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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.
| 1. |
`(dy)/(dx)` ज्ञात कीजिए| `x=a(cost+logtan.(t)/(2)),y=a sin t` |
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Answer» `x=a(cost+log tan(t)/(2))` और `y=a sin t` `rArr" "(dx)/(dt)=a[-sin t+(sec^(2).(t)/(2))/(2 tan.(t)/(2))]" और "(dy)/(dx)=a cos t` `=a[-sint+(1)/(2 sin.(t)/(2)cos.(t)/(2))]` `=a(-sin t+(1)/(sint))` `=a((1-sin^(2)t)/(sint))=a(cos^(2)t)/(sint)` अब `(dy)/(dx)=(dy//dt)/(dx//dt)=(a cos t)/((a cos^(2)t)/(sint))=tant` |
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| 2. |
`(dy)/(dx)` ज्ञात कीजिए - `x=2 cost - cost2t, y= 2 sin t- sin 2x` |
| Answer» Correct Answer - `tan.(3t)/(2)` | |
| 3. |
`(dy)/(dx)` ज्ञात कीजिए - `x=(3at)/(1+t^(3)),y=(3at^(2))/(1+t^(3))` |
| Answer» `(t(2-t^(3)))/(1-2t^(3))` | |
| 4. |
निम्नलिखित से `(dy)/(dx)` ज्ञात कीजिए - `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0` |
| Answer» `(-(ax+hy+g))/((hx+by+f))` | |
| 5. |
फलन tan x निम्न में सतत है -A. RB. `R-{npi, n in Z}`C. `R-[(npi)/(2), n in Z]`D. इनमें से कोई नहीं । |
| Answer» Correct Answer - d | |
| 6. |
यदि फलन `f(x)={{:(2ax",",xge3),(3x+1",",xlt3):},x=3` पर सतत है, तो a का मान है -A. `(5)/(3)`B. `(5)/(2)`C. `5`D. इनमें से कोई नहीं । |
| Answer» Correct Answer - a | |
| 7. |
निम्नलिखित फलनों का x के सापेक्ष अवकन गुणांक ज्ञात कीजिए - `tan3x` |
| Answer» Correct Answer - `3sec^(2)3x` | |
| 8. |
`(dy)/(dx)` ज्ञात कीजिए - `x=a cost, y=b sint` |
| Answer» Correct Answer - `-(b)/(a)cot t` | |
| 9. |
यदि `x=at^(2)` और y = 2 at, तो `(dy)/(dx)` ज्ञात कीजिए । |
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Answer» `x=at^(2)` `rArr" "(dx)/(dt)=a.(d)/(dt)(t^(2))=a.2t=2at` और `y=2` at `rArr" "(dy)/(dt)=2a.(d)/(dt)(t)=2a.(1)=2a` अब `" "(dy)/(dx)=(dy//dt)/(dx//dt)=(2a)/("2at")=(1)/(t)` |
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| 10. |
यदि `x=a(t+sint)` और `y=a(1-cost),` तो `(dy)/(dx)` ज्ञात कीजिए । |
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Answer» `x=a(t+sint)` `rArr" "(dx)/(dt)=a(1+cost)` और `" "y=a(1-cost)` `rArr" "(dy)/(dt)=a(0+sint)=asint` `therefore" "(dy)/(dx)=(dy//dt)/(dx//dt)=(a sint)/(a(1+cost))` `=(2sin.(t)/(2)cos.(t)/(2))/(2cos^(2).(t)/(2))=tan.(t)/(2)` |
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| 11. |
यदि `x=a cos^(3)t` और `y=a sin^(3)t` तो `(dy)/(dx)` ज्ञात कीजिए । |
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Answer» `x=acos^(3)t` `(dx)/(dt)=a. (d)/(dt)(cost)^(2)=3a cos^(2)t.(d)/(dt)(cost)` `=-3acos^(2)t.sint` और `" "y=asin^(3)t` `rArr" "(dy)/(dt)=a.(d)/(dt)(sint)^(3)` `=a.3sin^(2)t.(d)/(dt)sint=3asin^(2)t cost` अब `" "(dy)/(dx)=(dy//dt)/(dx//dt)=(3a sin^(2)t cost)/(-3a cos^(2)t.sint)=-tant` |
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| 12. |
यदि `x=sin^(-1).(2t)/(1+t^(2))` और `y=tan^(-1).(2t)/(1-t^(2)),` तो `(dy)/(dx)` ज्ञात कीजिए । |
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Answer» `x=sin^(-1).(2t)/(1+t^(2))` `=sin^(-1).(2tan theta)/(1+tan^(2)theta)" माना "t=tan theta` `=sin^(-1)(sin2theta)` `=2theta=2 tan^(-1)t` `rArr" "(dx)/(dt)=(2)/(1+t^(2))` और `" "y=tan^(-1).(2t)/(1-t^(2))=2tan^(-1)t` `rArr" "(dy)/(dt)=(2)/(1+t^(2))` `therefore" "(dy)/(dx)=(dy//dt)/(dx//dt)=(2(1+t^(2)))/(2//(1+t^(2)))=1` |
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| 13. |
`f(x)=|x|+|x-1|` की `x=0` और `x=1`पर सांतत्य का परीक्षण। कीजिए । |
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Answer» `" "|x|={{:(x",",xge0),(-x",",xlt0):}` और `" "|x-1|={{:(x-1",",xge1),(-(x-1)",",xlt1):}` `therefore f(x)=|x|+|x-1|={{:(1-2x",",xlt0),(" "1",",0lexlt1),(2x-1",",xle1):}` x = 0 पर f (0) = 1 `R.H.L.=underset(xrarr0^(+))(lim)f(x)=underset(hrarr0)(lim)f(0+h)=underset(hrarr0)(lim)1=1` `L.H.L.=underset(xrarr0^(-))(lim)f(x)=underset(hrarr0)(lim)f(0-h)` `=underset(hrarr0)(lim){1-2(-4)}=1+0=1` `because R.H.L.=f(0)=L.H.L.` `therefore f(x),x=0` पर सतत है।| x = 1 पर `f(1)=2(1)-1=1` `R.H.L.=underset(xrarr1^(+))(lim)f(x)=underset(hrarr0)(lim)f(1+h)` `=underset(hrarr0)(lim)2(1+h)-1=2(1+0)-1=1` `L.H.L.=underset(xrarr1^(-))(lim)2(1+h)-1=2(1+0)-1=1` `L.H.L.=underset(xrarr1^(-))(lim)f(x)=underset(hrarr0)(lim)f(1-h)=underset(hrarr0)(lim)(1)=1` `because R.H.L. =f(1) = L.H.L.` `therefore f(x),x=1` पर सतत है | |
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| 14. |
दिखाइए कि `f(x)=|x|,x=0` पर सतत है। |
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Answer» `f(x)=|x|={{:(x",",xge0),(-x",",xlt0):}` अब `f(0)=0` `R.H.L.=underset(xrarr0^(+))(lim)f(x)=underset(hrarr0)(lim)f(0+h)` `=underset(hrarr0)(lim)h=0` `L.H.L.=underset(xrarr0^(-))(lim)f(x)=underset(hrarr0)(lim)f(0-h)` `=underset(hrarr0)(lim){-(0-h)}=0` `because " R.H.L."=f(0)="L.H.L."` `therefore" "f(x),x=0` पर सतत है। |
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| 15. |
x के सापेक्ष अवकलन कीजिए । `(e^(x))/(sinx)` |
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Answer» माना `y=(e^(x))/(sinx)` `rArr(dy)/(dx)=(d)/(dx)((e^(x))/(sinx))=(sinx(d)/(dx)e^(x)-e^(x)(d)/(dx)sinx)/(sin^(2)x)` `=(sinx.e^(x)-e^(x).cosx)/(sin^(2)x)` `=(e^(x)(sinx-cosx))/(sin^(2)x)` |
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| 16. |
x के सापेक्ष अवकलन कीजिए । `e^(sin^(-1)x)` |
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Answer» माना `y=e^(sin^(-1)x)` `rArr" "(dy)/(dx)=(d)/(dx)e^(sin^(-1)x)` `=e^(sin^(-1)x).(d)/(dx)sin^(-1)x` `=e^(sin^(-1)x).(1)/(sqrt(1-x^(2)))=(e^(sin^(-1)x))/(sqrt(1-x^(2)))` |
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| 17. |
x के सापेक्ष अवकलन कीजिए । `e^(x^(3))` |
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Answer» माना `y=e^(x^(3))` `rArr " "(dy)/(dx)=(d)/(dx)e^(x^(3))` `=e^(x^(3))(d)/(dx)x^(3)=3x^(2).e^(x^(3))` |
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| 18. |
x के सापेक्ष अवकलन कीजिए । `sin(tan^(-1)e^(-x))` |
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Answer» माना `y=sin(tan^(-1)e^(-x))` `rArr" "(dy)/(dx)=(d)/(dx)sin(tan^(-1)e^(-x))` `=cos(tan^(-1)e^(-x))(d)/(dx)(tan^(-1)e^(-x))` `=cos(tan^(-1)e^(-x)).(1)/(1+(e^(-x))^(2))(d)/(dx)e^(-x)` `=(cos(tan^(-1)e^(-x)))/(1+e^(-2x)).e^(-x).(d)/(dx)(-x)` `=-(e^(-x).cos(tan^(-1)e^(-x)))/(1+e^(-2x))` |
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| 19. |
x के सापेक्ष अवकलन कीजिए । `log(cos e^(x))` |
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Answer» माना `y=log(cos e^(x))` `(dy)/(dx)=(d)/(dx)log(cos e^(x))` `=(1)/(cos e^(x))(d)/(dx)cos e^(x)` `=(1)/(cos e^(x)).(-sine^(x)).(d)/(dx)e^(x)` `=-e^(x).tane^(x)` |
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| 20. |
x के सापेक्ष अवकलन कीजिए । `e^(x)+e^(x^(2))+…+e^(x^(5))` |
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Answer» माना `y=e^(x)+e^(x^(2))+e^(x^(3))+e^(x^(4))+e^(x^(5))` `rArr" "(dy)/(dx)=(d)/(dx)(e^(x)+e^(x^(2))+e^(x^(3))+e^(x^(4))+e^(x^(5)))` `=e^(x)+e^(x^(2))(d)/(dx)x^(2)+e^(x^(3))(d)/(dx)x^(3)+e^(x^(4))(d)/(dx)x^(4)+e^(x^(4))(d)/(dx)x^(5)` `=(e^(x)+2x.e^(x^(2))+3x^(2).e^(x^(3))+4x^(3).e^(x^(4))+5x^(4).e^(x^(5)))` |
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| 21. |
x के सापेक्ष अवकलन कीजिए । `sqrt(e^(sqrt(x))), x gt0` |
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Answer» माना `y =sqrt(e^(sqrt(x)))` `rArr" "(dy)/(dx)=(d)/(dx)sqrt(e^(sqrtx))=(1)/(2sqrt(e^(sqrtx))).(d)/(dx)e^(sqrtx)` `=(e^(sqrtx))/(2sqrt(e^(sqrtx))).(d)/(dx)sqrtx` `=(e^(sqrtx))/(2sqrt(e^(sqrtx))).(1)/(2sqrtx)=(e^(sqrtx))/(4sqrt(x.e^(sqrtx)))` |
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| 22. |
x के सापेक्ष अवकलन कीजिए । `cos(logx+e^(x))` |
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Answer» माना `y=cos(logx+e^(x))` `rArr" "(dy)/(dx)=(d)/(dx)cos(logx+e^(x))` `=-sin(logx+e^(x))(d)/(dx)(logx+e^(x))` `=-sin(logx+e^(x))((1)/(x)+e^(x))` `=(-sin(logx+e^(x))(1+x.e^(x)))/(x)` |
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| 23. |
`(dy)/(dx)` ज्ञात कीजिए : `t=cos^(-1)((2x)/(1+x^(2))),-1 lt x lt 1` |
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Answer» `y=cos^(-1)((2x)/(1+x^(2)))" "{:("माना "x=tan theta),(rArr theta= tan^(-1)x):}` `rArr" "y=cos^(-1)((2 tan theta)/(1+tan^(2)theta))` `=cos^(-1)((2 tan theta)/(sec^(2) theta))=cos^(-1)(2 sin theta cos theta)` `=cos^(-1)(sin 2 theta)=cos^(-1)cos((pi)/(2)-2 theta)` `=(pi)/(2)-2 theta=(pi)/(2)-2 tan^(-1)x` `rArr" "(dy)/(dx)=(d)/(dx)((pi)/(2)-2 tan^(-1)x)` `=0-(2xx1)/(1+x^(2))=-(2)/(1+x^(2))` |
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| 24. |
`(dy)/(dx)` ज्ञात कीजिए : `y=sin^(-1)(2xsqrt(1-x^(2))),-(1)/(sqrt2) lt x lt (1)/(sqrt2)` |
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Answer» `y=sin^(-1)(2x sqrt(1-x^(2)))" "{:("माना "x=tan theta),(rArr theta= tan^(-1)x):}` `rArr" "y=sin^(-1)(2 sin thetasqrt(1-sin^(2) theta))` `=sin^(-1)(2sin theta cos theta)` `=sin^(-1)(sin 2theta)=2 theta=2sin^(-1)x` `rArr" "(dy)/(dx)=2(d)/(dx)sin^(-1)x=(2)/(sqrt(1-x^(2)))` |
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| 25. |
`(dy)/(dx)` ज्ञात कीजिए : `y=sin^(-1)((1-x^(2))/(1+x^(2))),0 lt x lt 1` |
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Answer» `y=sin^(-1)((1-x^(2))/(1+x^(2)))" "{:("माना "x=tan theta),(rArr theta= tan^(-1)x):}` `rArr" "y=sin^(-1)((1-tan^(2) theta)/(1+tan^(2) theta))` `=sin^(-1)((cos^(2) theta-sin^(2)theta)/(cos^(2)theta+sin^(2) theta))` `=sin^(-1)(cos 2 theta)=sin^(-1)sin((pi)/(2)-2 theta)` `=(pi)/(2)-2 theta=(pi)/(2)-2 tan^(-1)x` `rArr" "(dy)/(dx)=(d)/(dx)((pi)/(2)-2 tan^(-1)x)` `=0-(2xx1)/(1+x^(2))=-(2)/(1+x^(2))` |
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| 26. |
x के सापेक्ष अवकलन कीजिए । `(cosx)/(logx),x gt0` |
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Answer» माना `y=(cosx)/(logx)` `(dy)/(dx)=(d)/(dx)((cosx)/(logx))` `=(logx.(d)/(dx)cosx-cosx(d)/(dx)logx)/((logx)^(2))` `=(-sinx logx-cosx.(1)/(x))/((logx)^(2))` `=-((x sin x logx+cosx))/(x(logx)^(2))` |
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| 27. |
`sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))` |
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Answer» माना `y=sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))` `rArr" "logy=logsqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))` `=(1)/(2)[log(x-1)+log(x-2)-log(x-3)-log(x-4)-log(x-5)]` दोनों पक्षों का x के सापेक्ष अवकलन करने पर `(1)/(y)(dy)/(dx)=(1)/(2){(1)/(x-1)+(1)/(x-2)-(1)/(x-3)-(1)/(x-4)-(1)/(x-5)}` `rArr" "(dy)/(dx)=(1)/(2)sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))` `{(1)/(x-1)+(1)/(x-2)-(1)/(x-3)-(1)/(x-4)-(1)/(x-5)}` |
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| 28. |
`(dy)/(dx)` ज्ञात कीजिए : `y=sec^(-1)((1)/(2x^(2)-1)), 0 lt x lt (1)/(sqrt2)` |
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Answer» `y=sec^(-1)((1)/(2x^(2)-1))" "{:("माना "x=tan theta),(rArr theta= tan^(-1)x):}` `rArr" "y=sec^(-1)((1)/(2cos^(2) theta-1))` `=sec^(-1)((1)/(cos 2 theta))=sec^(-1)(sec 2 theta)` `=2 theta= 2 cos^(-1)x` `rArr" "(dy)/(dx)=2(d)/(dx)cos^(-1)x=(2)/(sqrt(1-x^(2)))` |
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| 29. |
`(dy)/(dx)` ज्ञात कीजिए : `y=cos^(-1)((1-x^(2))/(1+x^(2))),0 lt x lt 1` |
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Answer» `y=cos^(-1)((1-x^(2))/(1+x^(2)))" "{:("माना "x=tan theta),(rArr theta= tan^(-1)x):}` `rArr" "y=cos^(-1)((1-tan^(2) theta)/(1+tan^(2) theta))` `=cos^(-1)((cos^(2)theta-sin^(2)theta)/(cos^(2) theta+sin^(2) theta))` `=cos^(-1)(cos 2 theta)=2 theta= 2 tan^(-1)x` `rArr" "(dy)/(dx)=2(d)/(dx) tan^(-1)x=(2)/(1+x^(2))` |
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| 30. |
`(logx)^(cosx)` |
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Answer» माना `y=(logx)^(cosx)` `rArr" "logy=log{(logx)}^(cosx)` `=cosx. log(logx)` दोनों पक्षों का x के सापेक्ष अवकलन करने पर `(1)/(y)(dy)/(dx)=cosx. (d)/(dx)log(logx)+log(logx)(d)/(dx)cosx` `rArr" "(dy)/(dx)=y{(cosx)/(logx).(d)/(dx)(logx)+log(logx)(-sinx)}` `rArr" "(dy)/(dx)=(logx)^(cosx){(cosx)/(x.logx)-sinx.log(logx)}` |
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| 31. |
`cos(a cos x+b sinx),` किन्ही अचर a तथा b के लिए |
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Answer» माना `y=cos(acosx+b sinx)` `(dy)/(dx)=(d)/(dx)cos(a cosx +b sinx)` `=-sin(a cos x+b sinx).(d)/(dx)(a cos x+b sinx)` `=-sin(a cosx+b sinx).(-a sin x+b cosx)` |
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| 32. |
`(dy)/(dx)` ज्ञात कीजिए : `y=tan^(-1)((3x-x^(3))/(1-3x^(2))),-(1)/(sqrt3)lt x lt (1)/(sqrt3)` |
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Answer» `y=tan^(-1)((3x-x^(3))/(1-3x^(2)))" "{:("माना "x=tan theta),(rArr theta= tan^(-1)x):}` `rArr" "y=tan^(-1)((3 tan theta- tan^(3) theta)/(1-3 tan^(2) theta))` `=tan^(-1)(tan 3 theta)=3 theta = 3 tan^(-1)x` `rArr" "(dy)/(dx)=3(d)/(dx) tan^(-1)x=(3)/(1+x^(2))` |
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| 33. |
`x^(x)-2^(sinx)` |
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Answer» माना `y=x^(x)-2^(sinx)` माना `u=x^(x)` तथा `v=2^(sinx)` `therefore" "y=u-v" "rArr" "(dy)/(dx)=(du)/(dx)-(dv)/(dx)" …(1)"` अब `u=x^(x)` `rArr" "logu = log(x^(x))=xlogx` दोनों पक्षों का x के सापेक्ष अवकलन लेने पर `(1)/(u)(du)/(dx)=x.(d)/(dx)logx+logx.(d)/(dx)x` `rArr" "(du)/(dx)=u[x.(1)/(x)+logx.1]=x^(x)(1+logx)` तथा `v=2^(sinx)` `rArr" "(dv)/(dx)=(d)/(dx)2^(sinx)` `=2^(sinx)log2.(d)/(dx)sinx` `=2^(sinx).log2.cosx` अब समीकरण (1 ) से `(dy)/(dx)=x^(x)(1+logx)-2^(sinx).log2.cosx` |
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| 34. |
`(dy)/(dx)` ज्ञात कीजिए : `y=sin^(-1)((2x)/(1+x^(2)))` |
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Answer» `y=sin^(-1)((2x)/(1+x^(2)))" "{:("माना "x=tan theta),(rArr theta= tan^(-1)x):}` `rArr" "y=sin^(-1)((2 tan theta)/(1+tan^(2) theta))` `=sin^(-1)((2 tan theta)/(sec^(2) theta))=sin^(-1)(2 sin theta cos theta)` `=sin^(-1)(sin 2 theta)= 2 theta = 2 tan^(-1)x` |
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| 35. |
निम्नलिखित से `(dy)/(dx)` ज्ञात कीजिए - `ysecx-y^(2)cosx+2x=0` |
| Answer» `(y sec x tan x +y^(2)sinx+2)/(2y cos x-secx)` | |
| 36. |
निम्नलिखित से `(dy)/(dx)` ज्ञात कीजिए - `5x^(2)+5y^(2)-7y+3x+2=0` |
| Answer» Correct Answer - `(10x+3)/(7-10y)` | |
| 37. |
`(x+3)^(2).(x+4)^(3).(x+5)^(4)` |
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Answer» माना `y=(x+3)^(2)(x+4)^(3)(x+5)^(4)` `rArr" "logy=log{(x+3)^(2)(x+4)^(3)(x+5)^(4)}` `=2log(x+3)+3log(x+4)+4log(x+5)` दोनों पक्षों का x के सापेक्ष अवकलन लेने पर `(1)/(y)(dy)/(dx)=(2)/(x+3)+(3)/(x+4)+(4)/(x+5)` `rArr" "(dy)/(dx)=y[(2)/(x+3)+(3)/(x+4)+(4)/(x+5)]` `rArr" "(dy)/(dx)=(x+3)^(2)(x+4)^(3)(x+5)^(4)[(2)/(x+3)+(3)/(x+4)+(4)/(x+5)]` |
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| 38. |
`(dy)/(dx)` ज्ञात कीजिए| `x^(2)+3x+2` |
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Answer» माना `y=x^(2)+3x+2` `rArr" "(dy)/(dx)=(d)/(dx)(x^(2)+3x+2)=2x+3` `rArr" "(d^(2)y)/(dx^(2))=(d)/(dx)(2x+3)=2` |
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| 39. |
`(dy)/(dx)` ज्ञात कीजिए : `sin^(2)x+cos^(2)y=1` |
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Answer» `sin^(2)x+cos^(2)y=1` दोनों पक्षों का x के सापेक्ष अवकलन करने पर `2sin x cos x+2cosy(-siny)(dy)/(dx)=0` `rArr " "sin 2x-sin 2y(dy)/(dx)=0 rArr (dy)/(dx)=(sin 2x)/(sin 2y)` |
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| 40. |
निम्नलिखित से `(dy)/(dx)` ज्ञात कीजिए - `sin(xy)+(x)/(y)=x^(2)-y` |
| Answer» `(2xy^(2)-y-y^(3)cos(xy))/(xy^(2)cos(xy)-x+y^(2))` | |
| 41. |
निम्नलिखित से `(dy)/(dx)` ज्ञात कीजिए - `xsin2y=ycos2x` |
| Answer» `(2y sin 2x+sin 2y)/(cos 2x-2xcos 2y)` | |
| 42. |
`x^(x cosx)+(x^(2)+1)/(x^(2)-1)` |
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Answer» माना `y=x^(x cos x)+(x^(2)+1)/(x^(2)-1)` माना `u=x^(x cosx)" तथा "v=(x^(2)+1)/(x^(2)-1)` `rArr" "(dy)/(dx)=(du)/(dx)+(dv)/(dx)" …(1)"` अब `u=x^(xcosx)` `rArr" "logu=log(x^(x cosx))=x cos x. logx` `rArr" "(1)/(u)(du)/(dx)=x cosx.(d)/(dx)logx+x logx.(d)/(dx)cosx+cosx.logx.(d)/(dx)x` `rArr" "(du)/(dx)=u(x cosx.(1)/(x)-x log x.sinx+cosx.logx)` `=x^(x cosx)(cosx-x log x sin x+cos x logx)` तथा `v=(x^(2)+1)/(x^(2)-1)` `rArr (dy)/(dx)=((x^(2)-1)(d)/(dx)(x^(2)+1)-(x^(2)+1)(d)/(dx)(x^(2)-1))/((x^(2)-1)^(2))` `=((x^(2)-1).2x-(x^(2)+1).2x)/((x^(2)-1)^(2))` `=-(4x)/((x^(2)-1)^(2))` समीकरण (1 ) से `(dy)/(dx)=x^(x cosx)(cosx- x log x sinx+cos x logx)-(4x)/((x^(2)-1)^(2))` |
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| 43. |
`(x+(1)/(x))^(x)+x^((1+(1)/(x)))` |
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Answer» माना `y=(x+(1)/(x))^(x)+x^((1+(1)/(x)))` माना `u=(x+(1)/(x))^(x)` तथा `v=x^(1+(1)/(x))` `therefore" "y=u+v" "rArr" "(dy)/(dx)=(du)/(dx)+(dv)/(dx)" …(1)"` अब `u=(x+(1)/(x))^(x)` `rArr" "logu=log(x+(1)/(x))^(x)=x log (x+(1)/(x))` `rArr" "(1)/(u)(du)/(dx)=x.(d)/(dx)log(x+(1)/(x))+log(x+(1)/(x))(d)/(dx)x` `rArr" "(du)/(dx)=u[(x)/(x+(1)/(x))(1-(1)/(x^(2)))+log(x+(1)/(x))]` `=(x+(1)/(x))^(x)[(x^(2)-1)/(x^(2)+1)+log(x+(1)/(x))]` तथा `v=x(1+(1))/(x)` `rArr" "logv=log{x^((1+(1)/(x)))}=(1+(1)/(x))logx` `rArr" "(1)/(v)(dv)/(dx)=(1+(1)/(x))(d)/(dx)logx+logx(d)/(dx)(1+(1)/(x))` `rArr" "(dv)/(dx)=v[(1+(1)/(x)).(1)/(x)+logx(-(1)/(x^(2)))]` `rArr" "(dv)/(dx)=x^((1+(1)/(x))).(1)/(x^(2))[x+1-logx]` `therefore` समीकरण (1 ) से `(dy)/(dx)=(x+(1)/(x))^(x)[(x^(2)-1)/(x^(2)+1)+log(x+(1)/(x))]+x((1+(1)/(x)).(1)/(x^(2)))[x+1-logx]` |
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| 44. |
`x(cosx)^(x)+(x sinx)^((1)/(x))` |
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Answer» माना `y=(x cos x)^(x)+(x sin x)^(1//x)` माना `u=(x cos x)^(x)` तथा `v=(x sin x)^(1//x)` `therefore" "y=u+v` `rArr" "(dy)/(dx)=(du)/(dx)+(dv)/(dx)` अब `u=(x cosx)^(x)` `rArr" "logu=log(xcos x)^(x)=x (logx +log cosx)` `rArr" "(1)/(u)(du)/(dx)=x.(d)/(dx)(logx+logcos x)+(log x+log cosx)(d)/(dx)x` `rArr" "(du)/(dx)=u[x((1)/(x)-(sinx)/(cosx))+(logx+logcos x).1]` `rArr" "(du)/(dx)=(x cos x)^(x)[1- x tanx +logx+log cos x]` तथा `" "v=(x sin x)^(1//x)` `rArr" "logv=log(x sin x)^(1//x)=(1)/(x)(logx+logsinx)` `rArr" "(1)/(v)(dv)/(dx)=(1)/(x)(d)/(dx)(logx+logsin x)+(logx+log sinx)(d)/(dx)((1)/(x))` `rArr" "(dv)/(dx)v[(1)/(x)((1)/(x)+(cosx)/(sinx))-(1)/(x^(2))(logx+log sin x)]` `rArr" "(dv)/(dx)=(x sinx)^(1//x)[(1+x cot x-log (x sinx))/(x^(2))]` समीकरण (1 ) से `(dy)/(dx)=(x cos x)^(x)[1-k x tanx +log x+log cosx]+(x sin x)^(1/x)[1-x tan x+log(x sin x)]` |
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| 45. |
`(dy)/(dx)` ज्ञात कीजिए : `sin^(2)y+cos xy=k` |
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Answer» `sin^(2)y +cos xy=k` दोनों पक्षों का x के सापेक्ष अवकलन करने पर `2siny cosy (dy)/(dx)+(-sin xy)(d)/(dx)(xy)=0` `rArr 2sin 2y (dy)/(dx)-sin xy(x(dy)/(dx)+y.1)=0` `rArr" "(dy)/(dx)(sin 2y-x sin xy)=y sin xy` `rArr" "(dy)/(dx)=(y sin xy)/(sin 2y - x sin xy)` |
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| 46. |
`(dy)/(dx)` ज्ञात कीजिए : `x^(3)+x^(2)y+xy^(2)+y^(3)=81` |
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Answer» `x^(3)+x^(2)y+xy^(2)+y^(3)=81` दोनों पक्षों का x के सापेक्ष अवकलन करने पर `3x^(2)+(x^(2)(dy)/(dx)+2xy)+(2xy(dy)/(dx)+y^(2))+3y^(2)(dy)/(dx)=0` `rArr (x^(2)+2xy+3y^(2))(dy)/(dx)=-(3x^(2)+2xy+y^(2))` `rArr" "(dy)/(dx)=-((3x^(2)+2xy+y^(2)))/((x^(2)+2xy+3y^(2)))` |
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| 47. |
निम्नलिखित से `(dy)/(dx)` ज्ञात कीजिए - `x^(2)+y^(2)=log(xy)` |
| Answer» `(y(1-2x^(2)))/(x(2y^(2)-1))` | |
| 48. |
`(logx)^(x)+x^(logx)` |
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Answer» `y=x^(logx)+(logx)^(x)" (माना )"` `=u+v` `rArr" "(dy)/(dx)=(du)/(dx)+(dv)/(dx)" …(1)"` अब `u=x^(logx)` `rArr" "logu=log(x^(logx))=logx.logx=(logx)^(2)` `rArr" "(1)/(u)(du)/(dx)=(2logx)/(x)` `rArr" "(du)/(dx)=(2ulogx)/(x)=(2logx)/(x).x^(logx)` और `v=(logx)^(x)` `rArr" "logv=log(logx)^(x)=xlog(logx)` `rArr" "(1)/(v)(dv)/(dx)=(x)/(xlogx)+log(logx)` `rArr" "(dv)/(x)=v[(1)/(logx)+log(logx)]` `=(logx)^(x)[(1)/(logx)+log(logx)]` समीकरण (1 ) से `(dy)/(dx)=x^(logx).(2logx)+(logx)^(x)[(1)/(logx)+log(logx)].` |
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| 49. |
`x^(sinx)+(sinx)^(cosx)` |
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Answer» माना `y=x^(sinx)+(sinx)^(cosx)` माना `u=x^(sinx)" तथा "v=(sinx)^(cosx)` `therefore" "y=u+v` `rArr" "(dy)/(dx)=(du)/(dx)+(dv)/(dx)" …(1)"` अब `u=x^(sinx)` `rArr" "logu=log(x^(sinx))=sinx. logx` `rArr" "(1)/(u)(du)/(dx)=sinx.(d)/(dx)logx+logx.(d)/(dx)sinx` `rArr" "(du)/(dx)=u[(sinx)/(x)+logx.cosx]` `rArr" "(du)/(dx)=x^(sinx)[(sinx)/(x)+logx.cosx]` तथा `v=(sinx)^(cosx)` `rArr" "logv=log(Sinx)^(cosx` `=cosx.log(sinx)` `rArr" "(1)/(y)(dv)/(dx)=cosx (d)/(dx)log(sinx)+log(sinx)(d)/(dx)(cosx)` `rArr" "(dv)/(dx)=v[cosx.(cosx)/(sinx)+log(sinx).(-sinx)]` `rArr" "(dv)/(dx)=(sinx)^(cosx)[cos. cot x-sinxlog (sinx)]` समीकरण (1 ) से `(dy)/(dx)=x^(sinx)[(sinx)/(x)+logx.cosx]+(sinx)^(cosx)[cosx cotx-sinxlog(sinx)]` |
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| 50. |
`(dy)/(dx)` ज्ञात कीजिए| `x^(20)` |
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Answer» माना `y=x^(20)` `rArr" "(dy)/(dx)=(d)/(dx)x^(20)=20x^(19)` `rArr" "(d^(2)y)/(dx^(2))=(d)/(dx)(20x^(19))` `" "=20xx19x^(18)=380x^(18)` |
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