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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.
| 1. |
`(sinsqrt(x))/(sqrt(x))` का समाकलन कीजिये - |
| Answer» Correct Answer - `-2cossqrt(x)` | |
| 2. |
`x^(n-1)cos x^(n)` का समाकलन कीजिये - |
| Answer» Correct Answer - `(1)/(n)sin x^(n)` | |
| 3. |
` (x^(7))/(1+x^(16))` का समाकलन कीजिये - |
| Answer» Correct Answer - `(1)/(8)tan^(-1)(x^(8))` | |
| 4. |
`x e^(x^(2))` का समाकलन कीजिये - |
| Answer» Correct Answer - `(1)/(2)e^(x^(2))` | |
| 5. |
`(sqrt(x) +(1)/(sqrt(x)))` का प्रति अवकलज है -A. `(1)/(3)x^(1//3) + 2x^(1//2)+c`B. `(2)/(3)x^(2//3) +(1)/(2)x^(2)+c`C. `(2)/(3)x^(2//3) + 2x^(1//2)+c`D. `(3)/(2)x^(3//2) +(1)/(2)x^(1//2)+c` |
| Answer» Correct Answer - C | |
| 6. |
`int cos mx cos nx dx` का मान ज्ञात कीजिए। |
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Answer» दिया है - ` int cos mx cos nx dx` यदि `m ne n` तब ` = (1)/(2)int[cos(m +n)x + cos(m -n)x]dx` ` = (1)/(2)intcos(m+n) xdx +(1)/(2)int cos (m-n)x dx` ` = (sin (m +n) x)/(2(m+n))+(sin(m-n)x)/(2(m-n))+c` अब यदि `m =n` तब `int cos mx cos nx dx = int cos^(2) nx dx ` ` = (1)/(2)int2cos^(2) nx dx` ` = (1)/(2)int dx + (1)/(2)int cos 2nx dx` ` = (x)/(2) +(sin 2nx)/(4n)+c` |
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| 7. |
`xsin x^(2)` का समाकलन कीजिये - |
| Answer» Correct Answer - `-(1)/(2)cos x^(2)` | |
| 8. |
`int(dx)/(x[1+logx]^(n))`का मान ज्ञात कीजिये । |
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Answer» माना ` I = int (dx)/(x[1+logx]^(n))` माना `1 + log x = t rArr (1)/(x) dx =dt` `therefore " "I = int(dt)/(t^(n)) = int t^(-n) dt = (t^(-n+1))/(-n+1) = (1)/((1-n)t^((n-1))` |
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| 9. |
`(x)/(1+x^(2))`का समाकलन कीजिये - |
| Answer» Correct Answer - `(1)/(2)log(1+x^(2))` | |
| 10. |
फलन `x^(3)sqrt((x^(2) - 4))` का x के सापेक्ष समाकलन कीजिये । |
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Answer» माना `I = intx^(3)sqrt((x^(2) - 4))dx " ".....(1)` माना `x^(2)- 4 = trArr x^(2) = (t +4)` तथा `2x dx = dt rArr x dx = (1)/(2) dt` अतः समीकरण (1 ) से `I = int x^(2) xsqrt((x^(2) - 4))dx` ` = (1)/(2)int(t + 4)t^(1//2) +dt = (1)/(2)int[t^(3//2) + 4t^(1//2)]dt` ` = (1)/(2)[(2)/(5) t^(5//2) +(4xx2)/(3) t^(3//2)] = (1)/(5) t^(5//2) +(4)/(3) t^(3//2)` ` = (1)/(5)(x^(2) - 4)^(5//2)+(4)/(3)(x^(2) - 4)^(3//2)` |
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| 11. |
`intsin^(2//3)x.cos^(3) x dx` के मान ज्ञात कीजिए |
| Answer» Correct Answer - `(3)/(5)sin^(5//3) x - (3)/(11)sin^(11//3)x+c` | |
| 12. |
`int(tan(sin^(-1)x))/(sqrt(1-x^(2)))` का मान ज्ञात कीजिए। |
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Answer» माना `I= int(tan (sin^(-1) x))/(sqrt(1-x^(2)))dx` माना `sin^(-1) x = t rArr (1)/(sqrt(1-x^(2)))dx = dt` `therefore " "I int tan t dt = logsec t + c` `=log sec (sin^(-1) x) +c` |
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| 13. |
`int(1)/((1tan x))dx` का मान ज्ञात कीजिए। |
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Answer» माना `I = int (1)/((1+tanx)) dx = int(1)/((1+(sinx)/(cosx))) dx` ` = int (1)/(((cosx + sinx)/(cosx)))dx = int(cosx)/((cosx + sinx))dx` ` =int ((cos x + sinx)+(cosx-sinx))/(2(cos x+sinx))dx` ` = (1)/(2)int((cosx + sinx))/((cos x+sinx))dx +(1)/(2)int((cosx - sinx))/((cos x + sinx))dx` ` = (1)/(2)int dx + (1)/(2)int((cos x - sin))/((cos x + sinx))dx` यदि `(cos x + sinx) = t` व `(cos x - sinx) dx = dt` अब `I = (1)/(2) int dx + (1)/(2) int(1)/(t) dt` ` =(1)/(2) x +(1)/(2)log |t| +c = (1)/(2) x +(1)/(2)log |cos x+ sin x|+c` |
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| 14. |
`int(cosec^(2)x)/(1+cotx)` का समाकलन कीजिये - |
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Answer» Correct Answer - `- log (1+cot x)` माना ` 1+cotx = t" "rArr" " rArr - cosec^(2)xdx = dt` |
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| 15. |
निम्न फलनों का समाकलन कीजिये - `(x^(8))/((1-x^(3))^(1/3))` |
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Answer» माना `I = int (x^(8))/((1-x^(3))^(1//3))dx` माना `1 -x^(3) = t rArr - 3x^(2)dx = dt` `rArr " " x^(2) dx = - (1)/(3)dt` तथा ` 1- x^(3) = t rArr 1 - t = x^(3)` `because" "I = int((x^(3))^(2)x^(2)dx)/((1-x^(3))^(1//3)) = - (1)/(3)int((1-t)^(2)dt)/(t^(3//2))` ` = - (1)/(3)int((1-2t +t^(2))/(t^(1//3)))dt` `= - (1)/(3)int(t^(-1//3) - 2t^(2//3) + t^(5//3))dt` ` =- (1)/(3)[(3)/(2)t^(2//3) -(6)/(5)t^(5//3) + (3)/(8)t^(8//3)]` `= - (1)/(3)[(3)/(2)(1-x^(3))^(2//3)- (6)/(5)(1-x^(3))^(5//3) + (3)/(8)(1-x^(3))^(8//3)]` |
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| 16. |
`int(x^(e-1) - e^(x-1))/(x^(e) - e^(x))dx` का मान ज्ञात कीजिये । |
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Answer» मान `I=int (x^(e-1) - e^(x-1))/(x^(e) - e^(x))dx` माना `x^(e) - e^(x) = t` `(ex^(e-1) - e^(x)) dx = dt ` `therefore " "e(x^(e-1) - e^(x-1))dx = dt` `rArr " "(x^(e-1) - e^(x-1))dx = (1)/(e) dt` `therefore" "I = (1)/(e)int(dt)/(t) = (1)/(e)logt` `therefore" "I =(1)/(e)int(dt)/(e) = (1)/(e) log t` `therefore" "I = (1)/(e) log(x^(e) - e^(x))` |
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| 17. |
`(logsqrt((x+1)))/(x+1)` का समाकलन कीजिये - |
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Answer» Correct Answer - `(1)/(4)[log (x+1)]^(2)` `I = int(logsqrt(x+1))/((x-1))dx=int(log(x+1)^(1//2))/((x+1))dx = int((1)/(2)log(x+1))/((x+1))` अब माना `log(x+1) = t " "rArr (1)/(x+1)dx = dt` |
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| 18. |
`int sec^(6//5)x coses ^(4//5) x dx` का माना ज्ञात कीजिये। |
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Answer» माना `I = int sec^(6//5)x cosec ^(4//5) x dx` `= int(sec^(2)x//dx)/(sin^(4//5)x) xx (sec^(4//5) x)/(sec^(4//5x)) dx` अश और हर को `sec ^(4//5)` x से गुणा करने पर `int (sec^(2)x dx)/(tan^(4//5)x)` माना `tan x = t rArr sec^(2)xdx = dt` `therefore" "I = int (dt)/(t^(4//5)) = int t^(4//5)dt = 5t^(1//5) 5(tan x)^(1//5)` |
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| 19. |
`int(dx)/(sin^(4)x + cos^(4)x)`का मान ज्ञात कीजिए। |
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Answer» माना `I = (dx)/(sin^(4) x+cos^(4)x)` अश व हर को `cos^(4)x` से भाग देंने पर `I =int(sec^(4)xdx)/(1+tan^(4)x)` `= int((1+tan^(2)x)sec^(2)xdx)/(1+tan^(4)x)` माना `x = t therefore" "sec^(2) xdx = dt` `therefore" "I int((1+t^(2))/(1+t^(4)))dt` अश व हर को `t^(2)` से भाग देने पर `I = int(1+1//t^(2))/(t^(2) + 1//t^(2))dt` माना `t - (1)/(t) =urArr (1+(1)/(t^(2)))dt= du` पुन: `t^(2) +(1)/(t^(2)) = t^(2) +(1)/(t^(2)) -2+2` ` = (t -(1)/(t))^(2) +2 = u^(2) + 2` `therefore I = int(du)/(2+u^(2))= (1)/(sqrt(2))tan^(-1)((u)/(sqrt(2)))` ` = (1)/(sqrt(2))tan^(-1)[(1)/(sqrt(2))(t-(1)/(t))] = (1)/(sqrt(2))tan^(-1).((t^(2)-1)/(tsqrt(2)))` ` = (1)/(sqrt(2))^(-1)((tan^(2)x-1)/(sqrt(2)tanx))` |
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| 20. |
`int x^(2) e^(x^(3)) dx` का मान ज्ञात कीजिए। |
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Answer» माना `x^(3) = t rArr 3xd = dt` `rArr" "x^(2)dx = (1)/(3)dt` `therefore intx^(2)e^(x^(3))dx = (1)/(3) inte^(t)dt = (1)/(3)e^(t) + c =(1)/(3)e^(x^(3))` |
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| 21. |
`(1)/((1+x^(2))tan^(-1)x)` का समाकलन कीजिये - |
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Answer» Correct Answer - `log tan^(-1)` माना `tan^(-1) x =rArr (1)/(1+x^(2))dt = dt` |
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| 22. |
`(sin x)/(sin(x -alpha))` का समाकलन कीजिये - |
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Answer» Correct Answer - `(x-alpha)cosalpha+sin alpha log sin(x-alpha)` `I = int(sinx)/(sin(x-alpha)) dx` माना `x - alpha = t rArr dx = dt` `I = int(sin(t +alpha))/(sin t) dt = int(sin t cos alpha + cos t sinalpha)/(sin t)dt` |
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| 23. |
`(x^(2) tan^(-1)x^(3))/((1+x^(6)))` का समाकलन कीजिये - |
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Answer» Correct Answer - `(1)/(6)(tan^(-1) x^(3))^(2)` माना `tan^(-1)x^(-3) = t rArr (1)/(1+(x^(3))2).3x^(2)dx =dt` |
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| 24. |
`(1)/(sin x cos^(2)x)` का समाकलन कीजिये - |
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Answer» Correct Answer - `log tan.(x)/(2)+sec x` `I = int(dx)/(sin xcos^(2)x) = int(sin^(2)x+cos^(2))/(sinx cos^(2)x)dx = int(sinx)/(cos^(2)x)dx + int(1)/(sinx) dx` |
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| 25. |
`int(cos^(4) x - sin^(4) x)dx` का मान ज्ञात कीजिए। |
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Answer» `int (cos^(4) x - sin^(4)x)dx` `= int[(cos^(2) x)^(2) - (sin^(2) x)^(2)]dx` `=int (cos^(2) x + sin^(2)x)(cos^(2) x - sin^(2)x)dx` ` =int1.(cos^(2) x-sin^(2)x)dx` ` = int cos 2x dx` ` = (sin2x)/(2)+c = (1)/(2)sin 2x + c` |
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| 26. |
`(na (1+alog x)^(n-1))/(x(l+a logx)^(n))` का समाकलन कीजिये - |
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Answer» Correct Answer - `log(1+a logx)^(n)` माना `(1+alog x)^(n) = t rArr(na)/(x)(1+alogx)^(n-1)dx = dt` |
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| 27. |
`intsqrt((1-sqrt(x))/(1+sqrt(x)))` का मान ज्ञात कीजिए। |
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Answer» माना `sqrt(x) = cos 2 theta` `rArr (1)/(2sqrt(x)) dx = -2sin 2theta d theta ` `rArr" "dx = - 2sqrt(x).2sin 2 theta d theta = - 4cos2 theta sin 2theta d theta` `thereforeint sqrt((1-sqrt(x))/(1+sqrt(x))) dx` `= int sqrt((1-cos2theta)/(1+cos 2 theta)).(-4 cos 2 theta sin 2 theta )d theta` `= int (sin theta)/(cos theta) (- 4 cos 2 theta . 2 sin theta cos theta) d theta` `= - 8 int cos 2theta sin ^(2) theta d theta` `8 int cos 2 theta.((1-cos 2theta))/(2) d theta` `= - 4((sin 2theta)/(2) - int (1+cos 4theta)/(2) d theta)+c` ` = - 2(sin 2 theta - 2theta - (sin 4 theta)/(4))+c` `=-2 sin 2 theta + 2theta +(1)/(2) . 2 sin 2 theta cos 2 theta + c` ` = - 2 sqrt(1-x)+cos^(-1)sqrt(x)+sqrt(1-x). sqrt(x+c)` |
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| 28. |
`(1)/((x-2)sqrt(x^(2) - 4+3))` का समाकलन कीजिये - |
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Answer» Correct Answer - `sec^(-1)(x-2)` माना x - 2 = t |
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| 29. |
`(sin x)/(a^(2) +b^(2) cos^(2)x)` का समाकलन कीजिये - |
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Answer» Correct Answer - `-(1)/(ab)tan^(-1)((b cos x)/(a))` `i = int(sinx)/(a^(2) +b^(2) cos^(2)x) dx = int (sinx)/(b^(2)[1+((b)/(a)cosx)^(2)])` अब माना `(b)/(a)cosx=t rArr sin x dx = - (a)/(b)dt` |
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| 30. |
`inttan^(-1)sqrt((1-sinx)/(1+sinx))dx`का मान ज्ञात कीजिए। |
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Answer» हम जानते है `(1-sinx)/(1+sinx) = (cos^(2) .(x)/(2)+ sin^(2).(x)/(2)-2cos.(x)/(2)sin.(x)/(2))/(cos^(2).(x)/(2)+sin^(2).(x)/(2)+ 2cos.(x)/(2)sin.(x)/(2))` `((cos.(x)/(2) - sin.(x)/(2))/(cos.(x)/(2)+sin.(x)/(2)))` `rArr sqrt((1-sinx)/(1+sinx))=(cos.(x)/(2)-sin.(x)/(2))/(cos.(x)/(2)+sin.(x)/(2))` ` = (1-(sin x//2)/(cos x//2))/(1+(sin//2)/(cosx//2)) = (1-tan.(x)/(2))/(1+tan.(x)/(2))= tan((pi)/(2) - (x)/(2))` `therefore int tan^(-1) sqrt((1-sinx)/(1+sin))dx = int tan^(-1) [ tan ((pi)/(4) - (pi)/(2))]dx` ` = int ((pi)/(4) - (x)/(2))dx = (pi)/(4)x - (x^2)/(4) + c` |
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| 31. |
`(sin (tan^(-1)x))/(1+x^(2))` फलनो का x के सापेक्ष समाकलन कीजिये । |
| Answer» Correct Answer - `-(1)/(4)cos^(4)` | |
| 32. |
`(1)/(sinxcosx)` का समाकलन कीजिये - |
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Answer» Correct Answer - `-(1)/(2)log cos 2x` `I = int tan 2xdx = int(sin 2x)/(cos 2x)dx` माना `cos 2x = t rArr -2sin2x dx = dt` |
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| 33. |
`(sin 2x)/(a^(2)+b^(2) sin^(2)x)` का समाकलन कीजिये - |
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Answer» Correct Answer - `(1)/(b^(2))log(a^(2) + b^(2)sin^(2)x)` माना `a^(2) + b^(2) sin^(2) x = t rArr b^(2).2sinx cos x dx = dt` |
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| 34. |
`int(1-tanx)/(1+tanx)dx` का मान ज्ञात कीजिए। |
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Answer» `int (1-tanx)/(1+tanx) dx = f int (1-(sinx)/(cosx))/(1+(sinx)/(cosx)) dx = int (cos x- sin x)/(cosx +sinx)` माना `cosx + sin x = t` `rArr (- sin x + cosx) dx = dt` `rArr " "(cosx - sin x)dx = dt` `therefore int(1-tan x)/(1+tan x) dx = int(cos x - sinx)/(cos x + sin)dx = int(1)/(dt) = log t = log (cosx + sinx)` |
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| 35. |
`(sinx)/(a^(2)+b^(2)cos^(2)x)` का समाकलन कीजिये - |
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Answer» Correct Answer - `log tan x` `I = int (dx)/(sinx cos x) = int(sec^(2)x)/(tanx)dx` माना `tan x = t " "rArr sec^(2)xdx = dt` |
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| 36. |
`cos^(3) x sinx ` फलनो का x के सापेक्ष समाकलन कीजिये । |
| Answer» Correct Answer - `(1)/(2)tan^(2)x` | |
| 37. |
`sin2x . sqrt(a^(2) - cos^(2)x)` का समाकलन कीजिये। |
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Answer» Correct Answer - `(2)/(3)(a^(2) - cos^(2)x)^(3//2)` `I = intsin 2xsqrt(a^(2) - cos^(2)x).dx` ` = intsqrt((a^(2) - cos^(2)x)). sin2x dx` `a^(2) - cos^(2) x = t^(2)` रखने पर `rArr " "sin2x dx = 2tdt` तब `I = intsqrt(t^(2)).2tdt` |
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| 38. |
`(1)/(1+9x^(2))` का समाकलन कीजिये - |
| Answer» Correct Answer - `(1)/(3)tan^(-1)(3x)` | |
| 39. |
`sqrt((25-9x^(2)))` का समाकलन कीजिये - |
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Answer» Correct Answer - `(x)/(2)sqrt(25 -9x^(2))+(25)/(6)sin^(-1)((3x)/(5))` `I = intsqrt((25 - 9x^(2))) =intsqrt((5)^(2)-(3x)^(2)).dx` अब माना `3x = t rArr 3d = dt` |
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| 40. |
`(x^(2))/(sqrt(x^(2)-a^(2)))` का समाकलन कीजिये - |
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Answer» Correct Answer - `(x)/(2)sqrt((x^(2)-a^(2)))+(1)/(2)a^(2)loglog [ x + sqrt((x^(2)-a^(2)))` `I = int(x^(2))/(sqrt(x^(2) -a^(2)))dx = int(x^(2)-a^(2) +a^(2))/(sqrt(x^(2) - a^(2)))dx` `intsqrt(x^(2) -a^(2))dx + a^(2) int(dx)/(sqrt(x^(2) -a^(2)))` |
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| 41. |
`(sin x)/((1+cosx)^(2))` का समाकलन कीजिए - |
| Answer» Correct Answer - `(1)/(1+cosx)+c` | |
| 42. |
`(1)/(1-tanx)` का समाकलन कीजिए - |
| Answer» Correct Answer - `(x)/(2)-(1)/(2)log|cosx - sin|+c` | |
| 43. |
`(2sin 2x)/(cos^(2)2x)` का समाकलन कीजिये - |
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Answer» Correct Answer - `sec 2x` माना `cos2x = t rArr -sin2x. 2dx = dt` `rArr 2 sin 2xdx = - dt` तब `I = - int(1)/(t^(2))dt = (1)/(t)= (1)/(cos2x)` |
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| 44. |
`int(sin^(8)x -cos^(8)x)/(1-2sin^(2)x cos^(2)x)` का मान ज्ञात कीजिए। |
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Answer» `int (sin^(8) x- cos^(8)x)/(1-2sin^(2) xcos^(2)x)dx` ` = int((sin^(4) x -cos^(4)x)(sin^(4) x + cos^(4)x))/(1-2 sin^(2)x cos^(2) x)` `int((sin^(2)x - cos^(2)x)(sin^(2) x+ cos^(2)x){(sin^(2) x + cos^(2) x)^(2) - 2 sin^(2) x cos^(2) x})/ (1-2 sin^(2) xcos^(2) x)` `int(-cos2x . 1{1 - 2 sin^(2) x cos ^(2) x})/(1-2sin^(2) x cos^(2) x)dx` ` = - int cos 2xdx = - (sin2x)/(2)+c` |
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| 45. |
`(1)/(b^(2) +a^(2)x^(2))` का समाकलन कीजिये - |
| Answer» Correct Answer - `(1)/(b)sin^(-1)(bx +c)` | |
| 46. |
`int sin^(4) 2xdx` का मान ज्ञात कीजिए। |
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Answer» `int sin^(4) 2x dx = int (sin^(2) 2x)^(2)dx` `int((1-cos4x)/(2))^(3)dx` ` = (1)/(4) int (1-2 cos4x + cos^(2) 4x)dx` ` = (1)/(4)int[1-2cos4x+ cos8x] dx` ` = (1)/(8) int [3-4cos4x + cos 8x]dx` ` = (1)/(8)[3x - sin 4x + (1)/(8) +(1)/(8) sin8x] + c` |
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| 47. |
`(sinx)/(sqrt(9-cos^(2)x))` का समाकलन कीजिये - |
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Answer» Correct Answer - `(x)/(2)sqrt(25 - 9x^(2))` माना `cos x = t` |
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| 48. |
`(1)/(sqrt(9-x^(2)))` का समाकलन कीजिये - |
| Answer» Correct Answer - `(1)/(ab)tan^(-1)((ax)/(b))` | |
| 49. |
`int(sin(x-alpha))/(sin(x+alpha))dx` का मान ज्ञात कीजिए। |
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Answer» माना `I = int(sin(x -alpha))/(sin(x+alpha))dx` माना ` x + alpha = t " "rArr x = t - alpha rArr dx = dt` `therefore" "I int (sin(x-alpha))/(sin t)dt` ` = int [(sin t cos 2alpha)/(sin t) - (cos t sin 2alpha)/(sin t)]dt ` ` = int cos 2alpha dt - int sin2alpha cot t dt` `=cos2alpha int dt - sin 2alpha int cot t dt ` `= t cos 2alpha - sn 2 alpha log|sin t|+c` `=(x+alpha) cos2alpha - sin 2 alpha log|sin (x +alpha)|+c` `= x cos 2 alpha - sin 2 alpha log|sin (x + alpha )|+c_(1)` जहाँ` c_(1) = c + alpha cos 2alpha` |
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| 50. |
`int ((tan^(-1)x)^(4))/(1+x^(2)) dx` का मान ज्ञात कीजिए। |
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Answer» माना `tan^(-1) x =t rArr " " (1)/(1+x^(2))dx = dt` ` therefore int((tan^(-1)x)^(4))/(1+x^(2)) dx = int t^(4) dt = (t)/(5) = (1)/(5)(tan^(-1) x)^(5)` |
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