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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. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए - `int_(0)^(pi//4)(1)/(1+cos2x)dx` |
| Answer» Correct Answer - `(1)/(2)` | |
| 2. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(1)/(1+cos2x)dx` |
| Answer» Correct Answer - `(1)/(2)tanx+c` | |
| 3. |
निम्नलिखित समाकलों को हल कीजिए । (i) `int(2x^(3))/(4+x^(8))dx` (ii) `int(1)/(x^(2))sin.(1)/(x)dx` |
| Answer» (i) `(1)/(4)tan^(-1).(x^(4))/(2)+c` (ii) `cos((1)/(x))+c` | |
| 4. |
निम्नलिखित समाकलों को हल कीजिए । `inte^(x).(1+be^(x))^(n)dx` |
| Answer» `((a+be^(x))^(n+1))/(b(n+1))+c` | |
| 5. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(1)/(1-cos2x)dx` |
| Answer» Correct Answer - `-(1)/(2)cotx+c` | |
| 6. |
निम्नलिखित समाकलों को हल कीजिए । `int(xdx)/((1+x^(2))^(3//2))` |
| Answer» `-(1)/(sqrt(1+x^(2)))+c` | |
| 7. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(1+cos2x)/(1-cos2x)dx` |
| Answer» Correct Answer - `-cotx-x+c` | |
| 8. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- (i) `sqrt(1+sin.(x)/(2))dx` |
| Answer» (i) `4sin.(x)/(4)-2cos.(x)/(4)+c` | |
| 9. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(cos^(3)x+sin^(3)x)/(sin^(2)x.cos^(2)x)dx` |
| Answer» (i) `secx-"cosec"x+c` (ii) `-cotx-tanx+c` | |
| 10. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int tan^(2)xdx` |
| Answer» Correct Answer - `"cosec" x-cotx+c` | |
| 11. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `inte^(4-3x)dx` |
| Answer» `-(1)/(3).e^(4-3x)+c` | |
| 12. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `inte^(2x+5)dx` |
| Answer» `(1)/(2).e^(2x+5)+c` | |
| 13. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int((x+1)(2x-3))/(x)dx` |
| Answer» Correct Answer - `x^(2)-x-3logx+c` | |
| 14. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(x^(2)+3)/(x^(2)+1)dx` |
| Answer» Correct Answer - `x+2tan^(-1)x+c` | |
| 15. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `inte^(x+3)dx` |
| Answer» Correct Answer - `e^(x+3)+c` | |
| 16. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(x)/(x+a)dx` |
| Answer» Correct Answer - `x-alog(x+a)+c` | |
| 17. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(sinx)/(1+sinx)dx` |
| Answer» Correct Answer - `secx-tanx+x+c` | |
| 18. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `intsqrt(1-cos2x)dx` |
| Answer» Correct Answer - `-sqrt2cosx+c` | |
| 19. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `intsqrt(2x+(1)/(3))dx` |
| Answer» `(1)/(3)(2x+(1)/(3))^(3//2)+c` | |
| 20. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `intsqrt(1+cos2x)dx` |
| Answer» Correct Answer - `sqrt2sinx+c` | |
| 21. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(x^(2)+2x-5)/(sqrtx)dx` |
| Answer» `(2)/(5)x^(5//2)+(4)/(3)x^(3//2)-10sqrtx+c` | |
| 22. |
निम्नलिखित समाकलों को हल कीजिए । `int(e^(sqrtx).sin e^(sqrtx))/(sqrtx)dx` |
| Answer» Correct Answer - `-2cos e^(sqrtx)+c` | |
| 23. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- (i) `int((1+x)^(3))/(sqrtx)dx` (ii) `int((1+x)^(3))/(x^(4))dx` |
| Answer» (i) `(2)/(7)x^(7//2)+(6)/(5)x^(5//2)+2x^(3//2)+2sqrtx+c` (ii) `-(1)/(3x^(3))-(3)/(2x^(2))-(3)/(x)+logx+c` | |
| 24. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `intsqrt(2x-1)dx` |
| Answer» `((2x-1)^(3//2))/(3)+c` | |
| 25. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(1-x)/(sqrtx)dx` |
| Answer» `2sqrtx-(2)/(3)x^(3//2)+c` | |
| 26. |
`(cos sqrtx)/(sqrt(1+sinx))` |
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Answer» `int(cosx)/(sqrt(1+sinx))dx" माना "1+sinx=t` `" "rArr" "cos x dx = dt` `=int(1)/(sqrtt)dt=int t^(-1//2)dt` `=(t^(1//2))/(1//2)+c=2sqrt(1+sinx)+c` |
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| 27. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- (i) `(1)/(5x+1)dx` (ii) `int(1)/(sqrt(x+1)+sqrtx)dx` |
| Answer» (i) `(1)/(5)log_(e)(5x+1)+c` (ii) `(2)/(3)[(x+1)^(3//2)-x^(3//2)]+c` | |
| 28. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(3-7x)^(5)dx` |
| Answer» `((3-7x)^(6))/(-42)+c` | |
| 29. |
`int(x^(3)-x^(2)+x-1)/(x-1)dx` का मान होगा -A. `(x^(2))/(3)+x+c`B. `(x^(3))/(3)-x+c`C. `(x^(3))/(3)+x+c`D. `(x^(5))/(3)+2x+c` |
| Answer» Correct Answer - C | |
| 30. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(ax+b)^(3)dx` |
| Answer» `((ax+b)^(4))/(4a)+c` | |
| 31. |
`int(x^(3)-1)/(x^(2))dx` का मान होगा -A. `(x^(2))/(4)+(1)/(2)x+c`B. `(x^(3))/(3)-x^(2)+c`C. `(x^(2))/(2)+(1)/(2)+c`D. `x^(2)+c` |
| Answer» Correct Answer - C | |
| 32. |
`sqrt((1-sqrtx)/(1+sqrtx))` |
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Answer» माना `I=intsqrt((1-sqrtx)/(1+sqrtx))` माना `sqrtx=cost` `rArr" "x=cos^(2)t rArr dx=2cos t (-sin t)dt` `therefore" "I=intsqrt((1-cost)/(1+cost))(-2sint cost)dt` `I=-2intsqrt((2sin^(2).(t)/(2))/(2cos^(2).(t)/(2))).2sin.(t)/(2)cos.(t)/(2).cost dt` `=-4intsin^(2).(t)/(2).cost dt` `=-4int(1-cost)/(2).costdt` `=-2int(cost-cos^(2)t)dt` `=-2int(cost-(1+cos2t)/(2))dt` `=-2sint+t+(1)/(2)sin2t+C` `=-2sqrt(1-cos^(2)t)+t+(1)/(2)(2sqrt(1-cos^(2)t).cost)+C` `=-2sqrt(1-x)+cos^(-1)sqrtx+sqrtxsqrt(1-x)+C` `=-2sqrt(1-x)+cos^(-1)sqrtx+sqrt(x-x^(2))+C` |
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| 33. |
`int(x^(2//3)+1)dx` का मान होगा -A. `(3)/(5)x^(5//3)+x+c`B. `(3)/(5)x^(2//3) +c`C. `(3)/(7)x^(5//3) + 2x + c`D. `2x^(2)+c` |
| Answer» Correct Answer - A | |
| 34. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(3x-2)^(3)dx` |
| Answer» `(27x^(4))/(4)-18x^(3)+18x^(2)-8x+c` | |
| 35. |
मान ज्ञात कीजिए - `int x^(2)(1-(1)/(x^(2)))dx` |
| Answer» Correct Answer - `(x^(3))/(3)-x+c` | |
| 36. |
`int(x^(2))/((a+bx)^(2))dx` का मान है -A. `(x^(2))/(b(a+bx))+(2)/(b^(2))x+c`B. `(-x^(2))/(b(a+bx))+(2)/(b^(2))[x-(a)/(b)log(a+bx)]+c`C. `(-x^(2))/(b(a+bx))-(2)/(b^(2))[x-(1)/(b)log(a+bx)]+c`D. इनमे से कोई नहीं | |
| Answer» Correct Answer - B | |
| 37. |
मान ज्ञात कीजिए -`int (2x^(2)+e^(x))dx` |
| Answer» `(2)/(3)x^(3)+e^(x)+c` | |
| 38. |
`intx^(2) (1-(1)/(x^(2)))dx` का मान होगा -A. `(x^(3))/(3)+x+c`B. `(x^(2))/(2)+x+c`C. `(x^(3))/(3)-x+c`D. `(x^(2))/(2)-x+c` |
| Answer» Correct Answer - C | |
| 39. |
`int(ax^(2)+bx+c)dx` का मान होगा -A. `ax^(2)+bx+c`B. `(ax^(3))/(3)+(bx^(2))/(2)+cx+d`C. `ax^(3)+bx^(2)+cx`D. `ax+b` |
| Answer» Correct Answer - B | |
| 40. |
`(sin^(-1)sqrtx-cos^(-1)sqrtx)/(sin^(-1)sqrtx+cos^(-1)sqrtx),(x in [0,1])` |
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Answer» माना `I=int(sin^(-1)sqrtx-cos^(-1)sqrtx)/(sin^(-1)sqrtx+cos^(-1)sqrtx)dx` हम जानते हैं कि `sin^(-1)sqrtx+cos^(-1)sqrtx=(pi)/(2)` `rArr cos^(-1)sqrtx=(pi)/(2)-sin^(-1)sqrtx` `therefore" "I=int(sin^(-1)sqrtx-((pi)/(2)-sin^(-1)sqrtx))/((pi)/(2))dx` `" "=int(2sin^(-1)sqrtx-(pi)/(2))/((pi)/(2))dx` `" "=(2)/(pi)int(2sin^(-1)sqrtx-(pi)/(2))dx` `" "=(4)/(pi)intsin^(-1)sqrtxdx-int1dx` `" "=(4)/(pi)intsin^(-1)sqrtx dx-x` `rArr" "I=(4)/(pi)I_(1)-x+C" ...(1)"` जहाँ `" "I_(1)=intsin^(-1)sqrtx dx` माना `sqrtx=t" "rArr" "x=t^(2)" "rArr" "dx=2tdt` `I_(1)=intsin^(-1)t 2t dt = 2 int underset("I ")(sin^(-1))underset("II")(t.t) dt ` `=2(sin^(-1)t.(t^(2))/(2)-int(1)/(sqrt(1-t^(2))).(t^(2))/(2)dt)` `=t^(2)sin^(-1)t-int(t^(2))/(sqrt(1-t^(2)))dt` `=t^(2)sin^(-1)t-int(-(1-t^(2))+1)/(sqrt(1-t^(2)))dt` `=t^(2)sin^(-1)t+intsqrt(1-t^(2))dt-int(1)/(sqrt(1-t^(2)))dt` `=t^(2)sin6(-1)t+(tsqrt(1-t^(2)))/(2)+(1)/(2)sin^(-1)t-sin^(-1)t` `=(t^(2)-(1)/(2))sin^(-1)t+(1)/(2)tsqrt(1-t^(2))` `=(1)/(2)[(2x-1)sin^(-1)sqrtx+sqrtxsqrt(1-x)]` `=(1)/(2)[(2x-1)sin^(-1)sqrtx+sqrt(x-x^(2))]` `I_(1)` का मान समीकरण (1 ) में रखने पर, `int(sin^(-1)sqrtx-cos^(-1)sqrtx)/(sin^(-1)sqrtx+cos^(-1)sqrtx)dx` `=(2)/(pi)[(2x-1)sin^(-1)sqrtx+sqrt(x-x^(2))]-x+C` |
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| 41. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(ax^(2)+bx+c)dx` |
| Answer» `(ax^(3))/(3)+(bx^(2))/(2)+cx+k` | |
| 42. |
मान ज्ञात कीजिए -`int (ax^(2)+bx+c)dx` |
| Answer» `(ax^(3))/(3)+(bx^(2))/(2)+cx+c_(1)` | |
| 43. |
`int((ax^(4)+bx^(2)+c)/(x^(4)))dx` का मान ज्ञात कीजिए । |
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Answer» `int((ax^(4)+bx^(2)+c)/(x^(4)))dx` `=int((ax^(4))/(x^(4))+(bx^(2))/(x^(4))+(c)/(x^(4)))dx` `=aint1dx+b intx^(-2)dx+c intx^(-4)dx` `=ax+(b.x^(-1))/(-1)+(c.x^(-3))/(-3)+k` `" "` (k = समाकल अचर है ) `=ax-(b)/(x)-(c)/(3x^(3))+k` |
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| 44. |
`intx^(2)(1-(1)/(x^(2)))dx` |
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Answer» `intx^(2)(1-(1)/(x^(2)))dx=int(x^(2)-1)dx` `=intx^(2)dx-int1.dx=(x^(3))/(3)-x+c` |
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| 45. |
`int(ax^(2)+bx+c)dx` |
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Answer» `int(ax^(2)+bx+c)dx` `=aintx^(2)dx+b intxdx+cint1.dx` `=(ax^(3))/(3)+(bx^(2))/(2)+cx+k` जहाँ k = समाकल नियतांक है। |
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| 46. |
दर्शाइए कि `int_(0)^(a)f(x)g(x)dx=2int_(0)^(a)f(x)dx,` यदि f और g को `f(x)=f(a-x)` एवं `g(x)+g(a-x)=4` के रूप में परिभाषित किया गया है। |
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Answer» माना `I=int_(0)^(a)f(x)g(x)dx" …(1)"` `I=int_(0)^(a)f(a-x)g(a-x)dx` `" "[because int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]` `rArr" "I=int_(0)^(a)f(x){4-g(x)}dx" ..(2)"` `[because f(x)=f(a-x)" तथा "g(x)=g(a-x)=4" (दिया है)" ]` समीकरण (1 ) और (2 ) को जोड़ने पर `rArr " "@I=int_(0)^(a)4f(x)dx rArr I=2 int_(0)^(a)f(x)dx` |
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| 47. |
`int_(0)^(4)|x-1|dx` |
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Answer» माना `I=int_(0)^(4)|x-1|dx` `therefore" "I=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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| 48. |
`int_(0)^((pi)/(2))(2log sin x- log sin 2x)dx` |
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Answer» माना `" "I=int_(0)^(pi//2)(2log sin x-log sin 2x)dx` `=int_(0)^(pi//2)(log sin^(2)x-log sin 2x)dx` `=int_(0)^(pi//2)log((sin^(2)x)/(sin2x))dx` `=int_(0)^(pi//2)log((sin^(2)x)/(2sin x cos x))dx` `=int_(0)^(pi//2)log((tanx)/(2))dx` `=int_(0)^(pi//2)log(tanx)-log 2dx` `=int_(0)^(pi//2)log(tanx)dx-int_(0)^(pi//2)log2dx` `rArr" "I=I_(1)-log2[x]_(0)^((pi)/(2))=I_(1)-((pi)/(2)-0)log 2" ...(1)"` जहाँ,`" "I_(1)=int_(0)^(pi//2)log(tan x)dx" ...(2)"` `rArr" "I_(1)=int_(0)^(pi//2)log[tan((pi)/(2)-x)]dx` `[because int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]` `rArr" "I_(1)=int_(0)^(pi//2)log(cotx)dx" ...(3)"` समीकरण (2 ) और (3 ) को जोड़ने पर , `2I_(1)=int_(0)^(pi//2){(log(tanx)+log(cotx)}dx` `int_(0)^(pi//2)log(tanxcotx)dx` `=int_(0)^(pi//2)log 1dx=0` `rArr" "I_(1)=0` `I_(1)` का मान समीकरण (1 ) में रखने पर, `I=0-(pi)/(2)log 2=-(pi)/(2)log 2` |
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| 49. |
`int_(0)^(2)(dx)/(x+4-x^(2))` |
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Answer» `int_(0)^(2)(1)/(x+4-x^(2))dx` `=int_(0)^(2)(1)/(x+4-x^(2)-((1)/(2))^(2)+((1)/(2))^(2))dx` `=int_(0)^(2)(1)/(4+(1)/(4)-[(x^(2)-x+(1)/(4))])dx` `=int_(0)^(2)(1)/(((sqrt(17))/(2))^(2)-(x-(1)/(2))^(2))dx` `=(1)/(2.(sqrt(17))/(2))[log|((sqrt(17))/(2)+x-(1)/(2))/((sqrt(17))/(2)-x+(1)/(2))|]_(0)^(2)` `=(1)/(sqrt(17))[log|(sqrt(17)+2x-1)/(sqrt(17)-2x+1)|]_(0)^(2)` `=(1)/(sqrt(17))[log|(sqrt(17)+3)/(sqrt(17)-3)-log|(sqrt(17)-1)/(sqrt(17)+1)|]` `=(1)/(sqrt(17))log|(sqrt(17)+3)/(sqrt(17)-3)div(sqrt(17)-1)/(sqrt(17+1))|` `(1)/(sqrt(17))log|((sqrt(17)+3)(sqrt(17)+1))/((sqrt(17-3))(sqrt(17)-1))|` `=(1)/(sqrt(17))log|(20+4sqrt(17))/(20-4sqrt(17))|` `=(1)/(sqrt(17))log|(5+sqrt(17))/(5-sqrt(17))xx(5+sqrt(17))/(5+sqrt(17))|` `=(1)/(sqrt(17))log|((5+sqrt(17))^(2))/((5)^(2)-(sqrt(17))^(2))|` `" "[because (a-b)(a+b)=a^(2)-b^(2)]` `=(1)/(sqrt(17))log|(25+17+10sqrt(17))/(25-17)|` `=(1)/(sqrt(17))log|(42+10sqrt(17))/(8)|` `=(1)/(sqrt(17))log|(21+5sqrt(17))/(4)|` |
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| 50. |
`(1)/(x^(2)(x^(4)+1)^((3)/(4)))` |
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Answer» `int(1)/(x^(2)(x^(4)+1)^(3//4))dx` `" "=int(1)/(x^(2){x^(4)(1+(1)/(x^(4)))}^(3//4))dx` `" "=int(1)/(x^(5)(1+(1)/(x^(4)))^(3//4))dx` माना`" "1+(1)/(x^(4))=t" "rArr(-4)/(x^(5))dx=dtrArr(1)/(x^(5))dx=(dt)/(4)` `" "=int(1)/(x^(3//4)).(dt)/(-4)=-(1)/(4).intt^(-3//4)dt` `" "=-(1)/(4).(t^(1//4))/(1//4)+C` `" "=-(1+(1)/(x^(4)))^(1//4)+C` `" "=-((x^(4)+1)/(x^(4)))^(1//4)+C` `" "=-(1)/(x)(x^(4)+1)^(1//4)+C` |
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