-1)^(2)-1 ",",x le -sqrt(3) " or "x ge sqrt(3)"," x ge 1):}` `={(log_(e)x+1",",0lt x lt 1),(x^(2)",",1 le x lt sqrt(3)),((x^(2)-1)^(2)-1 ",",x ge sqrt(3)):}` |
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| 78. |
Find the value of `x^(2)` for the following values of x: (a) `[-5,-1] " (b) "(3,6) " (c ) "(-2,3]` (d) `(-3,oo) " (e ) "(-oo,4)` |
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Answer» Correct Answer - (a) `[1,5] " (b) "(9,36)" (c ) " [0,9] " (d) " [0, oo) " (e ) "[0, oo)` (a) `-5 le x le -1` `implies x^(2) in [1,5]` (b) `3 lt x lt 6` (c ) `-2 lt x le 3` `implies -2 lt x le 3` For `-2 lt x lt 0, x^(2) in (0,4) " (1)" ` and for `0 le x le 3, x^(2) in [0,9] " (2) ` From (1) and (2), `x^(2) in [0,9] ` Alternatively, ` x in (-2,3],` now least value of `x^(2)` is 0 which occurs when `x=0` Greatest value of `x^(2)` is 9 for `x=3` `implies x^(2) in [0,9]` (d) `(-3,oo)` Here least value of `x^(2)` is 0 for `x = 0`, and when x goes up to infinity, `x^(2)` also goes up to infinity `implies x^(2) in [0,oo)` (e) `(-oo,4)` Here least value of `x^(2)` is 0 for `x = 0` and `x^(2) to oo`, when `x to -oo` Hence, `x^(2) in [0,oo)` |
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| 79. |
Let there be n circles no three of which have collinear centres. If the total number of radical axes is equal to total number of possible radical centres, then possible value of n is |
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Answer» ` ^n C_2 = ^n C_3 ` `(n(n-1))/2=(n(n-1)(n-2))/(3xx2)` `n=5` answer`(3)` |
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| 80. |
Find all possible values of the following expressions : `(i) sqrt(x^(2)-4) (ii) sqrt(9-x^(2)) (iii) sqrt(x^(2)-2x+10)` |
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Answer» (i) `sqrt(x^(2)-4)` Least value of square root is 0 when `x^(2)=4 " or " x= +-2.` Also `x^(2)-4 ge 0` Hence, `sqrt(x^(2)-4) in [0, oo).` (ii) `sqrt(9-x^(2))` Least value of square root is 0, when `9-x^(2)=0` or `x= +-3.` Also, the greatest value of `9-x^(2)` is 9, when `x = 0`. Hence, we have `0 le 9 - x^(2) le 9 impliessqrt(9-x^(2)) in [0,3].` (iii)`sqrt(x^(2)-2x+10)=sqrt((x-1)^(2)+9)` Here, the least value of `sqrt((x-1)^(2)+9)` is 3 , when `x-1=0`. Also `(x-1)^(2)+9 ge 9impliessqrt((x-1)^(2)+9) ge 3` Hence, `sqrt(x^(2)-2x+10) in [3, oo).` |
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| 81. |
Let `f(x)={(2x+a",",x ge -1),(bx^(2)+3",",x lt -1):}` and `g(x)={(x+4",",0 le x le 4),(-3x-2",",-2 lt x lt 0):}` `g(f(x))` is not defined ifA. `a in (10,oo), b in (5,oo)`B. `a in (4,10), b in (5,oo)`C. `a in (10,oo), b in (0,1)`D. `a in (4,10), b in (1, 5)` |
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Answer» Correct Answer - A `g(f(x))` is not defined if (i) `-2+a gt 8 and " (ii) "b+3 gt 8` `or a gt 10 and b gt 5` |
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| 82. |
Find the values of 1/x for the following values fo x: (a) `(2,5) " (b) "[-5,-1) " (c ) "(3,oo)` (d) `(-oo,-2] " (e ) "[-3,4]` |
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Answer» Correct Answer - (a) `((1)/(5),(1)/(2)) " (b) "(-1,-(1)/(5)]" (c ) " (0,(1)/(3)) " (d) " [-(1)/(2), 0) " (e ) "(- oo, -(1)/(3)] cup [(1)/(4),oo)` (a) `2 lt x lt 5` `implies (1)/(2) gt (1)/(x) gt (1)/(5)` `implies (1)/(x) in ((1)/(5),(1)/(2))` (b) `-5 le x lt -1` `implies -(1)/(5) ge (1)/(x) gt -1` `implies (1)/(x) in (-1,-(1)/(5)]` (c ) ` x gt 3 or 3 lt x lt oo` `implies(1)/(3) gt (1)/(x) gt 0` `(1)/(x) in (0,(1)/(3))` (d) `x le -2 or -oo lt x le -2` `implies 0 gt (1)/(x) ge -(1)/(2)` `implies (1)/(x) in [-(1)/(2),0)` (e) ` x in [-3,4] or -3 le x le 4` Now for `(1)/(x)` to get defined, we must have `-3 le x lt 0 or 0 lt x le 4` `implies -(1)/(3) ge (1)/(x) gt -oo or (1)/(4) le (1)/(x) lt oo` `implies (1)/(x) in (-oo,-(1)/(3)] cup [(1)/(4),oo)` |
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| 83. |
If ` f(x) ={(x^(3)", " x lt1),(2x-1", " x ge 1):} " and " g(x)={(3x", " x le2),(x^(2)", " x gt 2):} " then find " (f-g)(x).` |
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Answer» ` f(x) ={(x^(3)", " x lt1),(2x-1", " 1 le x le 2),(2x-1", " x gt 2):} " and " g(x)={(3x", " x lt1),(3x", "1 le x le2),(x^(2)", " x gt 2):}` ` implies (f+g)(x) ={(x^(3)+3x", " x lt1),(2x-1+3x", " 1 le x le 2),(2x-1+x^(2)", " x gt 2):}` `={(x^(3)+3x", " x lt1),(5x-1", " 1 le x le 2),(x^(2)+2x-1", " x gt 2):}` |
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| 84. |
8. Show that the points A (1,-2,-8), B 0,-2) and C (11, 3, 7) are collinear and find the ratio in which B divides AC. |
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Answer» `m/n=(x-1)/(11-x)=2/3` `m/n=(x-1)/(11-x)=2/3` `m/n=2/3=(x-1)/(11-x)` `22-2x=3x-3` `25=5x,x=5`. |
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| 85. |
Find the value of `x`for which following expressions are defined:`1/(sqrt(x-|x|))`(ii) `1/(sqrt(x+|x|))` |
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Answer» ` "(i) " x-|x|= {(x-x=0", if " x ge 0),(x+x=2x", if "x lt 0):}` `impliesx-|x| le 0,` for all `x` Thus, `(1)/(sqrt(x-|x|))` does not take real values for any `x in R` So, `(1)/(sqrt(x-|x|))` is not defined for any `x in R.` ` "(ii) " x+|x|= {(x+x=2x", if " x ge 0),(x-x=0", if "x lt 0):}` Thus, `(1)/(sqrt(x+|x|))` is defined only when `x gt 0` |
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| 86. |
Find all possible values of expressions `(2+x^2)/(4-x^2)` |
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Answer» Correct Answer - `(-oo, -1)cup[(1)/(2),oo)` `(2+x^(2))/(4-x^(2))=(6-(4-x^(2)))/(4-x^(2))=(6)/(4-x^(2))=1` Now, `x^(2) ge 0` ` :. -x^(2) le 0` `implies 4-x^(2) le 4` `implies (4-x^(2))/(6) le (2)/(3)` `implies -oo lt (4-x^(2))/(6) lt 0 " or "0 lt(4-x^(2))/(6) le (2)/(3)` `implies 0 gt (6)/(4-x^(2)) gt -oo gt (4-x^(2))/(6) ge (3)/(2)` `implies (6)/(4-x^(2)) in (-oo,0) cup [(3)/(2),oo)` `implies ((6)/(4-x^(2))-1) in (-oo,-1) cup [(1)/(2),oo)` |
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| 87. |
Find all possible values ( range) of the following quadratic expressions when `x in R` and when `x in [-3,2]` (a) `4x^2+28x+41` (b) `1+6x -x^2` |
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Answer» Correct Answer - (a) `[-8,oo)` when `x in R; [-7,113]` when `x in [-3,2]` (b) `(-oo,10]` when `x in R;[-26,9]` when ` x in [-3,2]` (a) `4x^(2)+28x+41=(2x+7)^(2)-8` Now,`(2x+7)^(2)-8 ge 0 AA x in R.` `implies (2x+7)^(2)-8 ge -8 AA x in R` So, the values of the expression are `[-8, oo).` For ` x in [-3,2]` `-3 le x le 2` `implies -6 le 2x le 4` `implies 1le 2x+7 le 11` `implies 1 le (2x+7)^(2) le 121` `implies -7 le (2x+7)^(2)-8 le 113` Thus, for `x in [-3,2],` the values of the expression are `[-7,113].` (b) `1+6x-x^(2)=10-(x-3)^(2)` Now, `(x-3)^(2) ge 0 AA x in R.` `implies -(x-3)^(2) le 0 AA x in R` `implies 10-(x-3)^(2) le 10 AA x in R` So, the value of the expression are `(-oo,10].` For ` x in [-3,2]` ` -3 le x le 2` `implies -6 le x -3 le -1` `implies 1 le (x-3)^(2) le 36` `implies -36 le -(x-3)^(2) le -1` `implies -26 le 10-(x-3)^(2) le 9` Thus, for ` x in [-3,2]` values of the expression are `[-26,9].` |
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| 88. |
Find the value of `x^(2)` for the given values of x. `(i) x lt 3 (ii) x gt -1 (iii) x ge 2 (iv) x lt -1` |
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Answer» (i) When `x lt 3`, we have `x in (-oo,0)cup[0,3)` for `x in [0,3),x^(2) in [0,9)` for `x in (- oo,0),x^(2) in (0,oo)` `implies "for " x lt 3, x^(2) in [0,9) cup(0,oo)` `implies x in [0,oo)` (ii) When `x gt -1`, we have `x in (-1,0)cup[0,oo)` for `x in [-1,0),x^(2) in [0,1)` for `x in [-0,oo),x^(2) in [0,oo)` `implies "for " x gt -1, x^(2) in (0,1) cup(0,oo)` `implies x in [0,oo)` (iii) Here `x in [2,oo)` `implies x^(2) in {4, oo)` Here `x in (-oo, -1)` `implies x^(2) in (1, oo)` |
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| 89. |
For `2 lt x lt 4` find the values of |x|. (ii) For `-3 le x le -1`, find the values of |x|. (iii) For `-3 le x lt 1,` find the values of |x| (iv) For `-5 lt x lt 7 ` find the values of |x-2| (v) For `1 le x le 5` find fthe values of |2x -7| |
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Answer» (i) `2 lt x lt 4` i.e., values on real number line whose distance from zero lies between 2 and 4. Here values of x are positive `implies|x| in (2,4)` (ii) `-3 le x le -1` Here values on real number line whose distance from zero lies between 1 and 3 or equal to 1 or 3. `implies 1 le |x| le 3` (iii) ` -3 le x lt 1` For `-3 le x lt 0, |x| in (0,3]` For ` 0 le x lt 1, |x| in [0,1)` So for `-3 le x lt 1,|x| in [0,1)cup(0,3]" or " |x| in [0,3]` (iv) `-5 lt x lt 7` or `-7 lt x-2 lt 5` `implies 0 le |x-2| le 7` (v) `1 le x le 5` or `2 le2x le10` `implies -5 le 2x-7 le 3` `implies |2x-7| in[0,5]` |
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| 90. |
maximize `Z=25x_1+20x_2` subject to constraints`3x_1+2x_2 |
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Answer» `2x_1+1050=1400` `x_1=175` `700+x_2=1400` `x_2=100` `Z=25x_1+20x_2`at A(0,150) Z=0+3000=3000 at B(175,150) Z=25*175+20*150=7375 at C(350,100) Z=350*25+100*20=10750 at D(350,0) Z=350*25=8750 Z is maximum at C |
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| 91. |
Solve the following : (i) `|x-2|=(x-2) " (ii) " |x+3| = -x-3` (iii) `|x^(2)-x|=x^(2)-x " (iv) " |x^(2)-x-2| =2+x-x^(2)` |
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Answer» (i) `|x-2|=(x-2), " if " x-2 ge 0 " or " x ge 2` (ii) `|x+3|= -x-3, " if " x+3 le 0 " or " x le -3` (iii) `|x^(2)-x|=x^(2)-x, " if " x^(2)-x ge 0` or `x(x-1) ge 0` `implies x in (-oo,0] cup [1,oo)` (iv) `|x^(2)-x-2|=2+x-x^(2)` or `x^(2) -x-2 le 0` or `(x-2)(x+1)le0` or `-1 le x le 2` |
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| 92. |
For `x in R ,`find all possible values of`|x-3|-2`(ii) `4-|2x+3|` |
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Answer» We know that `|x-3| ge 0 AA x in R` `implies|x-3|-2 ge -2` `implies |x-3|-2 in [-2, oo)` (ii) We know that `|2x+3| ge 0 AA x in R` `implies -|2x+3|le 0` `implies 4-|2x+3|le4` or `4-|2x+3| in (-oo,4]` |
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| 93. |
Prove that`sqrt(x^2+2x+1)-sqrt(x^2-2x+1)={-2, x1` |
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Answer» `sqrt(x^(2)+2x+1)-sqrt(x^(2)-2x+1)` `=sqrt((x+1)^(2))-sqrt((x-1)^(2))` `=|x+1|-|x-1|` `= {(-x-1-(1-x)", "x lt -1),(x+1-(1-x)", "-1le x le1),(x+1-(x-1)", "x gt1):}` `= {(-2", "x lt -1),(2x","-1le x le1),(2", "x gt1):}` |
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| 94. |
Find the inverse of`f(x)={x ,4` |
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Answer» Given `f(x)={(x",", x lt1),(x^(2)",",1le x le 4),(8sqrt(x)",", x gt 4):}` Let `f(x)=y` or `x=f^(-1)(y) " (1) " ` ` :. x={(y",", y lt1),(sqrt(y)",",1le sqrt(y) le 4),(y^(2)//64",", y^(2)//64 gt 4):}={(y",", y lt1),(sqrt(y)",",1le y le 16),(y^(2)//64",", y gt 16):},` `f^(-1)(y)={(y",", y lt1),(sqrt(y)",",1le y le 16),(y^(2)//64",", y gt 16):}, " [From (1)]" ` Hence, `f^(-1)(y)={(x",", x lt1),(sqrt(x)",",1le x le 16),(x^(2)//64",", x gt 16):},` |
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| 95. |
Find the inverse of the function `f:[-(pi)/(2)-"tan"^(_1)(3)/(2),(pi)/(2)-"tan"^(-1)(3)/(4)] to [-1,1],` `f(x)=3cosx+4sinx +7.` |
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Answer» We have, `f(x)= y=3cosx +4sinx+7` `implies y=sqrt(3^(2)+4^(2))[(3)/(sqrt(3^(2)+4^(2)))cosx+(4)/(sqrt(3^(2)+4^(2)))sinx]+7` `=5[(3)/(5)cosx+(4)/(5)sinx]+7` `=5[sin theta cos x+cos theta sinx]+7," where " tan theta=(3)/(4)` `=5 sin(x+theta)+7` `implies y-7=5sin(x+theta)` `implies (y-7)/(5)=sin(x+theta)` `implies "sin"^(-1)(y-7)/(5)=x+theta` `implies x="sin"^(-1)(y-7)/(5)-theta` `f^(-1)(x)="sin"^(-1)(x-7)/(5)-"tan"^(-1)(3)/(4)` |
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| 96. |
Find the inverse of the function `f: [-1,1] to [-1,1],f(x) =x^(2) xx sgn (x).` |
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Answer» `f(x) =x^(2) xx sgn (x)` `={(x^(2)(1)",", x gt 0),(0"," , x=0),(x^(2)(-1)",", x lt 0):}` ` :. f(x) ={(x^(2)",", 0le x le1),(-x^(2)",", -1 le x lt0):}` Now `y=x^(2),0 le x le 1` Since `y in [0,1],` we have `x=sqrt(y)` For `y=-x^(2),-1 le x le 0.` Since `y in [-1,0], " we have " x= -sqrt(-y)` Thus, `f^(-1)(x)={(sqrt(x)",", 0le x le1),(-sqrt(-x)",", -1 le x lt0):}` |
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| 97. |
Which of the following function/functions is/are periodic ? (a) `sgn(e^(-x)) " (b) " sinx + |sinx|` (c ) `min(sinx, |x|) " (d) " (x)/(x)` |
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Answer» Correct Answer - a,b,c (a) `f(x) = sgn (e^(-x))=1 " as " e^(-x) gt 0 AA x in R`. Therefore, `f(x)` is periodic function. (b) `g(x)=sinx +|sin x|` is periodic with period `2pi` as `f(x+2pi)=f(x).` (c ) `h(x) = min(sinx, |x|)=sinx,` which is periodic. (d) `p(x)=(x)/(x)=1, x ne 0, ` which is nonperiodic. |
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| 98. |
If `cos^(-1) (x/2) + cos^(-1) (y/3) =alpha` then prove that `9x^2-12xycosalpha+4y^2=36sin^2alpha`. |
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Answer» `cos^-1(x/2)+cos^-1(y/3) = alpha` `=>cos^-1((x/2)(y/3) - sqrt(1-(x^2)/4)sqrt(1-y^2/9)) = alpha` `=>((xy)/6) - sqrt((4-x^2)/4)sqrt(9-y^2)/9) = cosalpha` `=>xy - sqrt(4-x^2)sqrt(9-y^2) = 6cos alpha` `=>xy - 6cos alpha= sqrt(4-x^2)sqrt(9-y^2)` Squaring both sides, `=>x^2y^2+36 cos^2 alpha - 12xy cosalpha = (4-x^2)(9-y^2)` `=>x^2y^2+36 cos^2 alpha - 12xy cosalpha = 36-4y^2-9x^2+x^2y^2` `=> 36(1-sin^2 alpha) - 12xy cosalpha = 36-4y^2-9x^2+x^2y^2` `=>9x^2+4y^2- 12xy cosalpha = 36sin^2alpha` |
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| 99. |
Range ofthe function `f(x)=cos(Ksin x)` is `[-1,1]`, then the least positive integral value of K will beA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - D `y=cos(K sinx)` `(cos^(-1)y)/(K)=sinx` `implies -1 le (cos^(-1)y)/(K) le 1` `implies -K le cos^(-1)y le K` Now `cos ^(-1) y in [0,pi]` `implies K=4` |
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| 100. |
If `f(x)=(h_1(x)-h_1(-x))(h_2(x)-h_2(-x))(h_(2n+1)(-x)a n df(200)=0,`then prove that `f(x)`is many one function. |
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Answer» `f(-x)=(h_(1)(-x)-h_(1)(x))(h_(2)(-x)-h_(2)(x))…(h_(2n+1)(-x)-h_(2n+1)(x))` ` :. f(-x)=(-1)^(2n+1)f(x)=-f(x)` or `f(x)+f(-x)=0` So, f(x) is odd. Therefore, `f(-200)=-f(200)=0` So, f(x) is many-one. |
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