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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.
| 501. |
If the letter of the word BRING are permuted in all possible ways and the word thus formed are arranged in the dictionary order, then find 59th word. |
| Answer» `59^(th)` word=IGRBN | |
| 502. |
If `y=3[x]+1=4[x-1]-10,` then find the value of `[x+2y]`. |
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Answer» Correct Answer - 107 `y=3[x]+1=4[x-1]-10=4[x]-14` or `[x]=15 " and " y=3(15)+1=46` or `[x+2y]=2y+[x]=2(46)+15=107` |
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| 503. |
Period of `f(x)=sin((cosx)+x)` is |
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Answer» `f(x)=sin(x+cosx)` Let period be T. ` :. f(x+T)=f(x)` for all real x. Putting `x=0`, we have `f(T)=f(0)` or ` sin(T+cos T)=sin 1` By intelligent guessing, `T=2pi`, which is the least positive value. Also, `f(x+2pi)=sin(x+2pi +cos(x+2pi))` `=sin(2pi+(x+cosx))` `=sin(x+cosx)=f(x)` |
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| 504. |
Find the period of (i) `f(x)=sin pi x +{x//3}`, where {.} represents the fractional part. (ii) `f(x)=|sin 7x|-"cos"^(4)(3x)/(4)+"tan"(2x)/(3)` |
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Answer» (i) `f(x)=sin pi x +{x//3}`, where {.} represents the fractional part Period of ` sin pi x " is " (2pi)/(pi)=2` Period of `{x//3} " is " (1)/(1//3)=3` Therefore, period of f(x) is L.C.M. of `(2,3)=6` (ii) `f(x)=|sin 7x|-"cos"^(4)(3x)/(4)+"tan"(2x)/(3)` Period of `|sin 7x| " is " (pi)/(7)` Period of `"cos"^(4)(3x)/(4) " is " (pi)/(3//4)=(4pi)/(3)` Period of `tan(2x)/(3) " is " (pi)/(2//3)=(3pi)/(2)` Therefore, period of f(x) is L.C.M. of `((pi)/(7),(4pi)/(3),(3pi)/(2))=pi xx (L.C.M. of (1,4,3))/(H.C.F. of (7,3,2)` `=12 pi` |
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| 505. |
Find the range of `f(x)=sqrt(sin(cos x))+sqrt(cos(sin x))`. |
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Answer» `f(x)=sqrt(cos(sinx))+sqrt(sin(cosx))` Period of `f(x)` is `2 pi`. Also, `sin(cosx) ge 0 impliescosx in [0,1]` `implies x in [-(pi)/(2),(pi)/(2)]` ` :. x` lies in `1^(st)` and `4^(th)` quadrants. Also `f(-x)=f(x)` ` :. f(x)` is even. ` :. ` we need to find the range in `[0,(pi)/(2)]` only In `[0,(pi)/(2)], sinx` increases, but `cosx` decreases ` :. " both " cos(sinx) and sin(cosx)` decrease. Hence `f(x)` decreases, ` :. ` Range is `[f(pi//2),f(0)] -=[sqrt(cos 1),1+sqrt(sin1)]` |
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| 506. |
Solve the system of equations in `x,y and z` satisfying the following equations `x+[y]+{z}=3.1, y+[z]+{x}=4.3 and z+[x]+{y}=5.4` |
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Answer» Adding all the three equations, we get `2(x+y+z)=12.8 " or " x+y+z=6.4 " (1)" ` Adding the first two equations, we get `x+y+z+[y]+{x}=7.4 " (2)" ` Adding the second and third equations, we get `x+y+z+[z]+{y}=9.7 " (3) " ` Adding the first and third equations, we get `x+y+z+[x]+{z}=8.5 " (4)" ` From (1) and (2), `[y]+{x}=1.` From (1) and (3), `[z]+{y}=3.3.` From (1) and (4), `[x]+{z}=2.1.` So, `[x]=2,[y]=1,[z]=3`, `{x}=0,{y}=0.3, " and " {z}=0.1` ` :. x=2,y=1.3, z=3.1` |
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| 507. |
The sum of all real values of x satisfying the equation `(x^2-5x+5)^(x^2+4x-60)=1`is:(1) 3 (2) `-4`(3) 6(4) 5 |
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Answer» if `a^x = 1` then 3 conditions are `a= -1,0,1` &`x=0 or in` even no `a= 1 ; x in R` `a=-1 ; x in `even no `x=0, a=1` now, `x^2 - 5x + 5 =1` `x^2 - 5x + 4=0 ` `x^2 - 4x - x + 4 = 0` `(x- 4)(x-1) = 0` `x=4,1` now,`x^2 - 5x + 5= -1` `x^2 - 5x + 6= 0` `x^2 - 2x - 3x + 6= 0` `(x-2)(x-3) =0 ` `x= 2,3` `x^2 + 4x - 60` for ` x=2` `4 + 8 - 60 = 48 in ` even number for `x=3` `9 + 12 - 60 = 30 cancel( in) ` even no, so not possible `x^2 + 4x - 60 = 0` `x^2 + 10x - 6x - 60 = 0` `(x+10)(x-6) = 0` `x= -10,6` `x= 4,1,2,-10,6` sum=`4+1+2-10+6 = 3` option 4 is correctAnswer |
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| 508. |
Solve |4-|x-1||=3 |
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Answer» Correct Answer - `x= -6,0,2,8` `|4-|x-1||=3` `implies |x-1|-4= +-3` `implies |x-1|=7,|x-1|=1` `implies x= -6,0,2,8` |
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| 509. |
Solve `|3x-2| le (1)/(2)`. |
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Answer» `|3x-2| le (1)/(2)` or `-(1)/(2) le 3x -2 le (1)/(2)` or ` (3)/(2) le 3x le (5)/(2)` or `(1)/(2) le x le (5)/(6)` or `x in [(1)/(2),(5)/(6)]` |
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| 510. |
The locus of the intersection point of `x cos alpha +y sin alpha=a` and `x sin alpha- y cos alpha=b` is |
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Answer» let the locus be `P(H,K)` `hcos alpha + k sin alpha = a` `h sin alpha - kcos alpha = b` by solving bith, we get `h^2(sin^2 alpha + cos^2 alpha) + k^2(sin^2 alpha + cos^2 alpha) = a^2 + b^2` `sin^2 alpha + cos^2 alpha = 1` `h^2 + k^2 = a^2 + b^2 ` locus of point P(h,k)`x^2 + y^2= a^2 + b^2 ` Answer |
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| 511. |
If `f(x) = [x] , 0 |
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Answer» `f(-x)={([-x]",",0 le {-x}lt0.5),([-x]+1",",0.5 lt {-x} lt1):}` `={([-x]",",{-x}=0),([-x]",",0 lt {-x} lt 0.5),([-x]+1",",0.5 lt {-x} lt1):}` `={(-[x]",",{x}=0),(-1-[x]",",0 lt 1-{x} lt 0.5),(-1-[x]+1",",0.5 lt 1-{x} lt1):}` `={(-[x]",",{x}=0),(-1-[x]",",0.5 lt {x} lt 1),(-[x]",",0 lt {x} lt0.5):}` `={(-[x]",",0 le{x}lt0.5),(-1-[x]",",0.5 lt {x} lt 1):}` `= -f(x)` |
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| 512. |
In each of the following cases find the period of the function if it is periodic. (i) `f(x)="sin"(pi x)/(sqrt(2))+"cos"(pi x)/(sqrt(3)) " (ii) " f(x)="sin"(pi x)/(sqrt(3))+"cos"(pi x)/(2sqrt(3))` |
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Answer» (i) Period of `"sin"(pi x)/(sqrt(2))=(2pi)/(pi//sqrt(2))=2 sqrt(2)` Period of `"cos"(pi x)/(sqrt(3))=(2pi)/(pi//sqrt(3))=2 sqrt(3)` Now, L.C.M. of two different kinds of irrational number does not exist. Therefore, f(x) is not periodic. (ii) Period of `"sin"(pi x)/(sqrt(3))=(2pi)/(pi//sqrt(3))=2 sqrt(3)` Period of `"cos"(pi x)/(2sqrt(3))=(2pi)/(pi//2sqrt(3))=4 sqrt(3)` Now, L.C.M. of `(2sqrt(3),4 sqrt(3))` `=sqrt(3) xx L.C.M. " of " (2,4)=4sqrt(3)` |
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| 513. |
Solve `[x]^(2)-5[x]+6=0.` |
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Answer» Correct Answer - `[2,4)` `[x]^(2)-5[x]+6=0` or `[x]=2,3` or `x in [2,4)` |
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| 514. |
Let `f : X->Y`be a function. Define a relation R in X given by `R = {(a , b): f(a) = f(b)}`. Examine if R is an equivalence relation. |
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Answer» `R = {(a,b): f(a) = f(b)}` As `f(a) = f(a)` `:. (a,a) in R` `:. R` is reflexive. If `f(a) = f(b)` Then, `f(b) = f(a)` Thus, `(b,a) in R` So, if `(a,b) in R`, then `(b,a) in R` `:. R` is ymmetric. If `(a,b) in R`, Then, `f(a) = f(b)`->(1) If `(b,c) in R`, Then, `f(b) = f(c)`->(2) From (1) and (2), `f(a) = f(c)` `:. (a,c) in R` `:. R` is transitive. As `R` is reflexive, summetric and transitive, `R` is an equivalence relation. |
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| 515. |
The period of function `2^({x}) +sin pi x+3^({x//2})+cos pi x` (where {x} denotes the fractional part of x) isA. 2B. 1C. 3D. None of these |
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Answer» Correct Answer - A The period of `sin pi and cos pi x and cos pi x ` is 2 and 1, respectively. The period of `2^({x})` is 1. The period of `3({x//2})` is 2. Hence, the period of `f(x)` is LCM of 1 and 2, i.e., 2. |
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| 516. |
Find the possible values of`sqrt(|x|-2)`(ii) `sqrt(3-|x-1|)`(iii) `sqrt(4-sqrt(x^2))` |
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Answer» (i) `sqrt(|x|-2) ` We know that square roots are defined for non-negative values only. It implies that we must have ` |x|-2 ge 0`. Thus, `sqrt(|x|-2) ge 0` (ii) `sqrt(3-|x-1|)` is defined when `3-|x-1| ge 0` But the maximum value of `3-|x-1|` is 3, when `|x-1|` is 0. Hence, for `sqrt(3-|x-1|)` to get defined, ` 0 le 3-|x-1|le 3`. Thus, `sqrt(3-|x-1|) in [0, sqrt(3)]` Alternatively, ` |x-1| ge 0` `implies -|x-1| le 0` `implies 3-|x-1| le 3` But for `sqrt(3-|x-1|)` to get defined, we must have `0 le 3 -|x-1| le3` `implies 0 le sqrt(3-|x-1|) le sqrt(3)` (iii) `sqrt(4-sqrt(x^(2)))=sqrt(4-|x|)` `|x| ge 0` `implies -|x| le 0` `implies4-|x| le 4` But for `sqrt(4-|x|)` to get defined `0 le 4 -|x| le 4` ` :. 0 le sqrt(4-|x|) le 2` |
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| 517. |
Verify that`x sgnx=|x|``|x|sgnx=x``x(sgnx)(sgnx)=x` |
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Answer» (i) `x " sgn "x={(x*1",",x gt 0),(0",", x=0),(x*(-1)",",x lt 0):} = {(x",",x gt0),(0",",x=0),(-x",", x lt 0):}=|x|` (ii) `|x| " sgn "x={(x*1",",x gt 0),(0",", x=0),((-x)*(-1)",",x lt 0):} = {(x",",x gt0),(0",",x=0),(x",", x lt 0):}=x` (iii) `x ("sgn "x) ("sgn "x)=|x| " sgn " (x)=x` |
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| 518. |
Prove that function `f(x)=cos sqrt(x)` is non-periodic. |
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Answer» We have `f(x)=cos sqrt(x)` Let f(x) be periodic with period T, where `T gt 0.` ` :. f(x+T)=f(x)` `implies cos sqrt(x+T)=cos sqrt(x) " for " x ge 0.` In particular choosing ` x=0`, we have `cos sqrt(T)=cos sqrt(0)=1 " …(1)" ` For `x=T`, we have `cos sqrt(T+T)=cos sqrt(T)=1` or `cos sqrt(2T)=1 " ...(2)" ` From (1) , `sqrt(T) =2m pi, m in Z` From (2), `sqrt(2T)=2n pi, n in Z` ` :. (sqrt(2T))/(sqrt(T))=(2n pi)/(2m pi)` or ` sqrt(2)=(n)/(m),` which is not true. So, `cos sqrt(x)` is not periodic. |
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| 519. |
Solve `|x-3|+|x-2|=1.` |
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Answer» `|x-3|=|x-2|=1` `implies |x-3|+|x-2|=(3-x)+(x-2)` `implies x-3 le 0 " and " x-2 ge 0` `implies x le 3 " and " x ge 2` `implies 2 le x le 3` |
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| 520. |
Period of the function `f(x)=sin((x)/(2))cos [(x)/(2)]-cos((x)/(2))sin[(x)/(2)]`, where [.] denotes the greatest integer function, is _________. |
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Answer» Correct Answer - 2 `f(x)=sin((x)/(2))cos [(x)/(2)]-cos((x)/(2))sin[(x)/(2)]` ` =sin((x)/(2)-[(x)/(2)])` `=sin({(x)/(2)})` ` :. " period of " f(x), T=2` |
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| 521. |
Verify that the period of function `f(x) =sin^(10)x " is " pi.` |
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Answer» We have `f(x)=sin^(10)x` Now `f(x+pi)=sin^(10)(x+pi)` `=(sin(x+pi))^(10)` `=(-sinx)^(10)` `=sin^(10)x` `=f(x)` Thus, we have `f(x+pi)=f(x)` for all ` x in R.` So,period of f(x) is `pi`. |
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| 522. |
Solve the differential equation `2x^2(dy)/(dx)-2xy+y^2=0`, `y^2(e)=e` |
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Answer» `2x^2dy/dx-2xy+y^2=0` `2x^2*1/y^2dy/dx-2x/y+1=0` `2x^2(-dz/dx)-2xz+1=0` `2x^2dz/dx+2xz=1` `dz/dx+1/x^2=1/(2x^2)` `d/dx{ze^(int1/xdx)}=1/(2x^2)*e^(1/xdx` `d/dx(2x)=1/(2x^2)*x` `2x=int 1/(2x)dx` `=1/2log|x|+c` `x/y=1/2log|x|+c` `e/(y(e))=1/2log|e|+c` `e/(y(e))=1/2+c` `c=e/(y(e))-1/2` `c=pmsqrte-1/2` `x/y=1/2log|x|+c` `c=pmsqrte-1/2`. |
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| 523. |
Let `f(x)=sin(x/(n!))+cos((2x)/((n+1)!))`. Find the period of `f(x)`. |
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Answer» Period of `sin(x/n_1)=T_1` `T_1/(n!) =2pi` `T_1=n!(2pi)` Period of `cos((2x)/(n+1)!)=T_2` `(2T_2)/((n+1)!)=2pi` `T_2=(n+1)!pi` Period of`sin(x/n_1)+cos((2x)/((n+1)!))` is T such that`n_1T_1=n_2T_2=T` Such at `n_1n_2` are to each `(n!)2pi*n_1=n_2(n+1)!pi` `2n_1=(n+1)n_2` `n_1/n_2=(n+1)/2` when n is own,`n_1/n_2=(n+1)/2` `n_1=n+1,n_2=2` `T=2(n+1)!pi` When n is odd `n_1/n_2=(n+1)/2` `n_1=(n+1)/2,n_2=1` `T=(n+1)!pi=(n+1)!pi` `T=2(n+1)!pi` when n is even. `T=(n+1)!pi`, when n is odd. |
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| 524. |
The period of the function `|sin^3(x/2)|+|cos^5(x/5)|`isA. `2pi`B. `10pi`C. `8pi`D. `5pi` |
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Answer» Correct Answer - B `f(x)=|"sin"^(3)(x)/(2)|+|"cos"^(5)(x)/(5)|` The period of `sin^(3)x` is `2pi`. So, the period of ` "sin"^(3)(x)/(2) " is " (2pi)/(1//2)=4pi.` So, the period of `|"sin"^(3)(x)/(2)|" is " 2pi.` The period of `cos^(5) x " is " 2pi.` So, the period of ` "cos"^(5)(x)/(5) " is " (2pi)/(((1)/(5)))=10pi.` So, the period of `|"cos"^(5)(x)/(2)| " is " 5pi` So, the period of `|"cos"^(5)(x)/(2)| " is " 5pi` Now, period of `f(x)=LCM " of " {2pi,5pi}=10pi` |
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| 525. |
Prove that period of function `f(x)=sinx, x in R " is " 2pi.` |
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Answer» Let period of `f(x)=sin x " be " T` ` :. F(x+T)=f(x)` for all real x. or ` sin(x+T)=sin x` for all real x. ` :. Sin(0+T)=sin 0 " " ("putting" x=0)` ` :. T=n pi, n in Z.` The least value of T is `pi`. But for `T=pi, sin(x+pi)= -sinx ne f(x)` So, let `T=2pi` for which `sin(x+2pi)=sinx.` Thus, period of `sinx` is `2pi`. |
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| 526. |
Solve the differential equation: `dy/dx + y cotx = 4x cosec x`, given that y=0 at x = `pi/2` |
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Answer» Comparing the given equation with first order differential equation, `dy/dx+Py = Q(x)`, we get,`P = cotx and Q(x) = 4xcosecx` So, Integrating factor `(I.F) = e^(intcotxdx)` `I.F.= e^(ln|sinx|) = sinx` we know, solution of differential equation, `y(I.F.) = intQ(I.F.)dx` `:.`Our solution will be, `ysinx = int 4xcosecxsinxdx` As `sinx cosecx = 1`, `=>ysinx = int 4xdx` `=>ysinx = 2x^2+c` At `y = 0 and x = pi/2`, equation becomes `0 = 2(pi/2)^2 +c => c = -pi^2/2` So, solution will be, `ysinx = 2x^2-(pi)^2/2` |
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| 527. |
Solve: `|-2x^2+1+e^x+sinx|=2x^2-1|+e^x+|sinx|,x in [0,2pi]dot` |
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Answer» `|-2x^(2)+1+e^(x)+sinx|=|2x^(2)-1|+e^(x)+|sinx|, x in [0,2pi]` in the R.H.S., each term is positive and `e^(x) gt 0.` So, `1-2x^(2) ge 0` and ` sin x ge 0` or `x in [-(1)/(sqrt(2)),(1)/(sqrt(2))] and x in [0,pi]` ` :. x in [0,(1)/(sqrt(2))]` |
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| 528. |
Let `f(x)={(sinx+cosx",",0 lt x lt (pi)/(2)),(a",",x=pi//2),(tan^(2)x+"cosec"x",",pi//2 lt x lt pi):}` Then its odd extension isA. `{(-tan^(2)x-"cosec"x",",-pi lt x lt -(pi)/(2)),(-a",",x=-(pi)/(2)),(-sinx+cosx",",-(pi)/(2) lt x lt 0 ):}`B. `{(-tan^(2)x+"cosec"x",",-pi lt x lt -(pi)/(2)),(-a",",x=-(pi)/(2)),(sinx-cosx",",-(pi)/(2) lt x lt 0 ):}`C. `{(-tan^(2)x+"cosec"x",",-pi lt x lt -(pi)/(2)),(a",",x=-(pi)/(2)),(sinx-cosx",",-(pi)/(2) lt x lt 0 ):}`D. `{(tan^(2)x+"cosec"x",",-pi lt x lt -(pi)/(2)),(-a",",x=-(pi)/(2)),(sinx+cosx",",-(pi)/(2) lt x lt 0 ):}` |
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Answer» Correct Answer - B For odd function, `f(x)= -f(-x)` ` = -{(sin(-x)+cos(-x)",",0 lt -x lt pi//2),(a",",-x=pi//2),(tan^(2)(-x)+"cosec"(-x)",",pi//2 lt -x lt pi):}` ` ={(sinx-cosx",", -pi//2 lt x lt 0),(-a",",x= -pi//2),(-tan^(2)x+"cosec"x",",-pi lt x lt -pi//2 ):}` |
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| 529. |
Solve `|sinx +cos x |=|sinx|+|cosx|, x in [0,2pi]`. |
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Answer» The given relation holds only when sin x and cos x have the same sign or at least one of them is zero. Hence, `x in [ 0, pi//2] cup [pi,3pi//2] cup {2pi}.` |
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| 530. |
Evaluate `int (xtan^(-1)x)/(1+x^2)^(3/2) dx` |
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Answer» `Let x=tantheta` `dx=sec^2d theta` `1+x^2=1+tan^2theta=sec^2theta` `int(xtan^(-1)xdx)/(1+x^2)^(3/2)` `int(tantheta*theta*sec^2theta*dthet a)/(sec^2theta)^(3/2)` `inttheta*tantheta*sec^2theta/sec^3theta dthet a` `inttheta*sinthetad theta` `-thetacostheta+sintheta+c` `(-tna^(-1)x)/sqrt(1+x^2)+x/sqrt(1+x^2)+c`. |
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| 531. |
`int 1/((1+x^2) tan^(-1)x) dx` |
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Answer» `tan^(-1)x=t` `1/(1+x^2)dx=dt` `=int1/t dt` `=lnt+C` `=ln(tan^(-1)x)+C`. |
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| 532. |
Using integration, find the value of `m`. If the area bounded by parabola `y^2=16ax` and the line `y=mx` is `a^2/12` square units. |
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Answer» `y^2=16an` `y=mx` `(mx)^2=16an` `x=0,x=(16a)/m^2` `int_0^(16/m^2)(sqrt(16ax)-mx)dx` `sqrt(16a)int_0^(16/m^2)x^(1/2)dx-m int_0^((16a)/m^2)*xdx` `sqrt(16a)*2/3*(8*9sqrta)/m^2-m/2*(256a^2)/m^4` `64/3*a^2/m^3-(128a^2)/m^3` `a^2/m^3{(8*64-3*128)/3}` `a^2/m^3*128/3=a^2/12` `m^3=(12*128)/3=4*128` `m=8`. |
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| 533. |
If the functions `f(x) and g(x)` are defined on `R -> R` such that `f(x)={0, x in` retional and `x, x in` irrational ; `g(x)={0, x in` irratinal and `x,x in` rational then `(f-g)(x)` is |
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Answer» We have, `(f-g):R to R,` `(f-g)(x)={(-x" if "x in "rational"),(x" if "x in "irrational"):}` Clearly `(f-g)(x)` is one-one and onto. |
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| 534. |
If ` f:[0,oo) to [0,1), " and " f(x)=(x)/(1+x)` then check the nature of the function. |
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Answer» Given that `f:[0,oo) to [0, oo),f(x)=(x)/(x+1)` Let `f(x_(1))=f(x_(2))` `implies (x_(1))/(x_(1)+1)=(x_(2))/(x_(2)+1)` `implies x_(1)x_(2)+x_(1)=x_(1)x_(2)+x_(2)` `implies x_(1)=x_(2)`. Thus f(x) is one-one. Now let `y=(x)/(1+x)` `implies y+yx=x` `implies x=(y)/(1-y)` As ` x ge 0, (y)/(1-y) ge 0` `implies (y)/(1-y) le 0` `implies 0 le y lt 1` or range of f(x) is `[0,1).` Thus f(x) is onto. |
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| 535. |
If m n, k are rational and` m =k+k/n` then the roots of `x^2+mx+n=0` are |
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Answer» `x+mx+n=0` `x^2+(k+n/k)x+n=0` `x=(-(k+n/k)pmsqrt((k+n/k)^2-4n))/2` `x=(-(k+n/k)pmsqrt(k^2+n^2/k^2+2n-4n))/2` `x=(-(k+n/k)pm(k-n/k))/2` `x=-k,-n/k`. |
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| 536. |
If ` f:[0,oo) to [0,1), " and " f(x)=(x)/(1+x)` then check the nature of the function. |
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Answer» Given that `f:[0,oo) to [0, oo),f(x)=(x)/(x+1)` Let `f(x_(1))=f(x_(2))` `implies (x_(1))/(x_(1)+1)=(x_(2))/(x_(2)+1)` `implies x_(1)x_(2)+x_(1)=x_(1)x_(2)+x_(2)` `implies x_(1)=x_(2)`. Thus f(x) is one-one. Now let `y=(x)/(1+x)` `implies y+yx=x` `implies x=(y)/(1-y)` As ` x ge 0, (y)/(1-y) ge 0` `implies (y)/(1-y) le 0` `implies 0 le y lt 1` or range of f(x) is `[0,1).` Thus f(x) is onto. |
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| 537. |
If the function `f: RvecA`given by `f(x)=(x^2)/(x^2+1)`is surjection, then find `Adot` |
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Answer» `f:R to A,f(x)=(x^(2))/(x^(2)+1)` Here domain is all real numbers. Since `f(x)` is onto, range must be same as codomain (A). To find the range of the function, `f(x)=1-(1)/(x^(2)+1)` Now` x^(2)+1 ge 1 AA in R.` `implies 0 lt (1)/(x^(2)+1) le 1` `implies -1 le -(1)/(x^(2)+1) lt 0` `implies 0 le 1 -(1)/(x^(2)+1) lt 1` Thus, range is `[0,1).` Hence codomain `A=[0,1).` |
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| 538. |
Let `f: NvecZ`be a function defined as `f(x)=x-1000.`Show that `f`is an into function. |
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Answer» Let `f(x)=y=x-1000` `implies x=y+1000=g(y)` (say) Here g(y) is defined for each `y in Z,` but `g(y) notin N " for " y le -1000.` Hence `f` is into. |
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| 539. |
Let `f:R to R` where `f(x) =sin x.` Show that `f ` is into. Also find the codomain if `f` is onto. |
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Answer» `f:R to R, f(x) =sinx` Range of the function is `[-1,1],` which is subset of codomain R. So, `f(x)` is into. To make `f(x)` onto, we modify the codomain to `[-1,1].` |
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| 540. |
If `f(x)=1/x ,g(x)=1/(x^2),`and `h(x)=x^2,t h e n``f(x)=x^2,x!=0,(h(g(x))=1/(x^2)``h(g(x))=1/(x^2),x!=0,fog(x)=x^2``fog(x)=x^2,x!=0,h(g(x))=(g(x))^2,x!=0`none of theseA. `fog(x)=x^(2),x ne 0, h(g(x))=(1)/(x^(2))`B. `h(g(x))=(1)/(x^(2)),x ne 0, fog(x)=x^(2)`C. `fog(x)=x^(2), x ne 0, h(g(x))=(g(x))^(2), x ne 0`D. None of these |
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Answer» Correct Answer - C `f(x)=(1)/(x), g(x)=(1)/(x^(2)), and h(x)=x^(2)` `f(g(x))=x^(2), x ne 0` `h(g(x))=(1)/(x^(4))=(g(x))^(2), x ne 0` |
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| 541. |
If `f(x)={(x^(2)",","for "x ge0),(x",","for "x lt 0):}`, then fof(x) is given byA. `x^(2) " for " x ge 0, x " for " x lt 0`B. `x^(4) " for " x ge 0, x^(2) " for " x lt 0`C. `x^(4) " for " x ge 0, -x^(2) " for " x lt 0`D. `x^(4) " for " x ge 0, x " for " x lt 0` |
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Answer» Correct Answer - D `f(f(x))={((f(x))^(2)",", "for " f(x) ge 0),(f(x)",","for " f(x) lt 0):}` `={((x^(2))^(2)","x^(2) ge 0"," x ge 0),(x^(2)","x ge 0"," x lt 0),(x^(2)"," x^(2) lt 0"," x ge 0),(x"," x lt 0"," x lt 0):}` `={(x^(4)","x ge 0),(x","x lt 0):}` |
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| 542. |
Find fog and gof , if (i) () `f(x)= |x|` and `g(x)=|5x-2|`(ii) `f(x)=8x^3` and `g(x)=x^(1//3)` |
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Answer» (i)Here, `f(x) = |x|,g(x) = |5x-2|` `:. fog = f(g(x)) = f(|5x-2|) = ||5x-2|| =|5x-2|` `gof = g(f(x)) = g(|x|) = |5|x|-2|` (ii)Here, `f(x) = 8x^3,g(x) = x^(1/3)` `:. fog = f(g(x)) = f(x^(1/3)) = (8(x^(1/3))^3) =8x` `gof = g(f(x)) = g(8x^3) = (8x^3)^(1/3) = 2x` |
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| 543. |
State with reason whether following functions have inverse (i) `f:{1,2,3,4} ->{10} " with " f={(1, 10),(2, 10),(3, 10),(4, 10)}`(ii) `g:{5, 6, 7, 8}->{1,2,3,4}" with "g={(5, 4),(6, 3),(7, 4),(8, 2)}`(iii) `h : {2, 3, 4, 5} → {7, 9, 11, 13}" with "h = {(2, 7), (3, 9), (4, 11), (5, 13)}` |
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Answer» (i) `f: {1, 2, 3, 4} → {10}` defined as: `f = {(1, 10), (2, 10), (3, 10), (4, 10)}` From the given definition of f, we can see that f is a many one function as:` f(1) = f(2) = f(3) = f(4) = 10` `:.` f is not one-one. Hence, function f does not have an inverse. (ii) `g: {5, 6, 7, 8} → {1, 2, 3, 4}` defined as: `g = {(5, 4), (6, 3), (7, 4), (8, 2)}` From the given definition of `g`, it is seen that g is a many one function as: `g(5) = g(7) = 4`. `:. g` is not one-one, Hence, function g does not have an inverse. (iii) `h: {2, 3, 4, 5} → {7, 9, 11, 13}` defined as: `h = {(2, 7), (3, 9), (4, 11), (5, 13)}` It is seen that all distinct elements of the set `{2, 3, 4, 5}` have distinct images under ` h`. `:.` Function h is one-one. Also, h is onto since for every element `y` of the set `{7, 9, 11, 13}`, there exists an element `x` in the set `{2, 3, 4, 5}` such that `h(x) = y`. Thus, `h` is a one-one and onto function. Hence, `h` has an inverse. |
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| 544. |
Show that `f: [-1, 1 ] to R`, given by `f(x) = (x)/((x + 2))` is one-one. Find the inverse of the function `f: [-1, 1] to ` Range `f`. ( Hint: For `y in ` Range `f, y = f(x) = (x)/(x+ 2)`, for some `x` in `[-1, 1]`, i.e., `x = (2y)/((1-y))`) |
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Answer» In `f : [-1, 1] to R , f(x) = (x)/( x+ 2)` Let `x, y in [-1, 1]` and `f(x) = f(y)` `rArr (x)/(x + 2) = (y)/(y +2) rArr xy + 2y = xy = 2y` `rArr 2x = 2y rArr x =y ` `therefore f ` is one- one. Let `f(x) = y` where `y in R` `rArr (x)/(x + 2) = y rArr x = xy + 2y ` `rArr x (1 -y) = 2y rArr x = (2y )/(1-y)` `therefore ` Range of `f= R- {1}` Let, In `g : ` range of ` f to [-1, 1]` is defined as `g(y) = ( 2y )/(1-y), y ne 1`. Now `(gof) (x) = g [f(x) ] = g((x)/( x+2))` `" " = (2((x)/( x+ 2))) /( 1- ((x)/(x + 2))) = (2x)/(x + 2 - x ) = (2x)/( 2) = x` and `(fog) (x) = f [g(x) ] = f((2x)/( 1-x))` `" " ((2x)/( 1-x))/(( 2x)/(1-x)+ 2)= (2x)/( 2x + 2 - 2x) = (2x)/(2) = x ` `therefore gof = fog = I_R` `rArr f ^(-1) = g` `rArr f^(-1)(y) = (2y)/( 1-y ) , y ne 1` |
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| 545. |
Consider `f: R->R`given by `f(x) = 4x + 3`. Show that `f` is invertible. Find the inverse of `f`. |
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Answer» `f: R → R` is given by, `f(x) = 4x + 3` Let `f(x) = f(y)` `=>4x+3 = 4y+3` `=>4x=4y` `=>x = y` `:. f` is a one-one function. For `y in R`, Let `y = 4x+3` `:. x = (y-3)/4` Now, `f(x) = f((y-3)/4) = 4((y-3)/4)+3 = y` `:. f` is an onto function. Let `g:R->R` such that `g(x) = (y-3)/4` Then, `gof(x) = g(f(x)) = ((4x+3)-3 )/4 = x` `fog(y) = f(g(y)) = 4((y-3)/4)+3 = y` `:. gof = fog = I_R` Hence, `f` is invertible and the inverse of f is given by `f^-1(y)=g(y)=(y-3)/4`. |
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| 546. |
Consider `f: R->R`given by `f(x) = 4x + 3`. Show that f is invertible. Find the inverse of f. |
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Answer» In `f: R to R, f(x) = 4x + 3` Let `x, y in R and f(x) = f(y)` `rArr 4x + 3 = 4y + 3` `rArr 4x = 4y rArr x =y ` ` therefore f` is one-one. Again, let `f(x) = y ` where `y in R` `rArr " " 4x + 3 = y rArr 4x = y - 3 ` `rArr " "x = ( y -3)/( 4) ` Now for each ` y in R, x = (y - 3)/(4) in R ` in such that `f(x) = f((y -3)/(4))= 4((y-3)/( 4)) + 3 = y` `therefore f ` is onto. Therefore, `f` is one-one onto function ` rArr f` is invertible. `therefore f^(-1) : R to R ` is defined as `f^(-1)(y) = (y-3)/(4)`. |
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| 547. |
If `f: R->R`be given by `f(x)=(3-x^3)^(1//3)`, then `fof(x)`isA. `x^(1//3)`B. `x^(3)`C. `x`D. `(3-x^(3))` |
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Answer» Correct Answer - c `f: R to R and f(x) = (3-x^(3))^(1//3)` `" " (fof) (x) = f{f(x)}` `" " = f{(3-x^(3))^(1//3)}` `" " = [ 3-{(3-x^(3))^(1//3) }^(3)]^(1//3) = x ` |
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| 548. |
Let `f: R-{-4/3}->R`be a function as `f(x)=(4x)/(3x+4)`. The inverse of f is map, `g: R a ngef->R-{-4/3}`given by.(a) `g(y)=(3y)/(3-4y)` (b) `g(y)=(4y)/(4-3y)`(c) `g(y)=(4y)/(3-4y)` (d) `g(y)=(3y)/(4-3y)`A. `g(y) = ( 3y)/( 3-4y)`B. `g(y)= ( 4y )/( 4-3y)`C. `g(y) = ( 4y)/( 4-3y)`D. `g(y) = ( 4y)/( 3-4y) ` |
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Answer» Correct Answer - b In `f: R - {- (4)/(3) } to R`, ` f(x) = ( 4x)/( 3x + 4) AA x in R - {-(4)/(3)}` Let for `y in R, x in R - {-(4)/(3)}` is such that `" " f(x)=y ` `rArr " " (4x)/( 3x + 4) = y rArr 4x = 3xy + 4y` `rArr x ( 4-3y) = 4y rArr x = (4y)/( 3-4y) ` Let, in `g: f ` range of `f to R - {- (4)/(3)}, g(y) = ( 4y )/( 3-4y )` Now `(gof) (x) = g{f(x)} = g((4x)/( 3x + 4))` `" " = (4((4x)/( 3x+ 4)))/( 4-3((4x )/( 3x + 4))) = ( 16x )/(12 x + 16 - 12x` `" " = ( 16x)/( 16) = x` and `(fog) (y) = f[g(y) ] = f[ ( 4y)/( 4-3y)]` `" " = ( 4(( 4y)/(3x + 4)))/( 3(( 4y )/( 4- 3y ))+4)` `" " = ( 16y )/( 12y + 16 - 12y ) = ( 16y)/(16) = y` `therefore " " gof = I_(R- {- (4)/(3)} ) and fog = I_R` ` therefore f^(-1) = g`. |
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| 549. |
Two person `A and B` take turns in throwing a pair of dice. The first person to through 9 from both dice will win the game. If A throwns fisrt then the probability that B wins the game is. |
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Answer» Total number of ways sum of both dice can come `9 = 4` Total number of combinations of both dice `= 6**6 = 36` `:.` Probability of winnining, `P(W) = 4/36 = 1/9` Probability of losing, `P(L) = 1-1/9 = 8/9` As A starts first probability of B winning in first attempt`= 1/9**8/9` Probability of B winning in second attempt `= 8/9**8/9**8/9**1/9=(8/9)^3**1/9` Probability of B winning in third attempt `= (8/9)^5**1/9` `:.` Probability of B winning `P(B)= 1/9(8/9+(8/9)^3+(8/9)^5+...)` `P(B)= 1/9((8/9)/(1-(8/9)^2)) =(81**8)/(17**81) = 8/17` |
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| 550. |
Find the range of`f(x)=tan^(-1)sqrt((x^2-2x+2))` |
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Answer» Correct Answer - `[pi//4,pi//2)` `f(x)=tan^(-1)(sqrt((x-1)^(2)+1))` Now, `(x-1)^(2) +1 in [1,oo)` or `tan^(-1)(sqrt((x-1)^(2)+1))in [(pi)/(4),(pi)/(2))` |
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