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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.
| 551. |
A dice is thrown repeatedly then probability getting 7 times digit 5 in `12^(th)` trial is equal to :- |
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Answer» Required probability+ `.^11C_6(1/6)^6*(5/6)^5*(1/6)^` `= .^11C_6(1/6)^7*(5/6)^5` Answer |
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| 552. |
The range of the values of p for which the equation ` sin cos^(-1) ( cos (tan^(-1) x)) = p` has a solution is |
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Answer» We know, `-pi/2 lt tan^-1x lt pi/2` `:. 0 lt cos(tan^-1x) le 1` So, ` 0 le cos^-1(cos(tan^-1x)) lt pi/2` ` 0 le sin(cos^-1(cos(tan^-1x))) lt 1` So, range of given expression is `[0,1)`. |
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| 553. |
Determine the minimum number of dice when we throw all of them we have at least a 50% chance of getting 6. |
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Answer» P(1 time6)+P(2 time 6)...+P(n times 6)`>=1/2` `1-P(no6)>=1/2` `1-(1-1/6)^n>=1/2` `1-1/2>=(5/6)^n` `1/2>=(5/6)^n` `1/2=(5/6)^n` `-log2=n(log5-log6)` `-3.0=n(log10-log2-log2-log3)` `30=n(1-0.3-0.3-0.479)` `n=3.79` `n>=3.79` `n=4` |
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| 554. |
If `f(x)=ax+b` , where a and b are integers, f(-1)=-5 and f(3)=3 then a and b are equal toA. a=-3,b=-1B. a=2,b=-3C. a=0,b=2D. a=2,b=3 |
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Answer» Correct Answer - B We have f(x)=ax+b f(-1)=a(-1)+b -5=-a+b `" "....(i)` and, f(3)=a(3)+b 3=3a+b `" "....(ii)` On solving Eqs. (i) and (ii), we get a=2 and b=-3 |
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| 555. |
`y = f(x) = (ax-b)/(cx-a)` then f(y) = ? |
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Answer» We have,`f(x)=y=(ax-b)/(cx-a)` `:. f(y) =(ay-b)/(cy-a)=(a((ax-b)/(cx-a))-b)/(c((ax-b)/(cx-a))-a)` `=(a(ax-b)-b(cx-a))/(c(ax-b)-a(cx-a))=(a^(2)x-ab-bcx+ab)/(acx-bc-acx+a^(2))` `=(a^(2)x-bcx)/(a^(2)-bc)=(x(a^(2)-bc))/((a^(2)-bc))=x` `:. f(y) =x` |
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| 556. |
Domain of `sqrt(a^(2)-x^(2))(agt0)` isA. (a-,a)B. [a-,a]C. [0-,a]D. [a-,0] |
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Answer» Correct Answer - B Let `sqrt(a^(2)-x^(2))` f(x) is defined, if `a^(2)-x^(2)ge0` `rArrx^(2)-a^(2)le0` `rArr (x-a)(x+a)le-0` `rArr -alex lea` `:.` Domain of f=[-a,a] `" " [ :. Agt0]` |
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| 557. |
Let `A={x_1,x_2,......X_m},B={y_1,y_2,.....,y_n}` then total number of non-empty relations that can bedefined from A to B, isA. `m^(n)`B. `n^(m)-1`C. mn-1D. `2^(mn)-1` |
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Answer» Correct Answer - D We have, `n(A)=m and n (B) =n` `n(AxxB)=n(A)*n(B)` Total number of releation from A to B =`2^(mn)-1=2^(n(AxxB)^(-1))-1` |
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| 558. |
If `[x]^2- 5[x]+6= 0`, [x] denote the greatest integer function, thenA. `x in[3,4]`B. `x in (2,3]`C. `x in [2,3]`D. `x in [2,4]` |
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Answer» Correct Answer - C We have `[x]^(2)-5[x]+6=0` `rArr [x]^(2)-3[x]-2[x]+6=0` `rArr [x]([x]-3)-2([x]-3)=0` `rArr([x)-3)([x]-2)=0` `rArr [x]=2,3]` `:. "x" in[2,3]` |
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| 559. |
let `f(x)=sqrt(1+x^2)` thenA. ` f(xy)=f(x)*f(y)`B. `f(xy)gef(x)*f(y)`C. `f(xy)lef(x)*f(y)`D. None of these |
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Answer» Correct Answer - C We have, `f(x)=sqrt(1+x^(2))` `f(xy)=sqrt(1+x^(2)y^(2))` `f(x)*f(y)=sqrt(1+x^(2))*sqrt(1+y^(2))` `=sqrt((1+x^(2))(1+y^(2)))` `sqrt(1+x^(2)+y^(2)+x^(2)+y^(2))` `:. sqrt(1+x^(2)y^(2))lesqrt(1+x^(2)+y^(2)+x^(2)+y^(2))` `rArr f(xy)lef(x)*f(y)` |
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| 560. |
Find the domain and range of the function `f(x)="sin"^(-1)(x^(2))/(2)` |
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Answer» `f(x)="sin"^(-1)(x^(2))/(2)` We must have `o le (x^(2))/(2) le 1` `implies 0le x^(2) le 2` `implies -sqrt(2) le x le sqrt(2)` So, domain is `[ -sqrt(2),sqrt(2)]` Now, ` 0 le (x^(2))/(2) le 1` `implies sin^(-1) 0 le "sin"^(-1)(x^(2))/(2) le sin^(-1)1` `implies 0 le "sin"^(-1)(x^(2))/(2) le pi//2` Hence range is `[0,pi//2]`. |
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| 561. |
Range of `f(x) =1/(1-2cosx)` isA. `[(1)/(3),1]`B. `[-1,(1)/(3)]`C. `(-oo,-1)uu[(1)/(3),oo)`D. `[-(1)/(3),1]` |
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Answer» Correct Answer - B We known that , `-1le-cosxle1` `rArr-2le-2cosxle2` `rArr 1-2le1-2cos le 1+2` `rArr -1le1-2cos xle3` `rArr-1le(1)/(1-2cosx)le(1)/(3)` `rArr-1lef(x)le(1)/(3)` `:.` Range of `f=[-1,(1)/(3)]` |
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| 562. |
Domain of the function f(x)=`sin^(- 1)(1+3x+2x^2)` |
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Answer» `f(x) = sin^(-1)(1+3x+2x^(2))` We must have ` -1 le 1 +3x +2x^(2) le 1` Now `2x^(2)+3x +1 ge -1` `implies 2x^(2)+3x +2 ge 0`, which is always true as discriminant ` le 0`. Also, `2x^(2) +3x +1 le 1` `implies 2x^(2) +3x le 0` `implies x(2x+3) le 0` `implies (-3)/(2) le x le 0` `implies x in [-(3)/(2),0]` |
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| 563. |
Find the domain and range of the function `f(x) = sin^(-1)((1+e^(x))^(-1))`. |
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Answer» `f(x) = sin^(-1)((1+e^(x))^(-1))` We know that `e^(x) gt 0` for all real x. ` :. e^(x) +1 gt 1` `implies 0 lt (1)/(1+e^(x)) lt 1` So, f(x) is defined for all real values of x. Hence domain is R. Also `0 lt (1)/(1+e^(x)) lt 1` `implies sin^(-1)0 lt "sin"^(-1)(1)/(1+e^(x)) lt sin^(-1)1` `implies 0 lt "sin"^(-1)(1)/(1+e^(x)) lt (pi)/(2)` Therefore, range of the function is `(0, (pi)/(2))` |
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| 564. |
Find the values of x for which the following pair of functions are identical. (i) `f(x)=tan^(-1)x+cot^(-1)x " and " g(x)=sin^(-1)x +cos^(-1)x` (ii) `f(x)=cos(cos^(-1)x) " and " g(x)=cos^(-1)(cosx)` |
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Answer» (i) `f(x)=tan^(-1)x+cot^(-1)x ,x in R.` ` " and " g(x)=sin^(-1)x +cos^(-1)x=(pi)/(2) " for " x in [-1,1]` Hence functions are identical if ` x in [-1,1].` (ii) `f(x)=cos(cos^(-1)x)` is defined if `cos^(-1)x` is defined for which ` x in [-1,1].` ` g(x)=cos^(-1)(cosx)` is defined if `cosx in [-1,1],` which is true `AA x in R.` So, domain is R. Thus, common domain of two functions is `[-1,1].` ` :. f(x) =x " and " g(x) =x " for " [-1,1].` |
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| 565. |
Find the range of `f(x)=1/(2cosx-1)` |
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Answer» `-1 le cosx le 1` or ` -2 le 2 cosx le 2` or ` -3 le 2cos x -1 le 1` For `(1)/(2cosx-1),` `-3 le 2 cosx -1 lt 0 " or " 0 lt 2 cosx -1 le1` i.e., `-oo lt (1)/(2cosx-1) le (-1)/(3) " or " 1 le (1)/(2cosx-1) lt oo` Hence, the range is `(-oo,-(1)/(3)] cup [1, oo).` |
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| 566. |
The range of function `f(x)=^(7-x)P_(x-3)i s`{1,2,3}(b) `{1,2,3,4,5,6}``{1,2,3,4}`(d) `{1,2,3,4,5}`A. `{1,2,3}`B. `{1,2,3,4,5,6}`C. `{1,2,3,4}`D. `{1,2,3,4,5}` |
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Answer» Correct Answer - A We have `f(x)=""^(7-x)P_(x-3)=((7-x)!)/((10-2x)!)` We must have `7-x gt 0, x ge 3, " and " 7-x ge x-3.` Therefore, `x lt 7, x ge 3, " and " x le 5` or ` 3 le x le 5` or ` x =3,4,5` Now, `f(3)=(4!)/(4!)=1, f(4)=(3!)/(2!)=3,f(5)=(2!)/(0!)=2.` Hence, `R_(f)={1,2,3}.` |
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| 567. |
Let R be a relation from A {1, 2, 3, 4) to B (1, 3, 5) such that `R{ (a, b):a < b, `where `a in A and b in B)` then what is `ROR^(-1)` equal to? |
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Answer» `A -> B` `R : { (1,3),(1,5),(2,3),(2,5),(3,5),(4,5)}` `R^-1 : {(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)}` `B= {1,3,5}` `y= ROR^-1` `x=1` `x=3` When x=3; `R(1) -> (3,3)(3,5)` `R(2) -> (3,3)(3,5)` when x=5 ; `R(1) -> (5,3),(5,5)` `R(2) -> (5,3),(5,5)` `R(3) -> (5,5)` `R(4) -> (5,5)` so, `ROR^-1 = {(3,3)(3,5)(5,3)(5,5)}` Answer |
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| 568. |
if:`f(x)=(sinx)/(sqrt(1+tan^2x))-(cosx)/(sqrt(1+cot^2x)),`then find the range of `f(x)` |
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Answer» `f(x)=(sin x)/(|sec x|)-(cos x)/(|"cosec"x|)=sin x |cos x|- cos x|sin x|` Clearly, the domain of f(x) is `R~{(n pi)/(2),n in Z}` and the period of `f(x)` is `2 pi`. `f(x)={(0", "x in (0,pi//2)),(-sin2x", "x in (pi//2,pi)),(0", "x in (pi,3pi//2)),(sin2x", "x in (3pi//2,2pi)):}` The range of `f(x)` is `[-1,1].` |
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| 569. |
The range of `sin^(-1)[x^2+1/2]+cos^(-1)[x^2-1/2]`, where [.] denotes the greatest integer function, is`{pi/2,pi}`(b) `{pi}`(c) `{pi/2}`(d) none of theseA. `{(pi)/(2),pi}`B. `{pi}`C. `{(pi)/(2)}`D. None of these |
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Answer» Correct Answer - B `[x^(2)+(1)/(2)]=[x^(2)-(1)/(2)+1]=1+[x^(2)-(1)/(2)].` Thus, from domain point of view, `[x^(2)-(1)/(2)]=0,-1 or [x^(2)+(1)/(2)]=1,0` i.e., `f(x)=sin^(-1)(1)+cos^(-1)(0) or sin^(-1) (0) + cos^(-1)(-1)` ` or f(x)={pi}` |
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| 570. |
The range of `f(x)=sin^(-1)((x^2+1)/(x^2+2))`is`[0,pi/2]`(b) `(0,pi/6)`(c) `[pi/6,pi/2]`(d) none of theseA. `[0,pi//2]`B. `(0,pi//6)`C. `[pi//6,pi//2)`D. None of these |
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Answer» Correct Answer - C Here, `(x^(2)+1)/(x^(2)+2)=1-(1)/(x^(2)+2)` Now, `2 le x^(2)+2 lt oo " for all " x in R` or ` (1)/(2) ge (1)/(x^(2)+2) gt 0` or `-(1)/(2) le (-1)/(x^(2)+2) lt 0` or ` (1)/(2) le 1- (1)/(x^(2)+2) lt 1` or ` (pi)/(6) le sin^(-1)(1- (1)/(x^(2)+2)) lt (pi)/(2)` |
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| 571. |
Find the range of the function `f(x)=3 sin (sqrt((pi^(2))/(16)-x^(2))).` |
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Answer» We must have `(pi^(2))/(16)-x^(2) ge0` Also, `(pi^(2))/(16)-x^(2) le (pi^(2))/(16)` `implies 0 le(pi^(2))/(16)-x^(2) le (pi^(2))/(16)` `implies 0 le sqrt((pi^(2))/(16)-x^(2)) le (pi)/(4)` `implies 0 le sinsqrt((pi^(2))/(16)-x^(2)) le (1)/(sqrt(2))` `implies 0 le 3sinsqrt((pi^(2))/(16)-x^(2)) le (3)/(sqrt(2))` Hence, range is `[0,(3)/(sqrt(2))]` |
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| 572. |
Find the value of `x`for which function are identical.`f(x)=cosxa n dg(x)=1/(sqrt(1+tan^2x))` |
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Answer» `f(x)=cos x` has domain R and range `[-1,1]` `g(x)=(1)/(sqrt(1+tan^(2)x))=(1)/(sqrt(sec^(2)x))=|cosx|,` has domain `R-{(2n+1)pi//2, n in Z` as `tan x` is not defined for `x=(2n+1)pi//2, n in Z` Also range of `g(x)=|cosx|` is `[0,1]` Hence f(x) and g(x) are identical if x lies in first and fourth quadrant. `implies x in underset(n in Z)(cup)(-(pi)/(2)+2n pi,(pi)/(2)+2n pi)` |
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| 573. |
Let `f(x)={(x^(2)-4x+3",",x lt 3),(x-4",",x ge 3):}`and `g(x)={(x-3",",x lt 4),(x^(2)+2x+2",",x ge 4):}`. Describe the function `f//g` and find its domain. |
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Answer» Correct Answer - `(f(x))/(g(x))={((x^(2)-4x+3)/(x-3)",",x lt 3),((x-4)/(x-3)",",3lt x lt 4),((x-4)/(x^(2)+2x+2)",",x ge 4):} ` `f(x)={(x^(2)-4x+3",",x lt 3),(x-4",",x ge 3):}` `={(x^(2)-4x+3",",x lt 3),(x-4",",3le x lt 4),(x-4",",x ge 4):} " (1)" ` `g(x)={(x-3",",x lt 4),(x^(2)+2x+2",",x ge 4):}` `={(x-3",",x lt 3),(x-3",",3 le x lt 4),(x^(2)+2x+2",",x ge 4):} " (2)" ` From (1) and (2), we have `(f(x))/(g(x))={((x^(2)-4x+3)/(x-3)",",x lt 3),((x-4)/(x-3)",",3lt x lt 4),((x-4)/(x^(2)+2x+2)",",x ge 4):} ` Clearly, `f(x)//g(x)` is not defined at `x=3`. Hence, the domain is `R-{3}.` |
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| 574. |
A quadrilateral has vertices `(4,1), (1,7),(-6,0)` AND `(-1,-9)`. Show that mid-points of the sides of this quadrilateral form a parallelogram. |
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Answer» Mid point of PR`=(((5/2)+(-7/2))/2,((4+(-9/2))/2)` `=(-1/2,-1/4)` MId point of SQ`=(((3/2)-(5/2))/2),((-4+(7/2))/2)` `=(-1/2,-1/4)` `PQ=sqrt(5^2+(1/2)^2)=sqrt(25+1/4)=sqrt(101/4)` `SR=sqrt(5^2+(1/2)^2)=sqrt(25+1/4)+sqrt(101/4)` `RQ=sqrt((-7/2+5/2)^2+(-9/2-7/2)^2)=sqrt65` `SP=sqrt((3/2-5/2)^2+(-8)^2)=sqrt65` `PQ=SR` `RQ=SP`. |
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| 575. |
If `x_1, x_2, x_3, x_n` are the roots of `x^n+ax+b=0` then the value of `(x_1-x_2)(x_1 -x_3) (x_1-x_4)...... (x_1-x_n)` is equal to |
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Answer» `P(x)=(x-x_1)(x-x_2)(x-x_3)...(x-x_n)=x^n+ax+b` `lim_(x->x_1)(x-x_2)(x-x_3)...(x-x_n)=lim_(x->x_1)((x^n+ax+b)/(x-x_1))` `(x-x_2)(x-x_3)...(x_1-x_n)=lim_(x->x_1)(x^n+ax+b)/(x-x_1)` `lim_(x->x_1)(nx^(n-1)+a)/1` `nx_1^(n-1)+a` option B is correct. |
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| 576. |
The range of `f(x)=[sinx|[cosx[tanx[secx]]]],x in (0,pi/4),w h e r e`[.] denotes the greatest integer function less than or equal to `x ,`is(0,1) (b) `-{1,0,1}``{1}`(d) none of these |
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Answer» Given `f(x)=[sin x+[cosx+[tanx+[secx]]]]` `=[sin+p], " where " P=[cosx +[tanx+[secx]]]]` `=[sinx]+p,` (as p is an integer) `=[sinx]+[cosx+[tanx+[secx]]]]` `=[sinx]+[cosx]+[tanx]+[secx]` Now, for ` x in(0,pi//4),sinx in(0, (1)/(sqrt(2))), cosx in((1)/(sqrt(2)),1), ` `tanx in(0,1), secx in (1, sqrt(2))` or `[sinx]=0,[cosx]=0,[tanx]=0, " and " [secx]=1` Therefore, the range of f(x) is 1. |
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| 577. |
A bag contains twelve pairs of socks and four socks are picked up at random. The probability that there is at least one pair is equal to - |
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Answer» P(atleast one pair) `= 1-`P(getting no pair) `= 1- (.^12C_4.^2C_1 .^2C_1 .^2C_1.^2C_1)/(.^24C_4)` `= 1- ((12xx11xx10xx9 xx2^4)/(4!))/((24 xx 23 xx 22xx21)/(4!))` `= 1-120/161` P( geeting atleast 1 pair) ` =41/161` option D is correct answer |
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| 578. |
Show that the relation R in the set R of real numbers, defined as `R={(a ,b): alt=b^2}`is neither reflexive nor symmetric nor transitive. |
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Answer» `R={(a,b):a le b^(2)}` Here `(1,1) in R " as " 1 le 1, " but " 0.5 notin R " as " 0.5 gt (0.5)^(2)` Therefore, R is not reflexive. Now, `(1,4) in R " as " 1 lt 4^(2)` But, 4 is not less than 1. Therefore, `(4,1) notin R` Therefore, R is not symmetric. Further, `(3,2),(2,1.5) in R ("as "3 lt 2^(2) " and " 2 lt (1.5)^(2))` But, ` 3 gt (1.5)^(2)=2.25` Therefore, `(3,1.5) notin R` Therefore, R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. |
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| 579. |
Determine whether each of the following relations are reflexive, symmetric and transitive : (i) Relation R in the set `A= {1, 2, 3, …, 13, 14}` defined as `" " R = { (x, y ) : 3x - y =0}` (ii) Relation R in the set N of natural numbers defined as `" " R = {(x, y) : y = x + 5 and x lt 4}` (iii) Relation R in the set `A= { 1, 2, 3, 4, 5, 6}` as `" "R = { (x, y) : y ` is divisible by x`}` (iv) Relation R in the set Z of all integers defined as `" " R = {(x,y) : x -y ` is an integer `}` (v) Relation R in the set A of human beings in a town at a particular time given by (a) `R= {(x, y) : and y` work at the same place `}` (b)` R = { (x, y) : x and y ` live in the same locality `}` (c) `R = {(x, y) : x ` is exactly 7 cm taller than y` }` (d) `R = {(x, y) : x ` is wife of `y}` (e) `R= { (x, y) :x ` is father of `y}` |
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Answer» (i) A =` { 1, 2, 3, …, 13, 14}` and `" " R = {(x, y) : 3x -y =0}` For reflexive `(x,x) in R AA x in A` but ` " " 3x -y =0 rArr y = 3x` `therfore " " (x, x) notin R ` if ` x = 2 in A` `rArr R `is not reflexive, For symmetricity `(x, y) in R rArr (y, x) in R AA x, y in R ` Now, `" " (x, y) in R rArr 3 x -y =0` `" " rArr 3y -x ne 0` `" " rArr (y, x) notin R` `therefore R` is not symmetric. e.g., `(1, 3) in R and (3, 1) notin R` For transitivity `(x, y) in R, (y, z) in R rArr (x, z) in R` `therefore (1, 3) in R and (3, 9) in R rArr (1, 9) in R` `rArr R` is not transitive. (ii) `R= {(x,y) : y = x + 5 and x lt 4 }` and N is the set of natural numbers. `rArr R = {(1, 6), (2, 7), (3, 8)}` For reflexive, `(1,1) notin R` `rArr R` is not reflexive. For transitivity, `(x,y) in R (y, z) in R rArr (x, z) in R`. No pair satisfies this condition. `therefore R` is not transitive. (iii) `A= {1, 2, 3, 4, 5, 6}` and `R = {(x, y) : y` is divisible by `x}` `rArr R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5, (6, 6)}` For each `x in A, (x, x) in R` `therefore R ` is reflexive. For each `x, y in R, (x, y) in R cancel(rArr) (y, x) in R` `therefore R` is not symmetric. For each `x, y, z in A` if `(x, y) in R, (y, z) in R` then `(x, z) in R` `therefore R` is transitive. (iv) In the set of all integers Z `" " R = {(x, y) : x -y ` is an integer. } Which is true. `therefore R` is reflexive. For symmetricity, `(x, y) in R rArr (x -y)` is an integer. `" " rArr (y-x) in R` ` therefore R` is symmetric. `therefore ` For transitivity. `(x, y) in R and (y, z) in R` `rArr (x -y) ` is an integer and `(y-z)` is an integer. `rArr (x-y)+ (y-z)` is an integer. `rArr (x, z) in R` `therefore R` is transitive. (v) (a) `R= {(x, y) : x and y` work at the same place } This relation is reflexive, symmetric and transitive. (b) `R= {(x, y) : x and y` live in the same locality}. This relation is reflexive, symmetric and transitive. (c) `R= {(x, y) : x` is exactly 7 cm taller than y } `(x, x) notin R` because x is no exactly 7 cm taller than y `therefore R ` is not reflexive. `(x, y) in R rArr x ` is exactly 7 cm taller than y. `" " cancel ( rArr) y` is exactly 7 cm taller than x. `" " cancel (rArr) (y, x) in R` `therefore R` is not symmetric. `therfore (x, y) in R and (y, z) in R rArr x ` is exactly 7 cm taller than y and y is exactly7 cm taller than z. |
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| 580. |
The domain of definition of the function`f(x)=sqrt(sin^(-1)(2x)+pi/6)`for real-valued `x`is`[-1/4,1/2]`(b) `[-1/2,1/2]`(c) `(-1/2,1/9)`(d) `[-1/4,1/4]` |
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Answer» Correct Answer - `[-1//4,1//2]` We have `f(x)=sqrt(sin^(-1)(2x)+(pi)/(6))` We must have `sin^(-1)(2x) +(pi)/(6) ge 0` `implies sin^(-1)(2x) ge -(pi)/(6) " …(1) " ` But `-(pi)/(2) le sin^(-1)(2x) le (pi)/(2) " (2)" ` From (1) and (2), we have ` -(pi)/(2) le sin^(-1) (2x) le (pi)/(2)` `implies "sin"(-(pi)/(6)) le 2x le "sin"(pi)/(2)` `implies -(1)/(2) le 2x le 1` `implies -(1)/(4) le x le (1)/(2)` Hence, domain is `[-(1)/(4),(1)/(2)]` |
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| 581. |
Find positive no. x which satisfy the equation ` log_3 x * log_4 x (log_5 x -1) = log_5 x*(log_4 x+ log_3 x)` |
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Answer» `log_3^xlog_4^x(log_5^x-1)=log_5^x(log_4^x+log_3^x)` `log_3^xlog_4^xlog_5^x-log_3^xlog_4^x=log_5^xlog_4^x+log_5^xlog_3^x` divide by`log_3^xlog_4^xlog_5^x` `1=1/log_5^x+1/log_3^x+1/log_4^x` `1=log_x^(5*4*3)` `1=log_x^60` `x=60,1`. |
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| 582. |
If `f(x)=(3x+2)/(5x-3)`, then `f[f(x)]` is equal to:A. `-x`B. `x`C. 0D. None of these |
| Answer» Correct Answer - b | |
| 583. |
show that the function `(i) f : N to N : f (x) =x^(2)` is one-one into (ii) `f : Z to Z : f (x) =x^(2)` is many -one into . |
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Answer» Let x,y `in` N and f (x) = f(y) `rArr x^(2)=y^(2)` `rArr x^(2)-y^(2)` = 0 `rArr (x-y)(x+y) = 0` `rArr x-y =0` (`because x + y ne 0`) `rArr` x=y Therefore, f is one one. Again , let f (x) = y, where y `in` N (co-domain) `rArr x^(2) = y` `rArrx=+-sqrt(y)in N is y -3 in N` Therefore , an element 3 belongs to co-domain such that it has no pre-image in domain N. `:.` f is into. Therefore, f is one-one into. |
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| 584. |
If `f:R->R and g:R-> R` are two mappings such that `f(x) = 2x and g(x)=x^2+ 2` then find `fog and gog`. |
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Answer» (i) (gof) (x) = g [f(x)] = g (2x) = `(2x)^(2)+2= 4x^(2) + 2.` and (fog)(x) = f[g(x)] = `f(x^(2)+2) = 2(x^(2)+2)= 2x^(2) + 4` `:.` gof `ne` fog. Hence Proved. (ii) (fog) (2) = f[g(2)] = f(2^(2) + 2) = f(6) = 12. (gog)(1) = g [g(1)] = `g (1^(2)+2) = g (3) = 3^(2) + 2 = 11`. and (fof) (3) = f[f(3)] = f`(2xx3) = f(6) = 2xx6 = 12` Ans. |
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