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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. |
`int(1)/(xlogx)dx` का मान ज्ञात कीजिए। |
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Answer» माना `I = int(1)/(x.logx)dx` माना `log x = t rArr (1)/(x)dx = dt` `therefore " " I = int(dt)/(t) = log (log x) + c` |
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| 52. |
`(1)/((a^(2)-x^(2))^(3//2))` का समाकलन कीजिये - |
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Answer» Correct Answer - `(1)/(a^(2))(x)/(sqrt(a^(2) - x^(2)))` `I = int(acos theta)/(a^(3)cos^(3)theta)d theta = (1)/(a^(2))intsec^(2)thetad theta` माना `x= a sin theta rArr dx = a cos theta d theta` तब ` =(1)/(a^(2))tan theta = (1)/(a^(2))(x)/(sqrt(a^(2)-x^(2)))` |
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| 53. |
`((sin^(-1)x)^(3))/(sqrt(1-x^(2)))` का समाकलन कीजिये - |
| Answer» Correct Answer - `sin^(-1)((x)/(3))` | |
| 54. |
`int(d theta)/(sin(theta - alpha)sin(theta-beta))`का मान ज्ञात कीजिए। |
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Answer» माना `I = int(d theta)/(sin(theta - alpha)sin(theta -beta))` ` int(sin(alpha - beta)d theta)/(sin(alpha-beta)sin(theta-alpha)sin(theta-beta))` ` = (1)/(sin(alpha - beta))int(sin[(theta - beta)- (theta - alpha)])/(sin(theta-alpha)sin(theta-beta))d theta` ` = (1)/(sin(alpha-beta))` ` int(sin(theta-beta) cos(theta-beta)-sin(theta-beta)cos(theta-beta))/(sin(theta -alpha)sin(theta-beta))` `=(1)/(sin(alpha -beta))int{(cos(theta - alpha))/(sin(theta-alpha))-(cos (theta - beta))/(sin(theta-beta))} d theta` `=(1)/(sin(alpha-beta))[int(cos (theta -alpha))/(sin(theta-alpha))d theta-int(cos(theta - beta))/(sin(theta-beta))d theta]` `=(1)/(sin(alpha-beta))[log sin(theta-alpha) - log sin (theta - beta)]` ` = (1)/(sin(alpha-beta))log[(sin(theta-alpha))/(sin(theta-beta))]" "(because log M - log N = log.(M)/(N))` |
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| 55. |
`int cot theta^(3) cosec^(2) theta d theta` का मान ज्ञात कीजिए। |
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Answer» माना `cot theta = t rArr - coses^(2) theta d theta dt` `therefore int cot^(3) theta cosec^(2) theta d theta = - int t^(3)dt = - (t^(4))/(4) - - (cot^(4) theta)/(4)` |
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| 56. |
`int(tan x)/((sec x + cos x))dx` का मान ज्ञात कीजिए। |
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Answer» दिया है - `I = int(tanx)/((sec x+cos x))` ` = int(sin)/((1+cos^(2)x))dx` यदि `cosx = t ` व `sin xdx = dt` अब `I = - int(1)/((1+t^(2)))dt = - tan^(-1)t +c` ` = - tan^(-1)(cos x)+c` |
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| 57. |
`(1)/(x(1-x))` का समाकलन कीजिये - |
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Answer» Correct Answer - `sin^(-1)(sqrt(x))` माना `x = sin^(2) theta rArr dx = 2 sin theta cos theta d theta` |
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| 58. |
`((a+bsin^(-1)x)^(n))/(sqrt((1-x^(2))))` का समाकलन कीजिये - |
| Answer» Correct Answer - `((sin^(-1)x)^(4))/(4)` | |
| 59. |
`int sec^(p) x tan x dx` का मान है -A. `(sec^(p+1)x)/(p+1) + c`B. `(sec^(p)x)/(p)+c`C. `(tan^(p+1)x)/(p+1)+c`D. `(tan^(p)x)/(p)+c` |
| Answer» Correct Answer - B | |
| 60. |
`int (sin 2 x dx)/(a cos^(2) x + b sin^(2))` का मान ज्ञात कीजिये । |
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Answer» माना `I = int (sin 2x dx)/(a cos^(2) x + b sin^(2)x)` माना `" "a cos^(2) x + b sin^(2) x = t` `rArr " "(-2a sin x cos x + 2b sin x cos x) dx = dt` `rArr " "(b-a) sin 2x dx = dt` `rArr" "sin2x dx = (dt)/(b-a)` `therefore" "I = ((1)/(b-a))int(dt)/(t) = (1)/(b-a) log t` `=(1)/(b-a)log[a cos^(2) x + bsin^(2) x]` |
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| 61. |
`int(d theta)/(sin theta cos^(3) theta)`A. `log tan theta + tan^(2) theta+ c`B. `log tan theta -(1)/(2)tan^(2) theta + c`C. `log tan theta +(1)/(2)tan^(2) theta+c`D. इनमे से कोई नहीं |
| Answer» Correct Answer - C | |
| 62. |
`(1)/(sqrt(1-x^(2))sin^(-1)x)` का समाकलन कीजिये - |
| Answer» Correct Answer - `((a+bsin^(-1)x)^(n+1))/(n+1)` | |
| 63. |
`int e^(2x + 5)dx`का मान ज्ञात कीजिये । |
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Answer» माना `I = int e^(3x +5) dx " "....(1)` माना `3x + 5 = t` `therefore" "3.1 + 0 = (dt)/(dx)` `rArr" "dt = 3dx " "rArr dx = (1)/(3)dt` अतः समय (1 ) से `I = int e^(t)((1)/(3))dt = (1)/(3) int e^(t) dt = (1)/(3) e^(t) = (1)/(3)e^(3x+5)` |
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| 64. |
`int (e^(x) (1 +x))/(cos^(2)(xe^(x)))dx` का मान ज्ञात कीजिये । |
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Answer» माना `I = int (e^(x) (1+x))/(cos^(2) (xe^(x)))dx` माना ` xe^(x) = t` `therefore" "(e^(x) + ex^(x)) dx = dt` `rArr" "e^(x)(1+ x)dx =dt` `therefore" " I= int(dt)/(cos^(2)t) = int sec^(2) t dt = tan t` `therefore " " I = tan (xe^(x))` |
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| 65. |
`int(sin(tan^(-1)x))/(1+x^(2))dx = - cos tan^(-1)x+c` सिद्ध कीजिये - |
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Answer» बाया पक्ष`= int(sin(tan^(-1)x))/(1+x^(2))dx` माना `tan^(-1) x = t` `rArr(1)/(1+x^(2))dx = dt` `therefore int(sin(tan^(-1)x))/(1+x^(2))dx = intsin t dt = - cos t + c` `= - cos (tan^(-1)x)+c=` दाया पक्ष |
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| 66. |
`int(x)/(a^(2) +x^(2))dx = (1)/(2)log(a^(2) +x)+c` सिद्ध कीजिये - |
| Answer» बाये पक्ष `10^(x) + x(7) = t` रखने पर , तथा तब समाकलन करने पर | |
| 67. |
`int(dx)/(sqrt(x+1)+sqrt(x+2)) = (2)/(3)[(x+2)^(3//2)-(x+1)^(3//2)]+c` सिद्ध कीजिये - |
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Answer» बाया पक्ष ` = int(dx)/(sqrt(x+1)+sqrt(x+2))` `= int(dx)/(sqrt(x+2)+sqrt(x+1))` `= int{(1)/(sqrt(x+2)+sqrt(x+1)).(sqrt(x+2)-sqrt(x+1))/(sqrt(x+2) - sqrt(x+1))}dx` `=int(sqrt(x+2)- sqrt(x+1))/(x+2-x-1)dx` `=int(sqrt(x+2)-sqrt(x+1))/(1)dx` `=int(x+2)^(1//2)dx -int(x+1)^(1//2)dx` ` =(2)/(3)(x+2)^(3//2)- (2)/(3)(x+1)^(3//2)+c` ` = (2)/(3)[(x+2)^(3//2)-(x+1)^(3//2)]+c`= दाया पक्ष |
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| 68. |
`int x^(3)e^(x^(2)) dx` का मान हैA. `x^(2)(e^(x^(2)) - 1) + c`B. `(1)/(2)x^(2)(e^(x^(2))-1)+c`C. `(1)/(2)e^(x^(2))(x^(2) -1)+c`D. `(1)/(2)(e^(x^(2))-1)+c` |
| Answer» Correct Answer - C | |
| 69. |
`int(7x^(6)+10^(x)log_(e) 10)/(10^(x) +x^(7))dx = log(10^(x) + 7^(x))+c` सिद्ध कीजिये |
| Answer» बाये पक्ष में `e^(x) = t rArr e^(x) dx = dt` रखने पर, तथा तब समाकलन करने पर | |
| 70. |
`int(2x^(3))/((x^(2)+1)^(2))dx = log (x^(2) +1)+(1)/(x^(2) +1)+c` सिद्ध कीजिये - |
| Answer» माना `x^(2) + 1 = trArr 2xdx = dt rArr int(2x^(3))/((x^(2)+1)^(2))dx = int(t-1)/(t^(2))dt = int[(1)/(t)-t^(-2)] dt = log t - (t^(-1))/(-1)+c` | |
| 71. |
`(xtan^(-1) x^(2))/(1+x^(4))` का समाकलन कीजिये - |
| Answer» Correct Answer - `log(sin^(-1)x)` | |
| 72. |
`(1)/(xlog x log log x)` का समाकलन कीजिये - |
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Answer» Correct Answer - `log log log x` माना `log logx = t rArr(1)/(logx) .(1)/(x)dx=dt` तब `I = int(1)/(t)dt = logt = log log log x ` |
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| 73. |
`int(x^(2)dx)/(sqrt(x+2))`का माना ज्ञात कीजिये | |
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Answer» माना ` I = int(x^(2)dx)/(sqrt((x+)))" "....(1)` माना `x + 2 = t rArr x = t - 2` तथा `dx = dt` अतः समी से `I = int((t-2))/((t)^(1//2)) = int ((t^(2) - 4t + 4)/(t^(1//2)))dt` ` = int(t^(3//2) -4t^(1//2) + 4t^(-1//2))dt ` `= (2)/(5)t^(5//2) - (8)/(3)t^(3//2) + 8t^(1//2)` `= (2)/(5)(x+2)^(5//2) - (8)/(3)(x+2)^(3//2)+8(x+2)^(1//2)` |
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| 74. |
`int(dx)/(e^(x) -1)` का मान ज्ञात कीजिये | |
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Answer» माना `I = int (dx)/(e^(x) -1)` `= int (dx)/(e^(x)( 1-e^(-x)))" "("नोट कीजिये ")` `= int (e^(-x) dx)/(1-e^(-x))` माना `1 -e^(-x)= t rArr e^(-x) dx = dt` `therefore" "I = int(dt)/(t) = log t = log (1- e^(x))` |
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| 75. |
`int(sin^(-1))/(sqrt(1-x^(2)))dx = (1)/(2)(sin^(-1)x)^(2)+c` सिद्ध कीजिये - |
| Answer» बाया पक्ष `= int(sin^(-1)x)/(sqrt(1-x^(2)))dx` माना `sin^(-1) x = t rArr (1)/(sqrt(1-x^(2)))dx = dt` | |
| 76. |
`int(sinx)/(sin(x-alpha))dx = sinalpha log sin (x-alpha)+(x-alpha)cos alpha + c` सिद्ध कीजिये - |
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Answer» बाया पक्ष `=int(sin(tan^(-1)x))/(1+x^(2))dx` माना `tan^(-1)x = t` `rArr " "(1)/(1+x^(2))dx = dt` `therefore int(sin(tan^(-1)x))/(1+x^(2))dx = int sin t dt =- cos t + c = - cos (tan^(-1)x)+c =` दाया पक्ष |
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| 77. |
`(1-(1)/(x^(2)))e^(x+(1)/(x))` का समाकलन कीजिये - |
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Answer» Correct Answer - `e^(x+(1)/(5))` माना `x +(1)/(x) = t rArr (1-x^(-2))dx = dt` |
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| 78. |
`int(xtan^(-1)x^(2))/(1+x^(4)) dx` का मान हैA. `(1)/(2)(tan^(-1) x^(2))^(2)+c`B. `(1)/(3)(tan^(-1)x^(2))+c`C. `(1)/(4)(tan^(-1)x^(2))^(2)+c`D. इनमे से कोई नहीं |
| Answer» Correct Answer - A | |
| 79. |
निम्न फलनों का समाकलन कीजिये - `cos^(2) x sin x` |
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Answer» माना `I = int cos^(2) x sin x dx " "....(i)` माना `cos x = t rArr - sin xdx = dt` `rArr " "sin x dx = - dt` अतः समीकरण (1 ) से `I = - intt^(2) dt = -(1)/(3)t^(3) = - (1)/(3) cos^(3) x` |
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| 80. |
निम्न फलनों का समाकलन कीजिये - `x^(2) sin x^(3)` |
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Answer» माना `I = intx^(2) sin x^(3) dx" "....(1)` माना `x^(3) = t therefore 3x^(2)dx = dt rArr " "x^(3)dx = (1)/(3)dt` अतः समी (1 ) से `I = (1)/(3)intsin t dt = - (1)/(3) cos t = (1)/(3) cos x ^(3)` |
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| 81. |
निम्न फलनों का समाकलन कीजिये - `x cos ^(2)` |
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Answer» माना `I = int x cos x^(2) dx` माना ` x^(2) = t " "therefore 2dx = dt` `rArr" "xdx= (1)/(2) dt` अतः समी (1 ) से `I = (1)/(2)int cos t dt = (1)/(2)sin t = (1)/(2) sin x^(2)` |
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| 82. |
`intsin^(2)x cos^(2) xdx = (1)/(8)[x - (sin 4x)/(4)]+c` सिद्ध कीजिये - |
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Answer» बाया पक्ष `=intsin^(2)x cos^(2)x dx` ` = int(sinx cos x)^(2)dx = (1)/(4)(2sin x cos x)^(2)dx` `=int(1)/(4)(sin 2x)^(2)dx = int(1)/(4)[(1-cos 4x)/(2)]dx` `= (1)/(8)(1-cos 4x)dx` |
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| 83. |
`int e^(log(sinx))dx` का मान हैA. `sin x+c`B. `-cos x + c`C. `e^(log(cosx))+c`D. इनमे से कोई नहीं |
| Answer» Correct Answer - B | |
| 84. |
निम्न फलनों के मान ज्ञात कीजिये - `int(e^(x))/(1+e)^(x)dx` |
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Answer» माना `I = int (e^(x))/(1+e^(x)) dx` माना `1+e^(x) = t " " therefore" "e^(x) dx = dt` अतः ` I = int (dt)/(t) = log t` `therefore " " I = log (1 + e^(x))` |
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| 85. |
`int(1+sinx)/(1+cos x)dx` का मान ज्ञात कीजिए। |
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Answer» दिया है - `int((1+sin x))/((1+cosx))dt` ` = int(1)/(1+cosx)dt + int(sinx)/((1+cosx))dx` `= int(1)/(2cos^(2)(x//2))dx + int(2sin(x//2)cos(x//2))/(2cos^(2)(x//2))` ` = (1)/(2)intsec^(2)((x)/(2))dx + int tan((x)/(2))dx` यदि `t = x//2` व `dx = 2dt` `=intsec^(2) t dt + 2 int tan tdt ` ` = tan t - 2 log (cos t) + c` ` = tan((x)/(2)) - 2 log cos((x)/(2))+c` |
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| 86. |
निम्न फलनों के मान ज्ञात कीजिये - `int(cotx)/(log sin x)dx` |
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Answer» माना ` I = int (cot x)/(log sinx)` माना `log sin x = t` `rArr (1)/(sinx) . cos x dx = dt` `rArr " " cot x dx = dt` अतः `I = int (dt)/(t) = log t` `therefore " " I = log (log sin x)` |
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| 87. |
निम्न फलनों का समाकलन कीजिये - `(1)/(x+sqrt(x))` |
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Answer» माना `I = int (dx)/(x+sqrt(x)) = int =(dx)/(sqrt(x)(sqrt(x) +1))` माना `sqrt(x) + 1 = t` `therefore (1)/(2sqrt(x))dx = dt rArr (dx)/(sqrt(x)) = 2dt` अतः समी (1 ) से `I = 2int (dt)/(t) = 2 log t = 2 log(sqrt(x+1))` |
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| 88. |
निम्न फलनों के मान ज्ञात कीजिये - `int(cos theta d theta)/(a+b sin theta)` |
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Answer» माना `I = int(cos theta d theta)/(a + b sin theta)` माना `a + b sin theta = t` `rArr " " bcos theta d theta = dt rArr cos theta = (1)/(b) dt` `therefore " "I = (1)/(b) int (dt)/(t) = (1)/(b) log t` ` therefore" "I = (1)/(b) log (a+b sin theta)` |
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| 89. |
निम्न फलनों के मान ज्ञात कीजिये - `int (x^(3))/(1+x^(4))dx` |
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Answer» माना `I = int(x^(3) dx)/(1+x^(4))` माना `1 +x^(4) = t " "(because 4x^(3) dx = dt)` `rArr x^(3) dx = (1)/(4)dt` अतः ` I = (1)/(4) int(dt)/(t) = (1)/(4)log t` `rArr" "I =(1)/(4)log(1+x^(4))` |
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| 90. |
`int((1+cos x)/(1-cos x))` का मान ज्ञात कीजिए। |
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Answer» माना `int ((1+cosx)/(1-cosx)) dx = int (2cos^(2)(x//2))/(2sin^(2)(x//2))` ` = int cot^(2)((x)/(2))dx - int (cosec^(2)(x)/(2)-1)` ` = int cosec^(2)((x)/(2))dx - int dx` यदि `(x)/(2) = t` व ` dx = 2dt` ` =- 2cot t -x + c` `= -2 cot ((x)/(2)) - x+c` |
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| 91. |
`int(x+sqrt(x+1))/(x+2)dx` का मान ज्ञात कीजिए। |
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Answer» माना `I =int(x+sqrt(x+1))/(x+2)dx` यदि `sqrt(x +1) = t, x + 1 = t^(2)`व` dx = 2tdt` अब `I = int(x+sqrt(x+1))/(x+2)dx = 2int((t^(2) -1 +t)t)/((t^(2) +1))dt` `=2int((t^(3) + t^(2) - t)/(t^(2) +1))dt = 2int(t+1-(2t+1)/(t^(2) +1))dt` ` = 2int(t+1-(2t)/(t^(2)+1) - (1)/(t^(2) +1))dt` ` = 2int t dt +2 int ddt - 2 int (2t)/(t^(2) +1)dt - 2 int (1)/(t^(2)+1)dt` `=t^(2) + 2t - 2 log(t^(2) +1) - 2tan^(-1) t+C` `=(x+1)+2sqrt(x+1)-2log (x+2) - 2tan^(-1) sqrt(x+1)+c` |
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| 92. |
`intsin^(3) x dx` का माना होगा -A. `sin^(2)x+1`B. `sin x^(2) + x^(2) + 1`C. `(cos^(3)x)/(3)-cosx`D. `(1)/(4)sin^(4)x - (3)/(4)sin^(2)x` |
| Answer» Correct Answer - C | |
| 93. |
यदि `int sin 5x cos 3x dx = - (cos 8x)/(16)+A` तब A का मान है -A. `(sin 2x)/(16)+` अचरB. `-(cos 2x)/(4)+` अचरC. अचरD. इनमे से कोई नहीं |
| Answer» Correct Answer - B | |
| 94. |
`int(sinx)/sqrt(1+sinx)dx` का मान ज्ञात कीजिए। |
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Answer» दिया है -` I = int(sinx)/(sqrt(1+sin x))dx = int((1+sin x) - 1)/(sqrt(1+sin x))` ` = int sqrt(1+sinx)dx - int(dx)/(sqrt(1+sin x))` `= intsqrt(cos^(2).(x)/(2)+sin^(2)+(x)/(2)+2sin.(x)/(2)cos.(x)/(2)dx)` `-int(dx)/(sqrt(cos^(2)(x//2)+sin^(2)(x//2)+2sin(x//2))cos(x//2))` `= int[cos((x)/(2))+sin((x)/(2))]dx - int(dx)/([cos (x//2)+sin(x//2)])` ` = (2sin.(x)/(2) - cos .(x)/(2)) -(1)/(2).int(dx)/((1)/(sqrt(2)cos.(x)/(2)+(1)/(sqrt(2))sin.(x)/(2)))` `=(2sin.(x)/(2)-2cos.(x)/(2)) - (1)/(sqrt(2)) . int(dx)/(sin((pi)/(2) +(pi)/(2)))` ` = (2sin.(x)/(2) - 2cos.(x)/(2)) -(1)/(sqrt(2)). intcosec((x)/(2) +(pi)/(4))dx` ` = 2 (sin.(x)/(2)-cos.(x)/(2)) - (1)/(sqrt(2)) xx 2 log[ tan ((x)/(4) +(pi)/(8))]+c` ` = 2(sin.(x)/(2) - cos.(x)/(2))- sqrt(2)[ tan ((pi)/(4) +(pi)/8)]+c` |
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| 95. |
निम्न फलनों के मान ज्ञात कीजिये - `int(e^(x) -sinx)/(e^(x) + cosx)dx` |
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Answer» माना `I = int(e^(x) - sinx)/(e^(x) + cosx) dx` माना `(e^(x) + cos x)= t` `rArr " "(e^(x) - sin x) dx = dt rArr I = int (dt)/(t) = log t` `therefore" " I = log (e^(x) + cosx)` |
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| 96. |
फलन `int x sin^(3) x^(2) cosx^(2) dx` का मान ज्ञात कीजिये | |
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Answer» माना `I = int x sin^(3) x^(2) cos x^(2) dx` माना `sin x^(2) = t rArr 2xcos x^(2) dx = dt` `rArr" "x cosx^(2) dx = (1)/(2) dt` `therefore " "I= (1)/(2) intt^(3) dt = (1)/(8) t^(4) = (1)/(8)sin^(4) x^(2)` |
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| 97. |
`int((3sinx - 2)cosx)/((5-cos^(2) x - 4sinx))` का मान ज्ञात कीजिए। |
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Answer» माना `I int ((3sinx - 2)cosx)/((5 - cos^(2) x - 4 sinx))` ` = int ((3sin x - 2)cos x)/((4+sin^(2)x - 4sinx))` ` =int ((3sinx - 2) cosx)/((2-sinx)^(2))dx` यदि `2-sin x = t, sin x = 2 - t `व` cos xdx = dt` अब ` I = - int({3(2-t)-2})/(t^(2))` ` = - int((4-3t))/(t^(2))dt = int((3t - 4))/(t^(2))dt` ` = int((3)/(t) - (4)/(t^(2)))dt = 3 log (t) +(4)/(t) +c` `= 3log (2-sinx)+(4)/((2-sinx))+c` |
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| 98. |
यदि `x in ((pi)/(4),(3pi)/(4))`तब `int(sinc - cosx)/(sqrt(1-sin 2x))e^(sin) cos x dx ` का मान है -A. `e^(sinx)+c`B. `e^(sinx-cosx)+c`C. `e^(sin x + cosx)+c`D. `e^(cosx - sin x)+c` |
| Answer» Correct Answer - A | |
| 99. |
`int(cosx)/((cos.(x)/(2)+sin.(x)/(2)))` का मान ज्ञात कीजिए। |
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Answer» दिया है - `I = int (cosx)/((cos.(x)/(2)+sin.(x)/(2)))` ` int (cos^(2)(x//2) -sin^(2)(x//2))/([cos(x//2) +sin(x//2)]^(3))` ` = int(cos (x//2) - sin^(2) (x//2))/([cos(x//2)+sin(x//2))]` ` = int (cos(x//2) - sin(x//2))/((cos.(x)/(2)+sin.(x)/(2)))` यदि `t= cos.(x)/(2)+sin.(x)/(2)rArr [cos((x)/(2)) -sin((x)/(2))]dx = 2dt` तब `I = 2int(1)/(t^(2)) dt = -(2)/(t) +c` ` = (-2)/(cos(x//2) +sin(x//2))+c` |
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`int cos x cos 2x cos 3x dx` का मान ज्ञात कीजिए। |
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Answer» दिया है - `int cos x cos 2xcos 3x dx` ` = (1)/(2)int(2cos x cos 2x) cos 3x dx` ` = (1)/(2)int (cos 3x + cos x)cos 3x dx` ` = (1)/(2)int(cos^(2) 3x + cos x cos 3x)dx` ` = (1)/(4)int(2cos^(2) 3x)dx+(1)/(4)int(2cos x cos 3x) dx` ` = (1)/(4)int(1+cos 6x )dx +(1)/(4) int(cos 4x + cos 2x)dx` ` =(1)/(4) intdx +(1)/(4) int cos 6x dx = (1)/(4) int cos 6dx +(1)/(4) int cos 4xdx +(1)/(4) int cos 2xdx` ` = (1)/(4) x+(1)/(4).(sin 6x)/(6)+(1)/(4).(sin 4x)/(4) +(1)/(4).(sin 2x)/(2)+c` ` = (x)/(4) +(sin 6x)/(24) +(sin 4x)/(16) +(sin 2x)/(8)+e` |
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