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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.
| 251. |
समाकलन कीजिए - `int (x^(2))/(1+x^(2))dx` |
| Answer» Correct Answer - `x-tan^(-1)x+c` | |
| 252. |
समाकलन कीजिए -`int secx (secx+tanx)dx` |
| Answer» Correct Answer - `tanx + sec x + c` | |
| 253. |
सिद्ध कीजिए कि `int_(-1)^(1)log((2-x)/(2+x))dx=0` |
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Answer» माना `f(x)=log((2-x)/(2+x))` `rArr" "f(-x)=log((2+x)/(2-x))` `" "=-log((2-x)/(2+x))=-f(x)` `rArr f(x)` एक विषम फलन है `rArr" "int_(-a)^(a)f(x)dx=0` `rArr" "int_(-1)^(1)log((2-x)/(2+x))dx=0` यही सिद्ध करना था । |
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| 254. |
निम्नलिखित को सिद्ध कीजिए । `int_(0)^(1)sin^(-1)xdx=(pi)/(2)-1` |
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Answer» माना `x=sin theta` `rArr " "dx=cos theta d theta` `x = 0` पर `" "sin theta=0" "rArr" "theta=0` `x=1` पर `" "sin theta=1" "rArr" "theta=pi//2` `therefore int_(0)^(1)sin^(-1)xdx=int_(0)^(pi//2)theta. cos theta d theta` `" "=[theta sin theta]_(0)^(pi//2)-int_(0)^(pi//2)1.sin theta d theta` `" "=(pi)/(2)-0+[cos theta]_(0)^(pi//2)` `" "=(pi)/(2)+cos.(pi)/(2)-cos 0` `" "=((pi)/(2)-1)" "` यही सिद्ध करना था । |
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| 255. |
`cos^(3)xe^(logsinx)` |
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Answer» माना `I=intcos^(3)xe^(logsinx)dx=int(cosx)^(3)sinx dx` `" "(because e^(logx)=x)` `=-int(cosx)^(3)(-sinx)dx` माना `cosx=t" "rArr" "-sinxdx=dt` `therefore" "I=-int t^(3)dt=-(t^(4))/(4)+C=-(cos^(4)x)/(4)+C` |
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| 256. |
`e^(3logx)(x^(4)+1)^(-1)` |
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Answer» माना `I=inte^(3logx)(x^(4)+1)^(-1)dx` `=int(e^(logx^(3)))/((x^(4)+1))dx=int(x^(3))/((x^(4)+))dx` `" "(because e^(logx)=x)` माना`" "x^(4)+1=t rArr 4x^(3)dx=dt" "rArr" "x^(3)dx=(dt)/(4)` `therefore" "I=(1)/(4)int(1)/(t)=(1)/(4)log|t|+C` `" "=(1)/(4)log|x^(4)+1|+C` |
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| 257. |
समाकलन कीजिए - `int[(cotx)/(sinx)-tan^(2)x-(tanx)/(cosx)+(2)/(cos^(2)x)]dx` |
| Answer» `-cosecx + tanx + x -secx +c` | |
| 258. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए - `int_(1)^(2)(1)/(x(1+x^(4)))dx` |
| Answer» `(1)/(4)log.(32)/(17)` | |
| 259. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए - `int_(0)^(pi//2)(1)/(4+3cosx)dx` |
| Answer» `(2)/(sqrt7)tan^(-1).(1)/(sqrt7)` | |
| 260. |
निम्नलिखित को सिद्ध कीजिए । `int_(1)^(3)(dx)/(x^(2)(x+1))=(2)/(3)+log.(2)/(3)` |
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Answer» `int_(1)^(3)(1)/(x^(2)(x+1))dx=(2)/(3)+log.(2)/(3)` माना `" "(1)/(x^(2)(x+1))=(A)/(x)+(B)/(x^(2))+(C)/(x+1)" …(1)"` `rArr" "1=Ax(x+1)+B(x+1)+Cx^(2)" …(2)"` x = 0 तो `1=0+B+0" "rArr" "B=1"` `x = -1` तो `1=0+0+C" "rArr" "C=1` `x^(2)` के गुणांकों की तुलना करने पर , `0=A+C" "rArr" "A=-C=-1` `therefore int_(1)^(3)(1)/(x^(2)(x+1))dx` `=int_(1)^(3)(-(1)/(x)+(1)/(x^(2))+(1)/(x+1))dx` `=[-log|x|-(1)/(x)+log|x+1|]_(1)^(3)` `=[log|(x+1)/(x)|-(1)/(x)]_(1)^(3)` `=(log|(4)/(3)|-(1)/(3))-(log|(2)/(1)|-(1)/(2))` `=(log.(4)/(3)-log2)+(1-(1)/(3))` `=log((4)/(3)xx(1)/(2)+(2)/(3)=log.(2)/(3)+(2)/(3))` `" "`यही सिद्ध करना था । |
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| 261. |
`int (1+cos4x)/(cotx-tanx)dx` का मान ज्ञात कीजिए | |
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Answer» `int (1+cos4x)/(cotx-tanx)dx=int(1+2cos^(2)2x-1)/((cosx)/(sinx)-(sinx)/(cosx))dx` `=int(2cos^(2)x2xcosxsinx)/(cos^(2)x-sin^(2)x)dx` `=int(2cos^(2)2xcosxsinx)/(cos2x)dx` `=intcos2x*2sinxcosxdx` `=intcos2xsin2xdx` `=(1)/(2)intsin4xdx=-(1)/(8)cos4x+c` |
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| 262. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए - `int_(pi//3)^(pi//4)(tanx+cotx)^(2)dx` |
| Answer» Correct Answer - `(-2)/(sqrt3)` | |
| 263. |
निम्नलिखित समाकलों को हल कीजिए । `int(sec^(2)x)/(sqrt(tanx))dx` |
| Answer» Correct Answer - `2sqrt(tanx)+c` | |
| 264. |
निम्नलिखित के मान ज्ञात कीजिए - `int(sqrt(tanx)+sqrt(cotx))dx` |
| Answer» `sqrt2 sin^(-1)(sinx - cosx)+c` | |
| 265. |
`int_(0)^(pi//4)log(1+tanx)dx` का मान है -A. `(pi)/(2)log2`B. `(pi)/(4)log2`C. `(pi)/(6)log2`D. `(pi)/(8)log2.` |
| Answer» Correct Answer - D | |
| 266. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए - `int_(0)^(pi//4)e^(x)(tanx+sec^(2)x)dx` |
| Answer» Correct Answer - `e^(pi//4)` | |
| 267. |
`int_(0)^(pi//2)log|tanx+cotx|dx` का मान है -A. `-pi log 2`B. `pi log 2`C. `(pi)/(2)log2`D. `-(pi)/(2)log2.` |
| Answer» Correct Answer - B | |
| 268. |
`int_(1)^(4)(|x-1|+|x-2|+|x-3|)dx` |
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Answer» `int_(1)^(4){(|x-1|+|x-2|+|x-3|)}dx` `=int_(1)^(2)(|x-1|+|x+2|+|x-3|)dx+int_(2)^(3){|x-1|+|x-2|+|x-3|}dx+int_(3)^(4){|x-1|+|x-2|+|x-3|}dx` `=int_(1)^(2){x-1-(x-2)-(x-3)}dx+int_(2)^(3){x-1+x-2-(x-3)}dx+int_(3)^(4)(x+1+x-2+x-3)dx` `=int_(1)^(2)(-x+4)dx+int_(2)^(3)xdx+int_(3)^(4)(3x-6)dx` `=[(-x^(2))/(2)+4x]_(1)^(2)+[(x^(2))/(2)]_(2)^(3)+[(3x^(2))/(2)-6x]_(3)^(4)` `=((-2^(2))/(2)+8)-((-1)/(2)+4)+(1)/(2)(3^(2)-2^(2))+((3)/(2)xx4^(2)-6xx4)-((3)/(2)xx3^(2)-6xx3)` `=6-(7)/(2)+(5)/(2)+(24-24)-(-(9)/(2))` `=(12-7+5)/(2)+0+(9)/(2)=(19)/(2)` |
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| 269. |
`int_(0)^((pi)/(2))sin 2x tan^(-1)(sinx)dx` |
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Answer» माना `I=int_(0)^(pi//2)sin 2x tan^(-1)(sinx)dx` `=int_(0)^(pi//2)2sinxcos x tan^(-1)(sinx)dx` माना ` sin x=t` `rArr cos x dx = dt` तथा `x=0" "rArr" "t=0,` `" "x=(pi)/(2)" "rArr" "t=sin.(pi)/(2)=1` `I=int_(0)^(1)2t tan^(-1)t dt = 2int_(0)^(1)underset("II")t.underset(" I ")(tan^(-1))tdt` `=2[{tan^(-1)t(t^(2))/(2)}_(0)^(1)-int_(0)^(1)(1)/(1+t^(2)).(t^(2))/(2)dt]` (खण्डशः समाकलन के प्रयोग से ) `=2((tan^(-1)1)/(2)-0)-int_(0)^(1)(t^(2))/(1+t^(2))dt` `=2((pi)/(8))-int_(0)^(1)((1+t^(2))-1)/(1+t^(2))dt` `=(pi)/(4)-int_(0)^(1)(1-(1)/(1_t^(2)))dt` `=(pi)/(4)-[t-tan^(-1)t]_(0)^(1)` `=(pi)/(4)-[1-tan^(-1)1-(0-0)]` `=(pi)/(4)-1+(pi)/(4)=(pi)/(2)-1` |
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| 270. |
`int_(0)^(pi)(x tanx)/(sec x +tan x)dx` |
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Answer» माना `I=int_(0)^(pi)(x tan x)/(sec x+ tanx)dx" …(1)"` `rArr" "I=int_(0)^(pi)((pi-x)tan(pi-x))/(sec(pi-x)+tan(pi-x))dx` `" "` [ प्रगुण (4 ) से ] `=int_(0)^(pi)(-(pi-x)tanx)/(-secx-tanx)dx` `rArr" "I=int_(0)^(pi)((pi-x)tanx)/(secx+tanx)dx" ...(2)"` समीकरण (1 ) और (2 ) को जोड़ने पर `2I=int_(0)^(pi)(x tanx +(pi-x)tanx)/(secx+tanx)dx` `=int_(0)^(pi)(pi tan x)/(secx+tanx)dx` `=pi int_(0)^(pi)(tanx(secx - tanx))/(sec^(2)x-tan^(2)x)dx` `=piint_(0)^(pi)(tanxsecx-tan^(2)x)dx` `=pi int_(0)^(pi)(tanx sec x - sec^(2)x+1)dx` `=pi[secx - tanx+x]_(0)^(pi)` `=pi[(secpi-tanpi+pi)-(sec 0 - tan0+0)]` `=pi[-1-0+pi-1-0+0]` `=pi(pi-2)` `rArr" "I=pi((pi)/(2)-1)` |
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| 271. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `ints^(-1//3)dx` |
| Answer» Correct Answer - `(3)/(2)z^(2//3)+c` | |
| 272. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int2^(x)dx` |
| Answer» `(2^(x))/(log_(e)2)+c` | |
| 273. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `intb^(x+a)dx` |
| Answer» `(b^(x+a))/(logb)+c` | |
| 274. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(1)/(sqrt1-y^(2))dy` |
| Answer» Correct Answer - `sin^(-1)y+c` | |
| 275. |
निम्नलिखित समाकलों के मान ज्ञात कीजिए- `int(1)/(tsqrt(t^(2)-1))dt` |
| Answer» Correct Answer - `sec^(-1)t+c` | |
| 276. |
`(1)/(sqrt(x^(2)+2x+2))` |
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Answer» `int(1)/(sqrt(x^(2)+2x+2))dx` `=int(1)/(sqrt((x+1)^(2)+(1)^(2)))dx` `=log|(x+1)+sqrt((x+1)^(2)+1)|+c` `=log|(x+1)+sqrt(x^(2)+2x+2)|+c` |
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| 277. |
`(sec^(2)x)/(sqrt(tan^(2)x+4))` |
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Answer» `int(sec^(2)x)/(sqrt(tan^(2)x+4))dx" माना "tanx=t` `=int(1)/(sqrt(t^(2)+2^(2)))dt" "rArr sec^(2)x dx = dt` `=log|t+sqrt(t^(2)+2^(2))|+c` `=log|tanx+sqrt(tan^(2)x+4)|+c` |
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| 278. |
`int_(0)^(1)(dx)/(1+x^(2))` |
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Answer» `int_(0)^(1)(1)/(1+x^(2))dx=[tan^(-1)x]_(0)^(1)` `" "=tan^(-1)1-tan^(-1)0=(pi)/(4)` |
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| 279. |
`int_(2)^(3)(dx)/(x^(2)-1)` |
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Answer» `int_(2)^(3)(1)/(x^(2)-1)dx=[(1)/(2)log|(x-1)/(x+1)|]_(2)^(3)` `=(1)/(2)log|(3-1)/(3+1)|-(1)/(2)log|(2-1)/(2+1)|` `=(1)/(2)log|(2)/(4)|-(1)/(2)log|(1)/(3)|` `=(1)/(2)[log(1)-log(2)-log(1)+log(3)]` `=(1)/(2)log((3)/(2))` |
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| 280. |
`(x)/(e^(x^(2)))` |
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Answer» `int(x)/(e^(x^(2)))dx" माना "-x^(2)=t` ltrgt `=intx.e^(-x^(2))dx=inte^(t).(dt)/(-2)" "rArr -2xdx=dt` `=-(1)/(2)e^(t)+c=-(1)/(2)e^(-x^(2))+c" "rArr xdx=(dt)/(-2)` |
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| 281. |
`(e^(tan^(-1))x)/(1+x^(2))` |
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Answer» `int(e^(tan^(-1))x)/(1+x^(2))" माना "tan^(-1)x=t` `=inte^(t)dt" "rArr (1)/(1+x^(2))dx=dt` `=e^(t)+c=e^(tan^(-1)x)+c` |
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| 282. |
`int(4e^(3x)+1)dx` |
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Answer» `int(4e^(3x)+1)dx=4inte^(3x)dx+int1.dx` `=(4)/(3)e^(3x)+x+c` |
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| 283. |
`sin2x-4e^(3x)` |
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Answer» `int(sin2x-4e^(3x))dx=intsin 2x dx-4 inte^(3x)dx` `=-(cos2x)/(2)-(4.e^(3x))/(3)+c` `=-(1)/(2)cos 2x-(4)/(3)e^(3x)+c` |
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| 284. |
`(1)/(cos^(2)x(1-tanx)^(2))` |
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Answer» `int(1)/(cos^(2)x(1-tanx)^(2))dx` माना `1-tanx=t` `rArr -3 sec^(2)x dx = dt` `rArr sec^(2)x dx = -dt` `=int(sec^(2)x)/((1-tanx)^(2))dx=int-(dt)/(t^(2))` `=(1)/(t)+c=(1)/(1-tanx)+c` |
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| 285. |
योगफल की सीमा के रूप में निम्नलिखित समाकलनों के मान ज्ञात कीजिए - `int_(0)^(2)e^(x)dx` |
| Answer» Correct Answer - `(e^(2)-1)` | |
| 286. |
`int(x)/(sqrt(x+4))dx` |
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Answer» `(x)/(sqrt(x+4))dx" माना "x+4=t^(2)` `=int((t^(2)-4).2tdt)/(sqrt(t^(2)))=2int(t^(2)-4)dt" "rArr dx=2tdt` `=2[(t^(3))/(3)-4t]+c` `=(2)/(3)(x+4)^(3//2)-8(x+4)^(1//2)+c` |
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| 287. |
`int(x^(3)-x^(2)+x-1)/(x-1)dx` |
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Answer» `int(x^(3)-x^(2)+x-1)/((x-1))dx=int((x^(2)+1)(x-1))/((x-1))dx` `=int(x^(2)+1)dx` `=intx^(2)dx+int1.dx =(x^(3))/(3)+x+c` |
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| 288. |
`int (1)/(x-sqrtx) dx` |
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Answer» `int(1)/(x-sqrtx)dx" माना "sqrtx-1=t` `=int(1)/(sqrtx(sqrtx-1))dx" "rArr (1)/(2sqrtx)dx=dt` `=int(2dt)/(t)" "rArr (1)/(sqrtx)dx=2dt` `=2log|t|+c=2log|sqrtx-1|+c` |
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| 289. |
`int(x^(3)+3x+4)/(sqrtx)dx` |
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Answer» `int(x^(3)+3x+4)/(sqrtx)` `=int((x^(3))/(sqrtx)+(3x)/(sqrtx)+(4)/(sqrtx))dx` `=intx^(5//2)dx+3intx^(1//2)dx+4 intx^(-1//2)dx` `=(x^(7//2))/(7//2)+3.(x^(3//2))/(3//2)+4.(x^(1//2))/(1//2)+c` `=(2)/(7)x^(7//2)+2x^(3//2)+8x^(1//2)+c` |
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| 290. |
`int(1-x)sqrtx dx` |
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Answer» `int(1-x)sqrtx dx = int (x^(1//2)-x^(3//2))dx` `=intx^(1//2)dx-intx^(3//2)dx` `=(x^(3//2))/(3//2)-(x^(5//2))/(5//2)+c` `=(2)/(3)x^(3//2)-(2)/(5)x^(5//2)+c` |
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| 291. |
`intsqrtx(3x^(2)+2x+3)dx` |
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Answer» `intsqrtx(3x^(2)+2x+3)dx` `=int(3x^(5//2)+2x^(3//2)+3x^(1//2))dx` `=3intx^(5//2)dx+2 intx^(3//2)dx+3 intx^(1//2)dx` `=3(x^(7//2))/(7//2)+2.(x^(5//2))/(5//2)+3.(x^(3//2))/(3//2)+c` `=(6)/(7)x^(7//2)+(4)/(5)x^(5//2)+2x^(3//2)+c` |
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| 292. |
`int(2x-3 cosx+e^(x))dx` |
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Answer» `int(2x-3 cosx+e^(x))dx` `=2intxdx-3 int cos x dx +inte^(x)dx` `=2(x^(2))/(2)-3sin x+e^(x)+c` `=x^(2)-3 sin x +e^(x)+c` |
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| 293. |
`int(2x^(2)-3 sin x+5sqrtx)dx` |
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Answer» `int(2x^(2)-3 sin x+5sqrtx)dx` `=2 intx^(2)dx-3int sin xdx+5intx^(1//2)dx` `=(2x^(3))/(3)-3(-cosx)+5(x^(3//2))/(3//2)+c` `=(2)/(3)x^(3)+3 cos x+(10)/(3)x^(3//2)+c` |
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| 294. |
`(cos sqrtx)/(sqrtx)` |
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Answer» `int(cos sqrtx)/(sqrtx)dx" माना "sqrtx=t` `=intcost-2dt" "rArr (1)/(2sqrtx)dx=dt` `=2 sin t+c" "rArr (1)/(sqrtx)dx=2dt` ltbgt `=2sin sqrtx+c` |
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| 295. |
`(e^(2x)-1)/(e^(2x)+1)` |
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Answer» `int(e^(2x-1)/(e^(2x)+1))dx" माना "e^(x)+e^(-x)=t` `=int(e^(x)-x^(-x))/(e^(x)+e^(-x))dx=int(1)/(t)dt" "rArr (e^(x)-e^(-x))dx=dt` `=log|t|+c=log|e^(x)+e^(-x)|+c` |
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| 296. |
`(e^(2x-e^(-2x)))/(e^(2x)+e^(-2x))` |
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Answer» `int(e^(2x-e^(-2x)))/(e^(2x)+e^(-2x))dx" माना "e^(2x)+e^(-2x)=-t` `" "rArr(2e^(2x)-2e^(-2x))dx=dt` `=int(1)/(t)(dt)/(2)" "rArr (e^(2x)-e^(-2x))dx=(dt)/(2)` `=(1)/(2)log|t|+c` `=(1)/(2)log|e^(2x)+e^(-2x)|+c` |
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| 297. |
`int(e^(msin^(-1)))/(sqrt(1-x^(2)))dx` का मान ज्ञात कीजिए । |
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Answer» माना `I=int(e^(msin^(-1)x))/(sqrt(1-x^(2)))dx` माना `m sin^(-1)x=t` `rArr" "(m)/(sqrt(1-x^(2)))=(dt)/(dx)` `rArr" "(dx)/(sqrt(1-x^(2)))=(dt)/(m)` `therefore" "I=inte^(t).(dt)/(x)=(1)/(m).e^(t)+c` `" "=(1)/(m).e^(msin^(-1)x)+c` |
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| 298. |
`int(dx)/(x+sqrtx)` का मान ज्ञात कीजिए । |
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Answer» माना `" "I=int(1)/(x+sqrtx)dx` `I=int(1)/(sqrtx(sqrtx+1))dx` माना `sqrtx+1=t` `rArr" "(1)/(2sqrtx)=(dt)/(dx)` `rArr" "(1)/(sqrtx)dx=2dt` `therefore" "I=int(1)/(t)(2dt)=2logt+c` `" "=2log(sqrtx+1)+c` |
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| 299. |
`int(x^(2)tan^(-1)x^(3))/(1+x^(6))` का मान ज्ञात कीजिए । |
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Answer» माना `I=int(x^(2).tan^(-1)x^(3))/(1+x^(6))dx` `" "=int tan^(-1)x^(3).((x^(2)dx)/(1+x^(6)))` माना `tan^(-1)x^(3)=t` `rArr" "(1)/(1+x^(6)).3x^(2)=(dt)/(dx)` `rArr" "(x^(2)dx)/(1+x^(6))=(dt)/(3)` ltBrgt `therefore" "I=intt.(dt)/(3)=(1)/(6)t^(2)+c` `" "=(1)/(6)(tan^(-1)x^(3))^(2)+c` |
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| 300. |
`int(x^(3))/(1+x^(8))` का मान ज्ञात कीजिए । |
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Answer» माना `I=int(x^(3))/(1+x^(8))dx` माना `" "x^(4)=t` `rArr" "4x^(3)dx=dt` `rArr" "x^(3)dx=(dt)/(4)` `therefore" "I=int(dt)/(4(1+t^(2)))=(1)/(4)tan^(-1)t+c` `" "=(1)/(4)tan^(-1)x^(4)+c` |
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