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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.
| 51. |
A curve `y=f(x)`passesthrough point `P(1,1)`. The normal to the curve at `P`is a `(y-1)+(x-1)=0`. If the slope of the tangent at any point on thecurve is proportional to the ordinate of the point, then the equation of thecurve is(a)`( b ) (c) y=( d ) e^(( e ) (f) K(( g ) (h) x-1( i ))( j ))( k ) (l)`(m) (b) `( n ) (o) y=( p ) e^(( q ) (r) K e (s))( t ) (u)`(v)(c)`( d ) (e) y=( f ) e^(( g ) (h) K(( i ) (j) x-2( k ))( l ))( m ) (n)`(o) (d) None of these |
| Answer» Correct Answer - `y = e^(a(x-1)), (1+(e^(-a))/(a) - (1)/(2a))` | |
| 52. |
Let `f(x) = int_(0)^(x)(dt)/(sqrt(1+t^(2)))` and `g(x)` be the inverse of `f(x)`, then which one of the following holds good ?A. `2g" = g^(2)`B. `2g" = 3g^(2)`C. `3g" = 2g^(2)`D. `3g" = g^(2)` |
| Answer» Correct Answer - B | |
| 53. |
The area bounded between the parabola `x^(2)=y/4` and `x^(2)=9y` and the straight line `y=2` isA. `20sqrt(2)`B. `(10sqrt(2))/(3)`C. `(20sqrt(2))/(3)`D. `10 sqrt(2)` |
| Answer» Correct Answer - C | |
| 54. |
If `f(x) = int_(0)^(x)(2cos^(2)3t+3sin^(2)3t)dt`, `f(x+pi)` is equal to :A. `f(x) + 2f(x)`B. `f(x) + 2f(pi/2)`C. `f(x)+4f(pi/4)`D. `2f(x)` |
| Answer» Correct Answer - B | |
| 55. |
If length of perpendicular drawn from points of a curve to a straight line approaches zero along an infinite branch of the curve, the line is said to be an asymptote to the curve. For example , y-axis is an asymptote to `y = lnx`& x-axis is an asymptote to ` y = e^(-x)`. If `lim_(xrarr0) f(x) = e` (a finite number) then `y = e` is an asymptote to `y = f(x)` . Similarly if `lim_(xrarr0)f(x) = alpha`, then `y = alpha` is also an asymptote. If `lim_(xrarra)f(x) = oo` or `lim_(xrarra) f(x) = -oo`, then `x = a` is a an asymptote to `y = f(x)`. Number of asymptotes parallel to co -ordinate aces for the function `f(x) = ((x+1)(x+2))/((x-1)(x-2))` is equal to :A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - C | |
| 56. |
Let `I_(n) = int_(0)^(1)x^(n)(tan^(1)x)dx, n in N`, thenA. `(n+1)I_(n)+(n-1)I_(n-2)= (pi)/(4)+(1)/(n) AA n ge 3`B. `(n+1)I_(n)+(n-1)I_(n-2)= (pi)/(2)-(1)/(n) AA n ge 3`C. `(n+1)I_(n)-(n-1)I_(n-2)= (pi)/(2)+(1)/(n) AA n ge 3`D. `(n+1)I_(n)-(n-1)I_(n-2)= (pi)/(2)-(1)/(n) AA n ge 3` |
| Answer» Correct Answer - B | |
| 57. |
Let `T_(n) = sum_(r=1)^(n) (n)/(r^(2)-2r.n+2n^(2)), S_(n) = sum_(r=0)^(n)(n)/(r^(2)-2r.n+2n^(2))` thenA. `T_(n) gt S_(n) AA n in N`B. `T_(n) gt pi/4`C. `S_(n) lt pi/4`D. `underset(nrarroo)("lim")S_(n) = pi/4` |
| Answer» Correct Answer - A::B::C::D | |
| 58. |
Find the area bounded by the parabola `x^(2) = y`, x-axis and the line `y = 1` |
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Answer» Graph of ` y = x^(2)` Area OEBD = Area OAEO `= underset(0)overset(1)int|x| dy = underset(0)overset(1)intsqrt(y) dy = 2/3` |
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| 59. |
Compute the area of the figure bounded by thestraight lines `=0,x=2`and the curves`y=2^x ,y=2x-x^2dot` |
| Answer» Correct Answer - `((3)/(log_(e)2)- (4)/(3))` sq. units | |
| 60. |
Area bounded by `y = x^(2)- 2|x|` and `y = -1` is equal toA. `2 underset(0)overset(1)int(2x-x^(2))dx`B. `2/3` sq. unitsC. `2/3` (Area of rectangle ABCD) where points A,B,C,D are `(-1,-1),(-1,0),(1,0)`&`(1,-1)`D. `2/3` (Area of rectangle ABCD) where points A,B,C,D are `(-1,-1),(-1,0),(1,0)`&`(1,-1)` |
| Answer» Correct Answer - B::D | |
| 61. |
For the area included between the parabolas `x=y^(2)` and `x = 3-2y^(2)`. |
| Answer» Correct Answer - `16/3` sq. units | |
| 62. |
Find the area bounded by the y-axis and the curve `x = e^(y) sin piy, y = 0, y = 1`. |
| Answer» Correct Answer - `((e+1)pi)/(1+pi^(2))` | |
| 63. |
The value of `int_(1//e)^(tanx)(t)/(1+t^(2))dt+int_(1//e)^(cotx)(1)/(t(1+t^(2)))dt`, where `x in (pi//6, pi//3`), is equal to : |
| Answer» Correct Answer - C | |
| 64. |
Find area bounded by `y = f^(-1)(x), x = 10, x = 4` and x-axis. Given that area bounded by `y = f(x) , x = 2, x = 6` and x -axis is `30` sq. units, where `f(2) = 4` and `f(6) = 10`. (given `f(x)` is an invertiable function). |
| Answer» Correct Answer - 22 sq. unit | |
| 65. |
If `u_(n) = int_(0)^(pi//2) x^(n)sinxdx`, then the value of `u_(10) + 90 u_(8)` is :A. `9(pi/2)^(8)`B. `(pi/2)^(9)`C. `10 (pi/2)^(9)`D. `9 (pi/2)^(9)` |
| Answer» Correct Answer - C | |
| 66. |
The area between the arms of the curve `|y|=x^(3)` from `x = 0` to `x = 2` isA. 2B. 4C. 8D. 16 |
| Answer» Correct Answer - C | |
| 67. |
The area bounded by the parabola `y=(x+1)^2` and `y=(x-1)^2` and the line `y=1/4` is (A) `4` sq. units (B) `1/6` sq. units (C) `3/4` sq. units (D) `1/3` sq. unitsA. 4 sq. unitsB. `1/6` sq. unitsC. `4/3` sq. unitsD. `1/3` sq. units |
| Answer» Correct Answer - D | |
| 68. |
For any real `t ,x=1/2(e^t+e^(-t)),y=1/2(e^t-e^(-t))`is a point on the hyperbola `x^2-y^2=1`Show that the area bounded by the hyperbola and the lines joining itscentre to the points corresponding to `t_1a n d-t_1`is`t_1dot` |
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Answer» It is a point on hyperbola `x^(2)-y^(2)=1`. Area `(PQRP) = 2 underset(1)overset(e^(t_(1))+e^(-t_(1)))overset(2)(int)ydx = 2 underset(1)overset(e^(t_(1))+e^(-t_(1)))overset(2)(int)sqrt(x^(2)-1)dx` `= 2[x/2sqrt(x^(2)-1)-1/2l(x+sqrt(x^(2)-1))]_(1)^(e^(t_(1))+e^(-t_(1))) = (e^(2t_(1))-e^(2t_(1)))/(4)-t_(1)` Area of `DeltaOPQ = 2 xx 1/2((e^(t_(1))+e^(t_(1)))/(2))((e^(t_(1))-e^(t_(1)))/(2)) = (e^(2t_(1))-e^(-2t_(1)))/(4)` `:.` Required area = area `DeltaOPQ` - area `(PQRP) = t_(1)`. (b) If `g(y) le 0` for `in [c,d]` then area bounded by curve `x = g(y)` and y-axis between abscissa `y = c`, and `-underset(y=c)overset(d)intg(y)dy` |
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| 69. |
Find the area between the x-axis and the curve ` y = sqrt(1+cos4x), 0 le x le pi`. |
| Answer» Correct Answer - `2sqrt(2)` sq. units | |
| 70. |
If `I_1=int_0^1(e^x)/(1+x) dx`aand `I_2=int_0^1 x^2/(e^(x^3)(2-x^3)) dx` then `I_1/I_2` isA. `3//e`B. `e//3`C. `3e`D. `1//3e` |
| Answer» Correct Answer - C | |
| 71. |
`int_(0)^(pi//2)(xsinxcosx)/(sin^(4)x+cos^(4)x)dx`. |
| Answer» Correct Answer - `(pi^(2))/(16)` | |
| 72. |
`int_(0)^(oo)((ln(1+x^(2)))/(1+x^(2)))dx`. |
| Answer» Correct Answer - `pi ln//2` | |
| 73. |
`int_(pi/12)^(pi/2)(dx)/(1+sqrt(cotx))` |
| Answer» Correct Answer - `(pi)/(6)` | |
| 74. |
Evaluate `int_(0)^(pi//4)(sinx+cosx)/(9+16sin 2x)dx`. |
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Answer» Correct Answer - `[(1)/(20)(log3)]` Let `I= int_(0)^(pi//4) ((sin x + cos x))/(9+ 16 sin 2 x)dx` ` I= int _(0)^(pi//4)(sin x + cos x)/(25- 16 (sin x - cos x)^(2))dx` Put `4 ( sin x - cos x ) = t rArr 4 (cos x + sin x ) dx = dt` `:. I= (1)/(4) int_(-4)^(0)(dt)/(25- t ^(2))= (1)/(4)* (1)/((5))log [ |(5+t)/(5-t)|]_(-4)^(0)` `I= (1)/(40) [ log |(5+0)/(5-0)| - log |(5-4)/(5+4)|]` `=(1)/(40) ( log 1 - " log " (1)/(9))=(1)/(40) log 9 = (1)/(20) (log 3)` |
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| 75. |
Evaluate `int_(0)^(1//2)(xsin^(-1)x)/(sqrt(1-x^(2)))dx`. |
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Answer» Correct Answer - `(-(sqrt(3))/(12)pi+(1)/(2))` ` = int _(0)^(1//2) (x sin ^(-1))/(sqrt(1-x^(2)))dx " put" sin^(-1) x = theta rArr x = sin theta` `rArr dx = cos theta d theta` ` :. I = int _(0)^(pi//6) (thetas sin theta)/(sqrt(1- sin ^(2)theta))* cos theta d theta = int _(0) ^(pi//6) theta sin d theta` `=[- theta cos theta]_(0)^(pi//6) + int_(0)^(pi//6)cos theta d theta` ` =(-(pi)/(6)"cos"(pi)/(6)+0)+("sin"(pi)/(6)-sin0)=-(sqrt(3)pi)/(12)+(1)/(2)` |
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| 76. |
Evaluate `int_(0)^(pi//2)(xsinxcosx)/(cos^(4)x+sin^(4)x)dx` |
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Answer» Correct Answer - `(pi^(2))/(16)` Let `I= int_(0)^(pi//2) (x sin x* cos x)/(cos^(4)x + sin ^(4) x)dx` `rArr I = int_(0)^(pi//2) (((pi)/(2)-x)sin ((pi)/(2)-x)* cos ((pi)/(2)-x))/(sin ^(4)((pi)/(2)-x)cos ^(4) ((pi)/(2)-x))dx` `rArr I= int _(0)^(pi//2) (((pi)/(2)x)* sin xcos x)/(cos^(4)x+ sin ^(4)x)dx` `rArr I = (pi)/(2) int_(0)^(pi//2)(sin x cos x)/(sin^(4) x + cos^(4) x)dx- int_(0)^(pi//2)(x sin x * cos x)/( sin^(4) x + cos^(4) x)dx` `= (pi)/(2) int_(0)^(pi//2) ( sin x * cosx)/( sin^(4) x + cos^(4)) dx-I` ` rArr 2 I = (pi)/(2) int _(0)^(pi//2) (tan x * sin ^(2)x)/(tan^(4) x + 1)dx` ` rArr 2 I = (pi)/(2)* (1)/(2) int_(0)^(pi//2) (1)/((tan ^(2) x)^(2))d (tan^(2)x)` ` rArr 2 I = (pi)/(4)* [ tan ^(-1) t]_(0)^(oo) = (pi)/(4)(tan ^(-1)oo - tan ^(-1) 0)` [ where , t ` tan^(2) x]` ` rArr I= (pi^(2))/(16)` |
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| 77. |
Let `f(x)={{:(-2",",-3lexle0),(x-2",",0ltxle3):}andg(x)=f(|x|)+|f(x)|` What is the value of the differential coefficient of g(x) at x=-2? |
| Answer» Correct Answer - `23/2` | |
| 78. |
A value of `alpha` such that `int_(alpha)^(alpha+1) (dx)/((x+a)(x+alpha+1))="loge"((9)/(8))` isA. `-2`B. `(1)/(2)`C. `-(1)/(2)`D. 2 |
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Answer» Correct Answer - A Let I `int_(alpha)^(alpha+1)(dx)/((x+a)(x+alpha1))` `int_(alpha)^(alpha+1)((x+alpha+1)-(x+alpha))/((x+alpha)(x+alpha+1))`dx `=int_(alpha)^(alpha+1)((1)/(x+alpha)-(1)/(x+alpha+1)) dx` `=[log_(e)(x+alpha)-log_(e)(x+alpha+1)]_(alpha)^(alpha+1)` `=[log_(e)((x+alpha)/(x+alpha+1))]_(alpha)^(alpha+1)` `="log"_(e)(2alpha+1)/(2alpha+2)-"log"_(e)(2alpha)/(2alpha+1)` `=log_(e)((2alpha+1)/(2alpha+2)xx(2alpha+1)/(2alpha))="log"_(e)((9)/(8))` (given) `rArr((2alpha+1)^(2))/(4alpha(alpha+1))=(9)/(8)rArr8 [4alpha+1] = 36 (alpha^(2)+alpha)` `rArr8alpha^(2)+8alpha+2=9alpha^(2)+9alpha` `rArralpha^(2)+alpha-2=0` ` rArr(alpha+2)(alpha-1)=0` `rArralpha=1,-2` From the options we get `alpha -=2` |
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| 79. |
Let `I_(n) = int_(0)^(1//2)(1)/(sqrt(1-x^(n))) dx` where `n gt 2`, thenA. `I_(n) lt pi/6`B. `I_(n) gt pi/6`C. `I_(n) lt 1/2`D. `I_(n) gt 1/2` |
| Answer» Correct Answer - A::D | |
| 80. |
`lim_(n -> oo) (((n+1)(n+2)(n+3).......3n) / n^(2n))^(1/n)`is equal toA. `27/(e^(2)0`B. `9/(e^(2))`C. `3log3-2`D. `18/e^(4)` |
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Answer» Correct Answer - A `L=int_(nto oo) (((n+1)(n+2)……….(n+2n))/(n^(2n)))^(1//n)` `:. log_(e)L=1/n(lim_(nto oo) sum_(r=1)^(2n)log(1+4/n))` `:.log_(e)L=int_(0)^(2)log(1+x)dx` `:.log_(e)L(xlog(1+x))_(0)^(2)-int_(0)^(2)x/(1+x)dx` `:.log_(e)L=2log_(e)3-int_(0)^(2)(1-1/(1+x))dx` `:. log_(e)L=2log3-(x-log(1+x))_(0)^(2)` `=log_(e)L=2log3-(2-log3)` `:.log_(e)L=3log3-2="log"27/(e^(2))` `:.L=27/(e^(2))` |
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| 81. |
Find the value of `int_(1/2)^(2)e^(|x-1/x|)dx`. |
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Answer» Correct Answer - `esqrt(e)-1` Let `I=int_(1/2)^(2)e^(|x-1/x|)dx`…………….1 Put `x=1/t`, `:. I=-int_(0)^(1/2)e^(|t- 1/t|)((dt)/(t^(2)))=int_(1/2)^(2)e^(|x-1/x|)(dx)/(x^(2))`……………2 Adding 1 and 2 we get `2I=int_(1/2)^(2)e^(|x - 1/x|),(1+1/(x^(2)))dx` `=int_(1/2)^(1)e^(-(x-1/x)),(1+1/(x^(2)))dx+int_(1)^(2)e^((x-1/x)),(1+1/(x^(2)))dx` `=[-e^(-(x-1/x))]_(1//2)^(1)+[e^((x-1/x))]_(1)^(2)` `=-1+e^(3//2)+e^(3//2)-1` `=2(esqrt(e)-1)` `impliesI=esqrt(e)-1` |
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| 82. |
The integral `int_(2)(4)(logx^(2))/(logx^(2)+log(36-12x+x^(2))) dx` is equal toA. `2`B. `4`C. `1`D. `6` |
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Answer» Correct Answer - C `I=int_(2)^(4)(logx^(2))/(logx^(2)+log(36-12x+x^(2))dx` `I=2/2 int_(2)^(4)(log|x|)/(log|x|+log||6-x|)dx` ……….i `I=int_(2)^(4)(log|6-x|)/(log|6-x|+log|x|) dx{int_(a)^(b)f(x)dx=int_(1)^(b)f(a+b-x)dx}` ……ii Adding i and ii `2I=int_(2)^(4)(log|x|+log|6-x|)/(log|x|+log|6-x|)dx=int_(2)^(4)dx=2` Hence `I=1` |
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| 83. |
If `f(x)=int_(-1)^(x)|t|dt`, then for any `xge0,f(x)` equalsA. `1/2(1-x^(2))`B. `1/2x^(2)`C. `1/2(1+x^(2))`D. none of these |
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Answer» Correct Answer - C `f(x)={(int_(-1)^(x)-tdt,-1lexlt0),(int_(-1)^(0)-tdt+int_(0)^(x)tdt, xge0):}={(1/2(1-x^(2)),-1lexlt0),(1/2(1+x^(2)),xge0):}` |
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| 84. |
Let `f` be a positive function. If `I_1 = int_(1-k)^k x f[x(1-x)] dx` and `I_2 = int_(1-k)^k f[x(1-x)] dx,` where `2k-1 gt 0.` Then `I_1/I_2` isA. 2B. kC. `1//2`D. 1 |
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Answer» Correct Answer - C Given , `I_(1)=int_(1-k)^(k)xf[x(1-x)]dx` `rArrI_(1)=int_(1-k)^(k)(1-x)f[(1-x)x]dx` `=int_(1-k)^(k)f[(1-x)]dx]int_(1-k)^(k)xf(1-x)]dx` `rArrI_(1)=I_(2)-I_(1)rArr(I_(1))/(I_(2))=(1)/(2)` |
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| 85. |
If `int_0^(pi/2)logsinthetadtheta=k ,`then find the value of `int_pi^(pi/2)(theta/(s intheta))^2dtheta`in terms of `k` |
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Answer» `I=int_(0)^(pi//2)((theta)/(sin theta))^(2)d theta` `=int_(0)^(pi//2) theta^(2) cosec^(2) theta d theta` `=[theta^(2) (-cot theta)]_(0)^(pi//2)-int_(0)^(pi//2) 2theta (-cot theta) d theta` (Integrating by parts) `=[lim_(theta to oo) theta^(2).cot theta]+2 int_(0)^(pi//2) theta cot theta d theta` `=0+2[[ theta log sin theta]_(0)^(pi//2)-int_(0)^(pi//2) log sin theta d theta]` (Integrating by parts) `=2[-lim_(theta to oo) theta n sin theta -k]` `=-2k` |
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| 86. |
If `g(x)=int_(0)^(x)cos^(4)` t dt , then ` (x+pi)` equalsA. `g(x) +g (pi)`B. `g(x)-g (pi)`C. `g(x)g(pi)`D. `(g(x))/(g(pi))` |
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Answer» Correct Answer - A Given , f (x) `=int_(0)^(x)cos^(4)t dt` `rArrg(x+pi)=int_(0)^(pi+x)cos^(4)t dt` ` = int_(0)^(pi)cos^(4)t dt+int_(pi)^(pi+x)cos^(4)t dt = I_(1)+I_(2)` where , `I_(1)=int_(0)^(pi)cos^(4)t dt=g(pi)` and `I_(2)=int_(pi)^(pi+x)cos^(4)t dt` Put `t=pi+y` `rArrdt = dy` `I_(2)=int_(0)^(x)cos^(4)(y+pi)dy` `=int_(0)^(x)(-cosy)^(4)dy=int_(0)^(x)cos^(4)ydy=g(x)` `:. g(x+pi)=g(pi)+g(x)` |
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| 87. |
The integral `int_(0)^(pi)sqrt(1+4"sin"^(2)x/2-4"sin"x/2)dx` equalsA. `pi-4`B. `(2pi)/3-4-sqrt(3)`C. `4sqrt(3)-4`D. `4sqrt(3)-4-(pi)/3` |
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Answer» Correct Answer - D `I=int_(0)^(pi)sqrt(1+4"sin"^(2)x/2-4"sin"x/2)dx` `=int_(0)^(x)|1-2"sin"x/2|dx` `=int_(0)^(pi//3)|1-2"sin"x/2|dx` `=int_(0)^(pi//3) (1-2"sin"x/2)dx+int_(pi//3)^(pi)(2"sin"x/2-1)dx` `=(x+4"cos"x/2)|._(0)^(pi//3)+(-4"cos"x/2-x)|_(pi//3)^(pi)` `=4sqrt(3)-4-(pi)/3` |
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| 88. |
If `f(x)` is a function satifying `f(1/x) + x^(2)f(x) = 0` for all non-zero x, then `int_(sintheta)^("cosec" theta)f(x) dx` is equals to :A. `sintheta+cosestheta`B. `sin^(2)theta`C. `cosec^(2)theta`D. none of these |
| Answer» Correct Answer - D | |
| 89. |
Evaluate:`int_(-pi/2)^(pi/2)log((a-s intheta)/(a+sintheta))dtheta,a >0` |
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Answer» `f(theta)=log((a-sintheta)/(a+sin theta))` `:.f(-theta)=log((a+sin theta)/(a-sin theta))` `=-log((a-sin theta)/(a+sin theta))` `=-f(theta)` Hence the integrand is anodd function. So, the given integral is zero. |
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| 90. |
The value of the definite integral `int_(-1)^(1)(1+x)^(1//2)(1-x)^(3//2)dx` equalsA. `pi`B. `(3pi)/4`C. `(pi)/4`D. `(pi)/2` |
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Answer» Correct Answer - D `I=int_(-1)^(1)(1+x)^(1//2)(1-x)^(3//2)dx` `:.I=int_(-1)^(1)(1-x)^(1//2)(1+x)^(3//2)dx` (Replacing `x` by` -x`) Adding we get `2I=int_(-1)^(1)(1+x)^(1//2)(1-x)^(1//2)[(1-x)+(1+x)]dx` or `I=int_(-1)^(1)sqrt(1-x^(2))dx` `:. I=2int_(0)^(1)sqrt(1-x^(2))dx` `=2[x/2sqrt(1-x^(2)+1/2sin^(-1)x)]_(0)^(1)` `=2[0+1/2 . (pi)/2-0-0]=(pi)/2` |
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| 91. |
Showthat: `int_0^(pi//2)f(sin2x)sinxdx=sqrt(2)int_0^(pi//4)f(cos2x)cosxdxdot` |
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Answer» Let `I=int_(0)^(pi//2)f(sin2x)sinx dx` `=int_(0)^(pi//2)f{"sin"2(1/2pi-x)}sin(1/2 pi-x)dx` `:. I=int_(0)^(pi//2)f(sin2x)cosx dx` Then adding 1 and 2 we have `2I=int_(0)^(pi//2)f(sin2x)(sinx+cosx)dx` `=sqrt(2)int_(0)^(pi//2)f(sin2x)sin(x+(pi)/4)dx` To get `f(cos 2x)` on RHS, we have to substitute `(pi)/2-2theta =2x`. `:. dx=d theta` Also, when `x=0, theta pi//4`. and when `x=pi//2, theta=-pi//4`. `:. 2I=sqrt(2)int_(-pi//4)^(pi//4)f(cos 2theta) cos theta d theta ` `=2sqrtint_(0)^(pi//4) f(cos 2 theta) cos theta d theta` [as `g(theta)=f(cos 2theta) cos theta` is an even function) `:.I=sqrt(2)int_(0)^(pi//4)f(cos2x)cosx dx` |
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| 92. |
The value of difinite integral `int_(0)^(1)=(dx)/(sqrt((x+1)^(3)(3x+1)))` equalsA. `sqrt2-1`B. `tan.(pi)/(12)`C. `tan.(5pi)/(12)`D. none of these |
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Answer» Correct Answer - A `I=int_(0)^(1)(dx)/((x+1)sqrt((x+1)(3(x+1)-2)))` Put `x+1=(1)/(t)` `therefore" "I=int_(1)^(1//2)(dt)/(sqrt(3-2t))=sqrt2-1` |
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| 93. |
If" f, is a continuous function with `int_0^x f(t) dt->oo` as `|x|->oo`then show that every line `y = mx` intersects the curve `y^2 + int_0^x f(t) dt = 2` |
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Answer» We are given that `f` is a continuous function and `int_(0)^(x)f(t)dt to oo` as `|x|to oo` We have to show that every line `y=mx` intgersects the curve `y^(2)+int_(0)^(x)f(t)dt=2` If possible, let `y=mx` intersects the given curve. Then substituting `y=mx` in the curves, we get `m^(2)x^(2)+int_(0)^(x)f(t)dt=2` Consider `F(x)=m^(2)x^(2)+int_(0)^(x)f(t)dt-2` Then `F(x)` is a continuous function as `f(x)` is given to be continuous. Also `F(x)to oo` as `|x|to oo` But `F(0)=-2` So the graph of `y=F(x)` must corss `x` -axis to reach infinity as `|x|` approaches infinity. Hence `y=mx` intersects the given curves. |
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| 94. |
`f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt` `f(x)` is not invertible forA. `x epsilon [-(pi)/2-tan^(-1)2,(pi)/2-tan^(-1)2]`B. `x epsilon[tan^(-1)1/2,pi+"tan"^(-1)1/2]`C. `x epsilon[pi+cot^(-1)2, 2pi+cot^(-1)2]`D. none of these |
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Answer» Correct Answer - D `f(x)=sinx+sin int_(-pi//2)^(pi//2) f(t)dt+cosx int_(-pi//2)^(pi//2) tf(t)dt` `=sinx (1+int_(-pi//2)^(pi//2) f(t)dt)+cosx int_(-pi//2)^(pi//2) tf(t)dt` `=Asinx+Bcosx` Thus, `A=1+int_(-pi//2)^(pi//2) f(t)dt` `=1+int_(-pi//2)^(pi//2) (Asint+Bcost)dt` `=1+2Bint_(0)^(pi//2) cost dt` `=A=1+2B` ............1 `B=int_(-pi//2)^(pi//2) tf(t)dt` `=int_(-pi//2)^(pi//2)t(Asint+Bcost)dt` `=2Aint_(0)^(pi//2) t sin tdt` `=2A[-tcost+sint]_(0)^(pi//2)` `:.B=2A` From equations 1 and 2 we get `A=-1//3, B=-2//3` `:.f(x)=-1/3(sinx+2cosx)` Thus, the range fo `f(x)` is `[-(sqrt(5))/3,(sqrt(5))/3]` `f(x)=-1/2(sinx+2cosx)` `=-(sqrt(5))/3 sin (x+tan^(-1)2)` `=-(sqrt(5))/3cos(x-"tan"^(-1)1/2)` `f(x)` in invertible if `-(pi)/2 le x=tan^(-1)2le(pi)/2` or `-(pi)/2-tan^(-1)2 le xle (pi)/2 -tan^(-1) 2` or `0lex-"tan"^(-1)1/2 le pi` or `"tan"^(-1)1/2 le x le pi+"tan"^(-1)1/2` or `pi le x-"tan"^(-1)1/2le 2pi` or `x epsilon [pi+cot^(-1)2, 2pi +cot^(-1) 2]` `int_(0)^(pi//2) f(x)dx=-1/3int_(0)^(pi//2) (sinx+2cosx)dx` `=-1/3[-cosx+2sinx]_(0)^(pi//2) =-1` |
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| 95. |
Evaluate: `int_(-pi//2)^(pi//2)(cosx)/(1+e^x)dx`A. `(pi^(2))/4-2`B. `(pi^(2))/4+2`C. `pi^(2)-e^((pi)/2)`D. `pi^(2)+e^(pi)/2` |
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Answer» Correct Answer - A `I=int_(-(pi)/2)^((pi)/2)(x^(2)cosx)/(1+e^(x))dx`………………1 `=int_(-(pi)/2)^((pi)/2)(x^(2)cosx)/(1+1/(e^(x))`(Replacing x by `-x`) `:. I=int_(-(pi)/2)^((pi)/2)(x^(2)cosx.e^(x))/(1+e^(x))dx`……………..2 Adding 1 and 2 we get `2I=int_(-(pi)/2)^((pi)/2)x^(2)cosx dx` `:. I=int_(0)^((pi)/2)x^(2)cosxdx` `=x^(2). sin x|_(0)^(pi//2)-int_(0)^((pi)/2)2x sin xdx` `=(pi^(2))/4-2[(-xcosx)_(0)^(pi//2)-int_(0)^(pi//2)( -cosx)dx]` `=(pi^(2))/4-2[1]=(pi^(2))/4-2` |
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| 96. |
If `I_n=int_(-pi)^(pi) (sinnx)/((1+pi^x) sinx) dx, n=0,1,2,......` then which one of the following is not true ?A. `I_(n)=I_(n+2)`B. `sum_(m=1)^(10)I_(2m+1)=10pi`C. `sum_(m=1)^(10)I_(2m)=0`D. `I_(n)=I_(n+1)` |
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Answer» Correct Answer - A::B::C `I_(n)=int_(-pi)^(pi)(sin nx)/((1+pi^(x))sinx)dx` `=int_(0)^(pi)((sin nx)/((1+pi^(x))sin x)+(pi^(x)sin nx)/((1+pi^(x))sinx))dx=int_(0)^(pi)(sin nx)/(sin x)dx` Now `I_(n_2)-I_(n)=int_(0)^(pi)(sin(n+2)x-sin nx)/(sinx)dx` `=int_(0)^(pi)(2cos(n+1)x sinx)/(sin x)dx=0` `:. I_(1)=pi,I_(2)=int_(0)^(pi)2cosx dx=0` |
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| 97. |
Find `a` for which `lim_(n->oo) (1^a+2^a+3^a+...+n^a)/((n+1)^(a-1)[(na+1)+(na+2)+...+(na+n)])=1/60`A. 5B. 7C. `(-15)/2`D. `(-17)/2` |
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Answer» Correct Answer - B::D Given limit `=(lim_(n to oo) 1/n sum_(r=1)^(n)(r/n)^(a))/(lim_(n to oo) (1+1/n)^(a-1)lim_(n to oo) 1/n sum_(r=1)^(n)(a+r/n))=(int_(0)^(1)x^(a)dx)/(int_(0)^(1)(a+x)dx)` `=2/((2a+1)(a+1))=2/120` `:.a=7` or `-17/2` |
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| 98. |
`lim_(n->oo)[(1+1/n^2)(1+2^2 /n^2)(1+3^2 /n^2)......(1+n^2 / n^2)]^(1/n)`A. `(e^(x//2))/(2e^(2))`B. `2e^(2)e^(pi//2)`C. `(2)/(e^(2))e^(x//2)`D. `2e^(x)` |
| Answer» Correct Answer - C | |
| 99. |
If the line `x = alpha` divides the area of region `R = {(x,y) in R^(2) : x^(0) le y le x, 0 le x le 1}` into two equal parts, thenA. `2alpha^(4)-4alpha^(2)+1=0`B. `alpha^(4)+4alpha^(2)-1=0`C. `1/2 lt alpha lt 1`D. `0 lt alpha le 1/2` |
| Answer» Correct Answer - AC | |
| 100. |
For each positive integer `n`, let `y_(n) = 1/n ((n+1)(n+2) "……"(n+n))^(1//n)` For `x in R`, let `[x]` be the greatest integer less than or equal to `x`. If `lim_(nrarroo) y_(n) = L`, then the value of `[L]` is `"______"`. |
| Answer» Correct Answer - 1 | |