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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. |
`underset(n to oo)lim(1)/(2)" " underset(r=+1)overset(2n)sum (r)/(sqrt(n^(2)+r^(2)))` equalsA. `1+sqrt5`B. `-1+sqrt5`C. `-1+sqrt2`D. `1+sqrt2` |
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Answer» Correct Answer - B Let `l=underset(nrarroo)(lim)(1)/(n)sum_(r=1)^(2n)(r)/(sqrt(n^(2)-r^(2)))` `=underset(nrarroo)(lim)(1)/(n)sum_(r=1)^(2n)(r)/(sqrt(1+((r)/(n))^(2)))` `=underset(nrarroo)(lim)(1)/(n)sum_(r=1)^(2n)((r)/(n))/(sqrt(1+((r)/(n))^(2)))` Put `(r)/(n)=x,(1)/(n)=dx, underset(nrarroo)(lim)sum_(r=1)^(2n)=int_(0)^(2)` `therefore" "l=int_(0)^(2)(x)/(sqrt(1+x^(2)))dx=[sqrt(1+x^(2))]_(0)^(2)` `=sqrt5-1` |
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| 52. |
`lim_(n->oo)1/n[1/(n+1)+2/(n+2)+....+(3 n)/(4 n)]`A. log4B. `-log4`C. `1-log4`D. None of these |
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Answer» Correct Answer - D `underset(nrarroo)(lim)(1)/(n)((1)/(n+1)+(2)/(n+2)+...+(3n)/(4n))` `=underset(nrarroo)(lim)(1)/(n)(((1)/(n))/(1+(1)/(n))+((2)/(n))/(1+(2)/(n))+...+((3n)/(n))/(1+(3n)/(n)))` `=underset(nrarroo)(lim)(1)/(n)sum_(r=1)^(3n)(((r)/(n))/(1+(r)/(n)))=int_(0)^(3)(x)/(1+x)dx` `=int_(0)^(3)(x+1-1)/((1+x))dx=int_(0)^(3)dx-int_(0)^(3)(1)/(1+x)dx` `=[x-log(x+1)]_(0)^(3)=3-log4` |
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| 53. |
The value `overset(2)underset(-2)int {p" In"((1+x)/(1-x))+q" In "((1-x)/(1+x))-2+r}dx` depends on the value ofA. the value of pB. the value of qC. the value of rD. The values of p and q |
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Answer» Correct Answer - C ` l =int_(-2)^(2)[plog((1+x)/(1-x))+qlog((1-x)/(1+x))^(-2)+r]dx` Since , log `((1+x)/(1-x))` is an odd function . `rArrint_(-2)^(2)log((1+x)/(1-x))dx=0` `therefore l = int_(-2)^(2)r dx =r (2+2)=4r` |
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| 54. |
The value of ` int _(0)^(12a) (f(x))/(f(x)+f(12a-x))dx ` isA. aB. 2aC. 3aD. 6a |
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Answer» Correct Answer - D Let `l =int_(0)^(12a)(f(x))/(f(x)+f(12a-x))dx` … (i) `rArrl=int_(0)^(12a)(f(12a-x))/(f(12a-x)+f(x))dx` … (ii) On adding Eqs . (i) and (ii) , we get `2l=int_(0)^(12a)1dx=[x]_(0)^(12a)` `rArrl=6a` |
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| 55. |
If `2f(x) - 3 f(1//x) = x," then " int_(1)^(2) f(x) dx` is equal toA. `(3//5)log2`B. `(-3//5)(1+log2)`C. `(-3//5)log2`D. None of these |
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Answer» Correct Answer - B Given , `2f(x)-3f((1)/(x))=x` Put `x=(1)/(x)` Eq . (i) , we get `2f((1)/(x))-3f(x)=(1)/(x)` On solving Eqs . (i) and (ii) , we get `f(x)=-(3+2x^(2))/(5x)` Now , `int_(1)^(2)f(x)dx=-int_(1)^(2)(3+2x^(2))/(5x)dx` `=-int_(1)^(2)((3)/(5x)+(2x)/(5))dx` `=(1)/(5)[3logx+(2x^(2))/(2)]_(1)^(2)` `=-(1)/(5)(3 log2+4-2log1-1)` `=-(1)/(5)(3log2+4-0-1]` `=-(3)/(5)(1+log2)` |
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| 56. |
The value of `overset(pi)underset(-pi)int(1-x^(2)) sin x cos^(2) x" dx"`, is |
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Answer» Correct Answer - A Let `f(x)=(1-x^(2)) sin x cos^(2)x` `f(-x)=[1-(-x)^(2)][sin(-x)]cos^(2)(-x)` `=-(1-x^(2))sin x cos^(2) x = - f(x)` `rArr f(x)` in an odd function. `therefore " " int_(-pi)^(pi)(1-x^(2))sin x cos^(2) x dx = 0` |
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| 57. |
If ` int _(0)^(x^(2)) f(t) dt = x cos pix` , then the value of f(4) isA. 1B. `1/4`C. `-1`D. `(-1)/4` |
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Answer» Correct Answer - B `int_(0)^(x^(2))f(t)dt=xcospix` On differentiating both sides , we get `2xf(x^(2))=(-xsinpix)/(pi)+cospix` `rArrf(x^(2))=-(xsinpix)/(2pix)+(cospix)/(2x)` `thereforef(4)=f(2^(2))=(-2sin2pi)/(4pi)+(cos^(2)pi)/(4)=(1)/(4)` |
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| 58. |
`int_(-2)^1|2x+1|` dxA. `(5)/(2)`B. `(7)/(2)`C. `(9)/(2)`D. `4` |
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Answer» Put `2x+1=t` and `dx=(1)/(2)dt`. `[x=-2impliest=-3]` and `[x=1impliest=3]` `:.I=(1)/(2)int_(-3)^(3)|t|dt=(1)/(2){int_(-3)^(0)(-t)dt+int_(0)^(3)t dt}` `=(1)/(2){[(-t^(2))/(2)]_(-3)^(0)+[(t^(2))/(2)]_(0)^(3)}=(1)/(2)[{0-((-9)/(2))}+((9)/(2)-0)]=(9)/(2)`. |
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| 59. |
`l isum_(n-gtoo)sum_(r=1)^n1/(sqrt(4n^2-r^2))`A. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(2)` |
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Answer» Correct Answer - A `S=underset(nrarroo)(lim)sum_(r=0)^(n-1)(1)/(sqrt(4n^(2)-r^(2)))` `rArr" "S=(1)/(n)underset(nrarroo)(lim)sum_(r=0)^(n-1)(1)/(sqrt(2^(2)-((r)/(n))^(2)))` `rArr" "S=int_(0)^(1)(dx)/(sqrt(2^(2)-x^(2)))` `rArr" "S=[sin^(-1).(x)/(2)]_(0)^(1)=sin^(-1).(1)/(2)-sin^(-1)` `rArr" "S=(pi)/(6)` |
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| 60. |
` lim _(n to oo) ( 1/(n^(2)+1^(2)) + 1/(n^(2)+2^(2)) +...1/(2n^(2)))` equalsA. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(2)` |
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Answer» Correct Answer - B Let `S=(1^(lim))/(n_(nrarroo))((n)/(n^(2)+1^(2))+(n)/(n^(2)+2^(2))+...+(n)/(n^(2)+n^(2)))` `S=(1)/(n)underset(nrarroo)(lim)sum_(r=1)^(n)(1)/(1+((r)/(n))^(2))` `S=int_(0)^(1)(dx)/(1=x^(2))=[tan^(-1)x]=(pi)/(4)` |
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| 61. |
` int_(0)^(1) 1/(x+sqrt(x))` is equal toA. `log3`B. `log1`C. `log4`D. `log2` |
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Answer» Correct Answer - C Let ` l = int _(0)^(1) (dx)/(x+sqrt(x))= int _(0)^(1) (dx)/(sqrt(x)(sqrt(x)+1))` Put ` sqrt(x) = t rArr 1/(2sqrt(x)) dx = dt rArr 1/(sqrt(x)) dx = 2 dt ` ` l = int _(0)^(1) (2dt)/(t+1) = 2 [ log (t+1)]_(0)^(1) = 2 ( log 2 - log 0 ) = log 4 ` |
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| 62. |
` int _(-1)^(1)| 1 - x| dx ` is equal toA. `-2`B. 0C. 2D. 4 |
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Answer» Correct Answer - C Let ` l = int _(-1)^(1) | 1-x| dx ` Here, ` -1 le xle 1 rArr 1 - x ge 0 ` , ` :. l = int_(-1)^(1) (1-x) dx = [ x - (x^(2))/2 ] _(-1)^(1)` ` = 1- 1/2 + 1/2 +1 = 2` |
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| 63. |
` int _(-1)^(1) (e^(x^(3)) +e^(-x^(3))) (e^(x)-e^(-x)) dx` is equal toA. `(e^(2))/2- 2e`B. ` e^(2) - 2e`C. ` (e^(2)-2e)`D. 0 |
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Answer» Correct Answer - D Let `f(x)=(e^(x^(3))+e^(-x^(3)))(e^(x)-e^(-x))` `rArrf(-x)=(e^(-x^(3))+e^(x^(3)))(e^(-x)-e^(x))` `=-(e^(x^(3))+e^(-x^(3)))(e^(x)-e^(-x))=-f(x)` `rArr`f(x) is an odd function. `thereforeint_(-1)^(1)f(x)dx=0` |
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| 64. |
`underset(n to oo)lim underset(r=1)overset(n)sum(1)/(n)e^(r//n)` isA. eB. `e-1`C. `1-e`D. `e+1` |
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Answer» Correct Answer - B `lim_(pitoinfty)sum_(r=1)^(n)(1)/(n)e^(r//n)=int_(0)^(1)e^(x)dx=[e^(x)]_(0)^(1)=e-1` |
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| 65. |
The value of ` int _(-1)^(1) x|x| dx` isA. 2B. 1C. 0D. None of these |
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Answer» Correct Answer - C Let `f(x) = x|x|` `f(-x)=-x|x|=-f(x)` `therefore f(x)` is an odd function. Hence, l = 0 |
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| 66. |
`underset(n to oo)lim((1)/(n)+(1)/(n+1)+...+(1)/(3n))` is equal toA. log 2B. log3C. log5D. 0 |
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Answer» Correct Answer - B `lim_(ntoinfty)((1)/(n)+(1)/(n+1)+...+(1)/(n+2n))` `=sum_(r=0)^(2n)(1)/(n+r)=(1)/(n)sum_(r=0)^(2n)(1)/(1+(r)/(n))` `=int_(0)^(2)(1)/(1+x)dx=[log(1+x)]_(0)^(2)-log3-log1=log3` |
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| 67. |
The value of ` int_(0)^(oo) (dx)/(a^(2)+x^(2))` is equal toA. `(pi)/(2)`B. `(pi)/(2a)`C. `(pi)/(a)`D. `(1)/(2a)` |
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Answer» Correct Answer - B ` int _(0)^(oo) (dx)/((a^(2)+x^(2))) = [ 1/a tan^(-1). x/a ] _(0)^(oo) = 1/a (pi/2 - 0) = pi/(2a) ` |
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| 68. |
`int_(-2)^(1)(|x|)/(x)dx=?`A. `3`B. `2.5`C. `1.5`D. none of these |
| Answer» `I=int_(-2)^(0)(-x)/(x)dx+int_(0)^(1)(x)/(x)dx=int_(-2)^(0)-dx+int_(0)^(1)dx=[-x]_(-2)^(0)+[x]_(0)^(1)=(-2+1)=-1`. | |
| 69. |
`int_(-a)^(a)x|x|dx=?`A. `0`B. `2a`C. `(2a^(3))/(3)`D. none of these |
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Answer» `I=int_(-a)^(0)x(-x)dx+int_(0)^(a)x*xdx=int_(-a)^(0)-x^(2)dx+int_(0)^(a)x^(2)dx` `=[(-x^(3))/(3)]_(-a)^(0)+[(x^(3))/(3)]_(0)^(a)=+(1)/(3)(-a)^(3)+(a^(3))/(3)=0`. |
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| 70. |
The value of integral ` int _(1//pi)^(2//pi)(sin(1/x))/(x^(2))dx=`A. 2B. `-1`C. 0D. 1 |
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Answer» Correct Answer - D Put `t=(1)/(x) rArr dt = -(1)/(x^(2))dx` at `t=(pi)/(2)` and `pi` `therefore int_(1//pi)^(2//pi)(sin((1)/(x)))/(x^(2))dx=-int_(pi)^(pi//2)sin t dt = [cos t]_(pi)^(pi//2)` `=[cos. (pi)/(2)-cos (pi)]=1` |
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| 71. |
Evaluate :`int_0^(pi/2)(sinx+cosx)dx` |
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Answer» `I = int_0^(pi/2) (sinx+cosx)dx` `=>I = [-cosx+sinx]_0^(pi/2)` `=>I = [-cos(pi/2)+sin(pi/2) + cos (0) - sin (0)]` `=> I = [0+1+1 - 0]` `=> I = 2` |
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| 72. |
Evaluate :`int_(-2)^3 1/(x+7)dx` |
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Answer» `I = int_-2^3 1/(x+7)dx` `=>I = [log |x+7|]_2^3` `=>I = log 10 - log5` `=>I = log(10/5)` `=>I = log2` |
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| 73. |
If `bgta,` then `int_(a)^(b)(dx)/(sqrt((x-a)(b-x)))` is equal toA. `(pi)/(2)`B. `pi`C. `(pi)/(2)(b-a)`D. `(pi)/(4)(b-a)` |
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Answer» Correct Answer - B `int _(a)^(b) (dx)/(sqrt(x-a)(b-x))=int_(a)^(b) 1/(sqrt(-x^(2) + (a+b) x - ab))dx` , ` = int _(a)^(b) 1/(sqrt(((b-a)/2)^(2) - (x-(a+b)/2)^(2)))dx = [ sin^(-1).(((x-(a+b))/2)/((b-2)/2))]_(a)^(b)` ` = sin^(-1) 1- sin^(-1) (-1)` ` = pi/2 + pi/2 = pi` |
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| 74. |
Show that `int_(-pi//2)^(pi//2)sin^(7)xdx=0`. |
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Answer» Let `f(x)=sin^(7)x`. Then, `f(-x)=[sin(-x)]^(7)=-sin^(7)x=-f(x)`. `:.f(x)` is an odd function of `x`. But, `int_(-a)^(a)f(x)dx=0`, when `f(x)` is odd `:.int_(-pi//2)^(pi//2)sin^(7)xdx=0`. |
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| 75. |
` int_(-1)^(1) (dx) / (x^(2)+2x+5)` is equal toA. `pi/2`B. `pi/4`C. `pi/8`D. `pi/3` |
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Answer» Correct Answer - C ` int _(-1)^(1) 1/(x^(2)+2x+5)dx= int_(-1)^(1)1/((x^(2)+2x+1)+4)dx` ` = int _(-1)^(1) 1/((x+1)^(2)+2^(2))dx=1/2[ tan^(-1).(x+1)/2]_(-1)^(1)` ` = 1/2 [ tan^(-1)(2/2)-tan^(-1)(0/2)]` ` = 1/2 [ tan^(-1)1] = 1/2 xx pi/4 = pi/8` |
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| 76. |
` int _(0)^(pi//2) x sin^(2) x cos^(2) x dx ` is equal toA. `(pi^(2))/(32)`B. `(pi^(2))/(16)`C. `(pi)/(32)`D. None of these |
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Answer» Correct Answer - D Let ` l = int _(0)^(pi//2) x sin^(2) x cos ^(2) x dx " "` …(i) ` rArr l = int _(0)^(pi//2)(pi/2-x) x sin^(2) x cos^(2) x dx " "` …(ii) On adding Eqs . (i) and (ii) , we get ` 2l = pi/2 int_(0)^(pi//2) sin^(2) x cos ^(2) x dx ` ` = pi/8 int _(0)^(pi//2) sin^(2) 2 x dx = pi/8 int _(0)^(pi//2) ((1-cos 4x)/2) dx` ` rArr 2l = pi/16 [ x - (sin 4x)/4]_(0)^(pi//2)` ` rArr 2l = pi/16 [ pi/2]` ` rArr l = (pi^(2))/64` |
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| 77. |
`int_-2^2 |x| dx` is equal toA. `4`B. `3.5`C. `2`D. `0` |
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Answer» `I=int_(-2)^(0)|x|dx+int_(0)^(2)|x|dx=int_(-2)^(0)-xdx+int_(0)^(2)xdx`. `=[-(x^(2))/(2)]_(-2)^(0)+[(x^(2))/(2)]_(0)^(2)=0-(-2)+(2-0)=4`. |
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| 78. |
Examples: `int_(-pi/2) ^(pi/2) sin^2x dx` |
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Answer» Let `f(x)=sin^(2)x`. Then, `f(-x)=[sin(-x)^(2)]=(-sinx)^(2)=sin^(2)x=f(x)`. `:.f(x)` is an even function. So, `int_(-pi//2)^(pi//2)sin^(2)xdx=2int_(0)^(pi//2)sin^(2)xdx=2int_(0)^(pi//2)((1-cos2x)/(2))dx` `=int_(0)^(pi//2)(1-cos2x)dx=[x-(sin2x)/(2)]_(0)^(pi//2)=(pi)/(2)`. |
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| 79. |
`int_0^1(cos^(- 1)x)^2dx` |
| Answer» Correct Answer - Put `x=cost`. | |
| 80. |
` int_(-1)^(2)f(x)dx " where " f(x) = |x+1| +|x|+ |x-1|` is equal toA. `7/2`B. `9/2`C. `13/2`D. `19/2` |
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Answer» Correct Answer - D We can redefine f as f as `f (x) = {:{(2-x,"if"-1 lt x le 0),(x+2,"if " 0 lt x le 1 ),(3x,"if " 1ltxle 2):}` Therefore, ` int_(-1)^(2) f(x) dx = int_(-1)^(0) (2-x) dx + int _(0)^(1) (x+2) dx + int_(1)^(2)3x dx` ` = [ 2x - (x^(2))/2]_(1)^(0)+[ (x^(2))/2+2x] _(0)^(1) + [ (3x^(2))/2]_(1)^(2)` ` = { 0- (-2-1/2)}+(1/2 +2 )+3 (4/2-1/2)= 5/2 +5/2 +9/2 = 19/2 ` |
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| 81. |
Prove that `int_- 1^1log((2-x)/(2+x))^(20)dx=0` |
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Answer» Let `f(x)=log((2-x)/(2+x))`. Then, `f(-x)=log((2+x)/(2-x))=log((2-x)/(2+x))^(-1)=-log((2-x)/(2+x))=-f(x)`. `:.f(x)` is an odd function of `x`. But , we know that `int_(-a)^(a)f(x)dx=0`, when `f(x)` I an odd function of `x`. `:.int_(-1)^(1)log((2-x)/(2+x))dx=0`. |
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| 82. |
`int_(0)^(sqrt(2))sqrt(2-x^(2))dx=?`A. `pi`B. `2pi`C. `(pi)/(2)`D. none of these |
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Answer» Put `x=sqrt(2)sint` and `dx=sqrt(2)cost dt`. `[x=0impliest=0]` and `[x=sqrt(2)impliest=(pi)/(2)]` `:.I=int_(0)^(pi//2)sqrt(2-2sin^(2)t)*sqrt(2)costdt=2int_(0)^(pi//2)cos^(2)tdt=2int_(0)^(pi//2)(1)/(2)(1+cos2t)dt` `=[t+(sin2t)/(2)]_(0)^(pi//2)=(pi)/(2)`. |
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| 83. |
If f(x) is defined on `[-2, 2]` by `f(x) = 4x^2 – 3x + 1 and g(x) = (f(-x)-f(x))/(x^2+3)` then `int_(-2)^2 g(x) dx` is equal toA. 64B. `-48`C. 0D. 24 |
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Answer» Correct Answer - C Given , ` f(x) = 4x^(2) - 3x + 1, g(x) = (f(-x)-f(x))/(x^(2)+3)` ` :. g(x) = ((4x^(2) +3x+1)-(4x^(2)-3x+1))/(x^(2)+3) = (6x)/(x^(2)+3)` Now, ` g(-x) = - (6x)/(x^(2)+3) = - g(x)` Which is an odd function ` :. int _(-2)^(2) g(x) dx = 0` |
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| 84. |
`int_(pi//3)^(pi//2) (sqrt(1+cosx))/((1-cosx)^(5//2)) dx` |
| Answer» `I=int_(pi//3)^(pi//2)(cos(x//2))/(sin^(5)(x//2))dx`. Put `sin.(x)/(2)=t`. | |
| 85. |
` int_(-1)^(1)log(x+sqrt(x^(2)+1))` dx is equal to |
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Answer» Correct Answer - A Let f(x) = log `(x+sqrt(1+x^(2)))`and replacing x by -x , we get ` f(-x) = log (sqrt(1+x^(2))-x)` ` = log [ (sqrt(1+x^(2))-x) ((sqrt(1+x^(2))+x))/((sqrt(1+x^(2))+x))]` ` = log. ([ (1+x^(2))-x^(2)])/((sqrt(1+x^(2))+x))= log 1 - log (sqrt(1+x^(2))+x)` ` = - log (sqrt(1+x^(2))+x)=f(x)` Hence, f(x) is an odd function . ` :. int _(-1)^(1) log (x+sqrt(1+x^(2)))dx = 0 ` |
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| 86. |
` int_(0)^(4)|x-1| dx` is equal toA. `5/2`B. `3/2`C. `1/2`D. 5 |
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Answer» Correct Answer - D Let ` l = int _(0)^(4) | x-1| dx` ` :. l = int_(0)^(1)| x-1| dx + int _(1)^(4) (x-1) dx` ` = int _(0)^(1) (1-x) dx + int _(1)^(4) (x-1) dx ` ` = [ x- (x^(2))/2]_(0)^(1) + [ (x^(2))/2 - x]_(1)^(4)` ` = (1- 1/2) - 0 + ((4^(2))/2-4)-(1/2-1)` ` = 1/2 + 4 + 1/2 =5` |
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| 87. |
Evaluate: `int_(-a)^a sqrt((a-x)/(a+x)) dx `A. `api`B. `(api)/(2)`C. `2api`D. none of these |
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Answer» Put `x=acostheta` and `dx=-asintheta d theta`. `[x=-aimpliestheta=pi]` and `[x=aimpliestheta=0]`. `:.I=int_(pi)^(0)sqrt((1-costheta)/(1+costheta))*(-asintheta)d theta=aint_(0)^(pi){(2sin^(2)(theta//2))/(2cos^(2)(theta//2))}^(1//2)2sin((theta)/(2))cos((theta)/(2))d theta` `=aint_(0)^(pi)2sin^(2).(theta)/(2)d theta=a int_(0)^(pi)(1-costheta)d theta=api`. |
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| 88. |
Evaluate : `(i) int_(0)^(1//2)(dx)/(sqrt(1-x))` `(ii) int_(0)^(1)((1-x)/(1+x))dx` |
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Answer» `(i) int_(0)^(1//2)(dx)/(sqrt(1-x))=int_(0)^(1//2)(1-x)^(-1//2)dx=[(2sqrt(1-x))/(-1)]_(0)^(1//2)=(2-sqrt(2))`. `(ii) int_(0)^(1)((1-x)/(1+x))dx=int_(0)^(1)(-1+(2)/(x+1))dx` [on dividing `(-x+1)` by `(x+1)`] `=[-x+2log|x+1|]_(0)^(1)=[(2log2)-1]`. |
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| 89. |
`int_0^a(dx)/(sqrt(a x-x^2))` |
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Answer» `(i) int_(0)^(a)(dx)/(sqrt(ax-x^(2)))=int_(0)^(a)(dx)/(sqrt(-(x^(2)-ax+(a^(2))/(4))+(a^(2))/(4)))` `=int_(0)^(a)(dx)/(sqrt(((a)/(2))^(2)-(x-(a)/(2))^(2)))` `=[sin^(-1).((x-(a)/(2)))/(((a)/(2)))]_(0)^(a)=[sin^(-1)((2x-a)/(a))]_(0)^(a)` `=[sin^(-1)(1)-sin^(-1)(-1)]` `=2sin^(-1)(1)=(2xx(pi)/(2))=pi`. `(ii) int_(0)^(sqrt(2))sqrt(2-x^(2))dx=int_(0)^(sqrt(2))sqrt((sqrt(2))^(2)-x^(2))dx` `=[(x)/(2)sqrt(2-x^(2))+((sqrt(2))^(2))/(2)*sin^(-1).(x)/(sqrt(2))]_(0)^(sqrt(2))` `=[0+sin^(-1)(1)]-[0+sin^(-1)0]=(pi)/(2)`. |
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| 90. |
` int_(-1)^(1)x^(17)cos^(4)` x dx is equal toA. `17/4`B. `13/2`C. 0D. -1 |
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Answer» Correct Answer - C Let `f(x) = x^(17) cos^(4) x` f(x) is an odd function ` :. int _(-1)^(1) x^(17) cos^(4) x dx = 0 ` |
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| 91. |
`int_0^(pi//2)(sin2x)/(sin^4x+cos^4x) dx` |
| Answer» Divide num. and denom. By `cos^(4)x` and put `tan^(2)x=t`. | |
| 92. |
` int _(-1)^(1)(x^(3)+|x|+1)/(x^(2)+2|x|+1)dx` is equal toA. log 2B. 2 log 2C. `1/2 log2`D. `4 log 2` |
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Answer» Correct Answer - B Let ` l = int _(-1)^(1) (x^(3) + |x |+1)/(x^(2) +2|x|+1)` `rArr l = int _(-1)^(1) (x^(3))/(x^(2)+2 |x| +1) + int _(-1)^(1) (|x|+1)/(x^(2) +2|x| +1)dx` ` = 0 +2 int _(0)^(1) (x+1)/((|x|+1)^(2))dx` [ odd function + even function ] ` = 2 int _(0)^(1) (x+1)/(x+1)^(2)dx= 2 int _(0)^(1) 1/(x+1) dx` ` = 2 [ log |x+1|]_(0)^(1)= 2 log 2 ` |
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| 93. |
The value of `int_(1)^(2) (dx)/(x(1+x^(4))` isA. `(1)/(4)log.(17)/(32)`B. `(1)/(4)log.(32)/(17)`C. `log.(17)/(2)`D. `(1)/(4)log.(17)/(2)` |
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Answer» Correct Answer - B Let ` l = int_(1)^(2) (dx)/(x(1+x^(4)) ) =int_(1)^(2) (x^(3))/(x^(4) (1+x^(4)))dx` Put ` x^(4) = t rArr 4x^(3) dx = dt` ` :. l = 1/4 int_(1)^(16)(dt)/(t(1+t))= 1/4 int_(1)^(16) (1/t - 1/(1+t))dt` ` = 1/4 (log. t/(1+t))_(1)^(16) = 1/4 (log. 16/17 - log. 1/2 )` ` = 1/4 log. 32/17 ` |
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| 94. |
Evaluate `int_(1)^(4)f(x)dx`, where `f(x)={{:(4x+3,"if" 1 le x le 2),(3x+5,"if" 2 le x le 4.):}` |
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Answer» `int_(1)^(4)f(x)dx=int_(1)^(2)f(x)dx+int_(2)^(4)f(x)dx` `=int_(1)^(2)(4x+3)dx+int_(2)^(4)(3x+5)dx` `=[2x^(2)+3x]_(1)^(2)+[(3x^(2))/(2)+5x]_(2)^(4)=(9+28)=37`. |
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| 95. |
Evaluate : `int_(0)^(4)(dx)/(sqrt(x^(2)+2x+3))` |
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Answer» `(i) int_(0)^(4)(dx)/(sqrt(x^(2)+2x+3))=int_(0)^(4)(dx)/(sqrt((x+1)^(2)+(sqrt(2))^(2)))` `=[log|(x+1)+sqrt(x^(2)+2x+3)|]_(0)^(4)` `={log|5+sqrt(27)|-log|1+sqrt(3)|}`. `(ii) (dx)/((1+x+x^(2)))=int_(0)^(1)(dx)/([(x^(2)+x+(1)/(4))+(3)/(4)])=int_(0)^(1)(dx)/([(x+(1)/(2))^(2)+((sqrt(3))/(2))^(2)])` `=[(2)/(sqrt(3))tan^(-1).((x+(1)/(2)))/((sqrt(3)//2))]_(0)^(1)=(2)/(sqrt(3))[tan^(-1)((2x+1)/(sqrt(3)))]_(0)^(1)` `=(2)/(sqrt(3))[tan^(-1)(sqrt(3))-tan^(-1)((1)/(sqrt(3)))]` `=(2)/(sqrt(3))*[(pi)/(3)-(pi)/(6)]=(pi)/(3sqrt(3))`. |
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| 96. |
`int_(0)^(pi)(dx)/((3+2sinx+cosx))` |
| Answer» Write `sinx=(2tan(x//2))/(1+tan^(2)(x//2))`, `cosx=(1-tan^(2)(x//2))/(1+tan^(2)(x//2))` and put `tan.(x)/(2)=t`. | |
| 97. |
`int(sin2x)/(cos^2x+3cosx+2)dx` |
| Answer» Correct Answer - Put `cosx=t`. | |
| 98. |
` int_(-2)^(3)| x^(2)-1|dx` is equal toA. 3B. `1/3`C. `17/3`D. `28/3` |
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Answer» Correct Answer - D `l = int _(-2)^(3) | x^(2) -1| dx ` ` :. int _(0)^(2) | x^(2) -1| dx + int _(-1)^(1) | x^(2) -1| dx + int_(1)^(3) (x^(2)-1) dx ` [ Here , modulus function will change at the points , when `x^(2) - 1 = 0 " i.e ., at " x = pm 1] ` So , `l = int _(-2)^(-1) (x^(2) -1) dx + int _(-1)^(1)(1-x^(2)) dx + int _(1)^(3) (x^(2)-1)dx` ` = [ (x^(3))/3 -x] _(2)^(-1) + [ x- (x^(3))/3]_(-1)^(1) + [ (x^(3))/3 - x] _(1)^(3)` ` = (2/3 +2/3) + (2/3 +2/3) + (6+2/3)=28/3` |
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| 99. |
Evaluate : `(i) int_(0)^(pi//4)sqrt(1+sin2x) dx` `(ii) int_(0)^(pi//2)sqrt(1+cos2x) dx` |
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Answer» `(i) int_(0)^(pi//4)sqrt(1+sin2x)dx=int_(0)^(pi//4)sqrt(cos^(2)x+sin^(2)x+2sinxcosx)dx` `=int_(0)^(pi//4)(cosx+sinx)dx=[sinx-cosx]_(0)^(pi//4)=1`. `(ii) int_(0)^(pi//2)sqrt(1+cos2x)dx=int_(0)^(pi//2)sqrt(2cos^(2)x)dx` `=sqrt(2)int_(0)^(pi//2)cosxdx=sqrt(2)[sinx]_(0)^(pi//2)=sqrt(2)`. |
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| 100. |
`int_(0)^(pi//4)(tan^(3)x)/((1+cos2x))dx` |
| Answer» Using `cos2x=(2cos^(2)x-1)`, we get `I=int_(0)^(pi//4)tan^(3)xsec^(2)xdx`. Now, put `tanx=t`. | |