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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.
| 151. |
निम्नलिखित फलन x के सापेक्ष अवलंकन गुणांक ज्ञात कीजिएः x log x |
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Answer» माना `y=log x` `therefore (dy)/(dx)=(d)/(dx)(x log x )` `=x(d)/(dx)log x+log x(d)/(dx)x` `=x.(1)/(x)+log x.1=1+logx` |
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| 152. |
निम्नलिखित फलन x के सापेक्ष अवलंकन गुणांक ज्ञात कीजिएः `7e^(x)sin x+10^(x)` |
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Answer» माना `7e^(x)sin x+10^(x)` `therefore (dy)/(dx)=7(d)/(dx)(e^(x)sin x)+(d)/(dx)10^(x)` `=7(e^(x)cos x+sin xe^(x))+10^(x)log_(e)10` `=7(sin x+cos x)+10^(x)log_e10` |
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| 153. |
निम्नलिखित फलन x के सापेक्ष अवलंकन गुणांक ज्ञात कीजिएः `5" "sqrt(x^(2))+9e^(x)+11log_(a)x-7log_(e)x` |
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Answer» माना `5" "3sqrt(x^(2))+9e^(x)+11log_(a)x-7log_(e)x` या `y=5x^(2//3)+9e^(x)+11log_(a)x-7log_ex` `therefore (dy)/(dx)=5(d)/(dx)x^(2//3)+9(d)/(dx)e^(x)+11log_(a)e(d)/(dx)log_ex-7 (d)/(dx)log_ex` `=(10)/(3)x^(-1//3)+9e^(x)+(11)/(x)log_(a)e-(7)/(x)` |
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| 154. |
निम्नलिखित फलन x के सापेक्ष अवलंकन गुणांक ज्ञात कीजिएः `x^(2)e^(x)` |
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Answer» माना `y=x^(2)e^(x)` `therefore (dy)/(dx)=(d)/(dx)(x^(2)e^(x))` `=x^(2)(d)/(dx)e^(x)+e^(x)(d)/(dx)x^(2)` `=x^(2)e^(x)+2xe^(x)` `=xe^(x)(x+2)` |
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| 155. |
फलन `tan x^(2)` का प्रथम सिद्धांत से अवलंकन गुणांक ज्ञात कीजिएः |
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Answer» `y=tan x^(2)` ........(i) `therefore y=deltay=tan (x+deltax)^(2)` ........(ii) `therefore y+deltay-y=tan(x+deltax)^2-tanx^(2)` `=(sin (x+deltax)^(2))/(cos(x+deltax)^(2))-(sin x^(2))/(cos x^(2))` `=(sin(x+deltax)^(2)cosx^(2)-cos(x+deltax)^(2)sinx^(2))/(cos (x+deltax)^(2).cosx^2)` `=sin [(x+deltax)^(2)-x^(2)]/(cos (x+deltax)^2cosx^(2))` `=(sin(x^2+deltax^(2)+2xdeltax-x^(2)))/(cos((x+deltax)^(2)cos))` दोनों पक्षों को `deltax` से भाग देकर `underset(deltax to 0)lim` लेने पर `underset(deltax to 0)lim (deltay)/(deltax)=(dy)/(dx)` `underset(deltax to 0)lim (sin deltax(2x+deltax))/(deltax(2x+deltax)).underset(deltax to 0)lim (2x+deltax)/(cos (x+deltax)^(2).cosx^(2))` `=1.underset(deltax to 0)lim (2x+deltax)/(cos (x+deltax)^(2).cosx^(2))` `=(2x+0)/(cos (x+0)^(2) cos x^(2))xx1` `=(2x)/(cos^(2)x^(2))=2x sec^(2)x^2` `therefore (d)/(dx)(tan x^(2))=2x sec^(2) x^(2)` |
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| 156. |
f(x)=sin 2x का मूल नियमो से प्रथम अवलंकन गुणांक ज्ञात कीजिएः |
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Answer» y=sin 2x ....(i) `therefore y+deltay=sin 2(x+deltax)` ....(i) `therefore y+deltay-y=sin 2(x+deltax)-sin 2x` `Rightarrow deltay=2cos (2x+delx)sin delta` `[therefore sin C-sin D =2cos"" (C+D)/(2)sin ""(C-D)/(2)]` दोनों पक्षों को `deltax` से भाग देकर `underset(deltax to 0)lim` लेने पर `underset(deltax to 0)lim (deltay)/(deltax)=(dy)/(dx)` `=underset(deltax to 0)lim 2cos(2x+deltax)(sin deltax)/(deltax)` `=2cos (2x+0)xx 1" "(therefore underset(theta to 0)lim (sin theta)/(theta)=1)` =2cos 2x `therefore (d)/(dx) sin 2x=2cos 2x` |
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| 157. |
निम्नलिखित फलन का अवलंकन गुणांक ज्ञात कीजिएः x |
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Answer» माना `y=x` `(dy)/(dx)=(d)/(dx)=x` या `(dy)/(dx)=1x^(1-1)=1.x^(0)=1.1" "(therefore x^(0)=1)` `therefore (d)/(dx)(x)=1` |
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| 158. |
फलन `sqrtx` का x के सापेक्ष प्रथम सिद्धांत से अवकलन गुणांक ज्ञात कीजिएः। |
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Answer» माना `y=sqrtx=(x)^(1//2)` `therefore y+deltay=(x+deltax)^(1//2)` अंत: `y+deltay-y=(x+deltax)^(1//2)-(x)^(1//2)` `Rightarrow deltay=x^(1//2) [1+(deltax)/(x)]^(-1//2)-x^(1//2)=x^(1//2)[(1+(deltax)/(x))^(1//2)-1]` `Rightarrow deltay=x^(1//2) [1+(1)/2((deltax)/(x))]+((1)/(2)((1)/(2)-1))/(2!)((deltax)/(x))^(1//2)+....oo-1]` `[therefore` द्विपद प्रमेय से, `(1+x)^(n)=1+nx+(n(n-1))/(2!)x^(2)+....]` `Rightarrow deltay=x^(1//2)((delta)/(x))[1+(((1)/(2)-1))/(2!)((deltax)/(x))+....oo]` `=(1)/(2sqrtx)deltax[1+(((1)/(2)-1))/(2!)((deltax)/(x))+......]` दोनों पक्षों को `deltax` से भाग लेकर `underset(x to 0)lim` लेने पर `Rightarrow underset(deltax to o)lim (deltay)/(deltax)=(dy)/(dx)=underset(deltax to 0)lim (1)/(2sqrt2) (deltax)/(deltax) [1+((1)/(2)-1))/(2!) ((deltax)/(x))+....]` `=(1)/(2sqrtx)[1+0+0+.....]` `therefore (d)/(dx) (x^(1//2))=(1)/(2sqrtx)` |
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| 159. |
`sqrt((3x-4))` का x के सापेक्ष प्रथम सिद्धांत से अवकल गुणांक ज्ञात कीजिएः |
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Answer» `y=sqrt(3x-4)=(3x-4)^(1//2)` .......(i) `Rightarrow y+deltay=[3(x+deltay)-4]^(1//2)` `y+deltay=[(3x-4)+3deltax]^(1//2)........(ii)` `y+deltay-y=[(3x-4)+3deltax]^(1//2)-(3x-4)^(1//2)` ` =(3x-4)^(1//2) [(1+(3deltax)/(3x-4))^(1//2)]-(3x-4)^(1//2)` ` Rightarrow deltay=(3x-4)^(1//2) [(1+(3deltax)/(3x-4))^(1//2)-1]` ` Rightarrow deltay=(3x-4)^(1//2) [1+(1)/(2)((3deltax)/(3x-4))+((1)/(2)((1)/(2)-1))/(2!)((3deltax)/(3x-4))^2+.....oo-1]` ` =(3x-4)^(1//2) [(1)/(2)((3deltax)/(3x-4))+((1)/(2)((1)/(2)-1))/(2!)((3deltax)/(3x-4))^2+.....oo]` ` =(3x-4)^(1//2) .(1)/(2)(3deltax)/(3x-4)[1+((1)/(2)((1)/(2)-1))/(2!)((3deltax)/(3x-4))+.....oo]` ` =(3deltax)/(2(3x-4)^(1//4))[1+((1)/(2)((1)/(2)-1))/(2!)((3deltax)/(3x-4))+.....oo]` दोनों पक्षों `deltax` को से भाग देकर `underset(deltax to 0)` लेने पर, `underset(deltax to 0)lim (deltay)/(deltax)=(dy)/(dx)=underset(deltax to 0)lim (3)/(2(3x-4)^(1//2))=(deltax)/(deltax)[1+(((1)/(2)-1))/(2!)(3deltax)/((3x-4))+......oo]` `=3/(2(3x-4)^(1//2)) [1+0+0+.....]` `=(3)/(2sqrt((3x-4)))` `therefore (d)/(dx)(3x-4)^(1//2)=(3)/(2sqrt((3x-4)))` |
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| 160. |
निम्नलिखित फलन का अवलंकन गुणांक ज्ञात कीजिएः `(lx^(4)+mx^(3)+nx^(2)+px+q)/(sqrtx)` |
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Answer» माना `y=(lx^(4)+mx^(3)+nx^(2)+px+q)/(sqrtx)` `=(lx^(4))/(x^(1//2))+(mx^3)/(x^(1//2))+(nx^(2))/(x^(1//2))+(px)/(x^(1//2))+(q)/(x^(1//2))` `=lx^(7//2)+mx^(5//2)+nx^(3//2)+px^(1//2)+qx^(-1//2)` `therefore (dy)/(dx)=l d/dx(x^(7//2))+m d/dx(x^(5//2))+nd/dx(x^(3//2))+p(d)/(dx))^(1//2)+q(-(1)/(2))x^(-3//2)` `=(7)/(2)lx^(5//2)+(5)/(2)mx^(3//2)+(3)/(2)nx^(1//2)+(p)/(2sqrtx)-(q)/(2xsqrtx)` |
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| 161. |
निम्नलिखित फलन का अवलंकन गुणांक ज्ञात कीजिएः `6x^(3)+9x^(2)-7x+15` |
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Answer» माना `6x^(3)+9x^(2)-7x+15` `therefore dy/dx=d/dx(6x^(3)+9x^(2)-7x+15)` `=6(d)/(dx)x^(3)+9(d)/(dx)x^2-7(d)/(dx)x+(d)/(dx)15` `=6(3x^(3-1))+9(2x^(2-1))-7(1x^(1-1))+0` `=18x^2+18x-7` |
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| 162. |
निम्नलिखित फलन का अवलंकन गुणांक ज्ञात कीजिएः `x^5-9x^(4)-15x^3+6x^(2)-17x+9` |
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Answer» माना `y=x^5-9x^(4)-15x^3+6x^(2)-17x+9` `therefore (dy)/(dx)=(d)/(dx)(x^(5)-9x^(4)-15x^(3)+6x^(2)-17x+9)` `=(d)/(dx)x^5-9(d)/(dx)x^(4)-15(d)/(dx)x^(3)+6(d)/(dx)x^(2)-17(d)/(dx)x+(d)/(dx)9` `=5x^(5-1)-9(4x^(4-1))-15(3x^(3-1))+6(2x^(2-1))-17(d)/(dx)x+(d)/(dx)9` `=5x^(5-1)-9(4x^(4-1))-15(3x^(3-1))+6(2x^(2-1))-17(1x^(1-1))+0` `=5x^(4)-36x^(3)-35x^(2)+12x-17` |
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| 163. |
यदि (If `y=logsqrt((1+cos^(2)x)/(1-e^(2x)))`,(find) `(dy)/(dx)` निकालें। |
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Answer» `y=logsqrt((1+cos^(2)x)/(1-e^(2x)))` `=(1)/(2)log((1+cos^(2)x)/(1-e^(2x)))=(1)/(2)[log(1+cos^(2)x)-log(1-e^(2x))]` `rArr(dy)/(dx)=(1)/(2)[(1)/(1+cos^(2)x)*(d)/(dx)(1+cos^(2)x)-(1)/(1-e^(2x))*(d)/(dx)(1-e^(2x))]` `=(1)/(2)[(1)/(1+cos^(2)x)*(-2cosxsinx)-(1)/(1-e^(2x))(-2e^(2x))]` `=(-sinxcosx)/(1+cos^(2)x)+(e^(2x))/(1-e^(2x))`. |
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| 164. |
`"tan"^(-1)(x)/(sqrt(1-x^(2)))` का `cos^(-1)(2x^(2)-1)` के सापेक्ष अवकल गुणांक निकालें। |
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Answer» माना कि `u="tan"^(-1)(x)/(sqrt(1-x^(2))),v=cos^(-1)(2x^(2)-1)` हमें `(du)/(dv)` निकालना है, `x=costheta`, रखें , तो `u="tan"^(-1)(x)/(sqrt(1-x^(2)))=tan^(-1)((costheta)/(sintheta))=tan^(-1)(cottheta)` `=tan^(-1)tan((pi)/(2)-theta)=(pi)/(2)-theta therefore(du)/( theta)=-1` तथा `v=cos^(-1)(2x^(2)-1)=cos^(-1)(2cos^(2)theta^(-1))` `=cos^(-1)(cos2theta)=2thetatherefore(dv)/(d theta)=2` अब `(du)/(dv)=(du//d theta)/(dv//d theta)=-(1)/(2)` |
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| 165. |
यदि (If) `y=sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))` , (find) `(dy)/(dx)` निकालें। |
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Answer» दिया है, `y=sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))` … (1) `thereforelogy=(1)/(2)log[((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)]` x के सापेक्ष दोनों तरफ अवकलित (differentiate) करने पर हमें मिलता है, `(1)/(y)(dy)/(dx)=(1)/(2)[(1)/(x-1)+(1)/(x-2)-(1)/(x-3)-(1)/(x-4)-(1)/(x-5)]` `therefore(dy)/(dx)=(1)/(2)y[(1)/(x-1)+(1)/(x-2)-(1)/(x-3)-(1)/(x-4)-(1)/(x-5)]` `=(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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| 166. |
यदि (If) y=tan (x+y), (find) `(dy)/(dx)` निकालें। |
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Answer» दिया है ,`y=tan(x+y)` x के सापेक्ष अवकलित करने पर हमें मिलता है, `(dy)/(dx)=(d)/(d(x+y))tan(x+y)*(d)/(dx)(x+y)` या `(dy)/(dx)=sec^(2)(x+y)(1+(dy)/(dx))` या `(dy)/(dx)=sec^(2)(x+y)+sec^(2)(x+y)(dy)/(dx)` या `[1-sec^(2)(x+y)](dy)/(dx)=sec^(2)(x+y)` या `(dy)/(dx)=(sec^(2)(x+y))/(1-sec^(2)(x+y))=(1+y^(2))/(1-(1+y^(2)))=(1+y^(2))/(y^(2))[because(1)` से, y =tan (x+y)] |
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| 167. |
यदि (If) `y=tan^(-1)sqrt((1+sinx)/(1-sinx)),` (find) `(dy)/(dx)` ज्ञात करें। |
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Answer» `y=tan^(-1)sqrt((1+sinx)/(1-sinx))=tan^(-1)sqrt(("cos"^(2)(x)/(2)+"sin"^(2)(x)/(2)+2" sin"(x)/(2) "cos"(x)/(2))/("cos"^(2)(x)/(2)+"sin"^(2)(x)/(2)-2"sin"(x)/(2)"cos"(x)/(2)))` `=tan^(-1)sqrt((("cos"(x)/(2)+"sin"(x)/(2))^(2))/(("cos"(x)/(2)-"sin"(x)/(2))^(2)))` `=tan^(-1)(("cos"(x)/(2)+"sin"(x)/(2))/("cos"(x)/(2)-"sin"(x)/(2)))=tan^(-1)((1+"tan"(x)/(2))/(1-"tan"(x)/(2)))` `thereforey=tan^(-1)[tan((pi)/(4)+(x)/(2))]=(pi)/(4)+(x)/(2)rArr(dy)/(dx)=(1)/(2)` |
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| 168. |
यदि (If) `y=logtan((pi)/(4)+(x)/(2))`, दिखाएँ कि (show that) `(dy)/(dx)-secx=0`. |
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Answer» दिया है, `y=logtan((pi)/(4)+(x)/(2))` `rArr(dy)/(dx)=(1)/(tan((pi)/(4)+(x)/(2)))*(d)/(dx)[tan((pi)/(4)+(x)/(2))]` `=cot((pi)/(4)+(x)/(2))*sec^(2)((pi)/(4)+(x)/(2))*(1)/(2)=(cos((pi)/(4)+(x)/(2)))/(sin((pi)/(4)+(x)/(2)))*(1)/(2)*(1)/(cos^(2)((pi)/(4)+(x)/(2)))` `=(1)/(2sin((pi)/(4)+(x)/(2))cos((pi)/(4)+(x)/(2)))=(1)/(sin2((pi)/(4)+(x)/(2)))=(1)/(sin((pi)/(2)+x))` `=(1)/(cosx)=secx` अतः `(dy)/(dx)-secx=0`. |
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| 169. |
यदि `sin(xy)+x/y=x^(2)-y` हो,तो `(dy)/(dx)` ज्ञात कीजिए । |
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Answer» `sin(xy)+x/y=x^(2)-y` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `d/(dx)sin(xy)+d/(dx)(x/y)=d/(dx)(x^(2))-(dy)/(dx)` `rArrcos(xy)d/(dx)(xy)+(y-x(dy)/(dx))/(y^(2))=2x-(dy)/(dx)` `rArrcos(xy)(x(dy)/(dx)+y)+1/y-x/(y^(2))(dy)/(dx)=2x-(dy)/(dx)` `rArrxcos(xy)(dy)/(dx)+ycos(xy)+1/y-x/(y^(2))(dy)/(dx)=2x-(dy)/(dx)` `rArr[xcos(xy)-x/(y^(2))+1](dy)/(dx)` `=2x-1/y-ycos(xy)` `rArr[xy^(2)cos(xy)-x+y^(2)](dy)/(dx)` `=[2x-1/y-ycos(xy)]y^(2)` `rArr[xy^(2)cos(xy)-x+y^(2)](dy)/(dx)` `=[2xy^(2)-y-y^(3)cos(xy)]` ltbr gt`rArr(dy)/(dx)=(2xy^(2)-y-y^(3)cos(xy))/(xy^(2)cos(xy)-x+y^(2))` |
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| 170. |
यदि (If) `y*sqrt(x^(2)+1)=log[sqrt(x^(2)+1)-x]` दिखाएँ कि `(x^(2)+1)(dy)/(dx)+xy+1=0` |
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Answer» यहाँ ,`ysqrt(x^(2)+1)=log[sqrt(x^(2)+1)-x]` दोनों पक्षों को x सापेक्ष अवकलित करने पर , हमें मिलता है `(dy)/(dx)*sqrt(x^(2)+1)+y*(1)/(2)(x^(2)+1)^(-1/(2))*2x=(1)/(sqrt(x^(2)+1)-x)(d)/(dx)(sqrt(x^(2)+1)-x)` `rArrsqrt(x^(2)+1)(dy)/(dx)+(xy)/(sqrt(x^(2)+1))=(1)/(sqrt(x^(2)+1)-x)((x)/(sqrt(x^(2)+1))-1)` `=(1)/(sqrt(x^(2)+1)-x)(x-sqrt(x^(2)+1))/(sqrt(x^(2)+1))=(-1)/(sqrt(x^(2)+1))` `rArr(x^(2)+1)(dy)/(dx)+xy=-1rArr(x^(2)+1)(dy)/(dx)+xy+1=0`. |
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| 171. |
यदि (If) `x=asin2t(1+cos2t)` तथा (and) `y=bcos2t(1-cos2t)`, दिखाएँ कि (show that ) `((dy)/(dx))_(t=(pi)/(4))=(b)/(a)` |
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Answer» यहाँ `x=asin2t(1+cos2t)` `rArr(dx)/(dt)=a.[2cos2t(1+cos2t)+sin2t(-2sin2t)]` `=2a[cos2t+cos^(2)2t-sin^(2)2t]` `=2a[cos2t+cos4t]` साथ ही `y=b cos2t(1-cos2t)` `rArr(dy)/(dt)=b[-2sin2t(1-cos2t)+cos2t.2sin2t]` `=2b[-sin2t+2sin2tcos2t]` `=2b[-sin2t+sin4t]` `therefore(dy)/(dx)=(dy//dt)/(dx//dt)=(2b[-sin2t+sin4t])/(2a[cos2t+cos4t])` `rArr((dy)/(dx))_(t=(pi)/(4))=(b)/(a)[(-"sin"(pi)/(2)+sinpi)/("cos"(pi)/(2)+"cos"pi)]=(b)/(a)[(-1+0)/(0-1)]=(b)/(a)`. |
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| 172. |
यदि `y=sin^(-1)((1)/(sqrt(1+x^(2))))+tan^(-1)((sqrt(1+x^(2))-1)/(x))`, `(dy)/(dx)` निकालें। |
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Answer» Correct Answer - `-(1)/(2(1+x^(2)))` `x=tantheta` रखने पर हमें मिलता है, `y=sin^(-1){(1)/(sqrt(1+tan^(2)theta))}+tan^(-1){(sqrt(1+tan^(2)theta)-1)/(tantheta)}` `=sin^(-1)(costheta)+"tan"^(-1)(1-costheta)/(sintheta)` `=sin^(-1)(costheta)+tan^(-1)("tan"(1)/(2)theta)` `=sin^(-1)[sin((1)/(2)pi-theta)]+tan^(-1)("tan"(1)/(2)theta)` `=(pi)/(2)-theta+(theta)/(2)=(pi)/(2)-(pi)/(2)-(1)/(2)tan^(-1)x` `(dy)/(dx)=-(1)/(2(1+x^(2)))` |
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| 173. |
यदि `y=log{e^(x)((x-2)/(x+2))^(3/4)}`, तब दर्शाइए की- `(dy)/(dx)=(x^(2)-1)/(x^(2)-4)` |
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Answer» दिया गया है- `y=log{e^(x)((x-2)/(x+2))^(3/4)}` `rArry=loge^(x)+log((x-2)/(x+2))^(3/4)` `rArry=loge^(x)+3/4log((x-2)/(x+2))` `rArry=x+3/4[log(x-2)-log(x+2)]` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=d/(dx)(x)+3/4[d/(dx)log(x-2)-d/(dx)log(x+2)]` `rArr(dy)/(dx)=1+3/4[1/(x-2)d/(dx)(x-2)-1/(x+2)d/(dx)(x+2)]` `rArr(dy)/(dx)=1+3/4[1/(x-2)xx1-1/(x+2)xx1]` `rArr(dy)/(dx)=1+3/4[1/(x-2)-1/(x+2)]` `rArr(dy)/(dx)=1+3/4[(x+2-x+2)/((x-2)(x+2))]` `rArr(dy)/(dx)=1+3/4xx4/(x^(2)-4)` `rArr(dy)/(dx)=1+3/(x^(2)-4)` `rArr(dy)/(dx)=((x^(2)-4)+3)/(x^(2)-4)` `rArr(dy)/(dx)=(x^(2)-1)/(x^(2)-4)` यही सिद्ध करना था |
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| 174. |
`(dy)/(dx)` ज्ञात कीजिए, यदि- `y=tan^(-1){(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))}` |
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Answer» `y=tan^(-1){(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))}` `=tan^(-1){(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))xx(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)sqrt(1-sinx))}` `=tan^(-1){((sqrt(1+sinx)+sqrt(1-sinx))^(2))/((sqrt(1+sinx))^(2)-(sqrt(1-sinx))^(2))}` `=tan^(-1){(1+sinx+1-sinx+2(sqrt(1+sinx))(sqrt(1-sinx)))/(1+sinx-(1-sinx))}` `=tan^(-1){(2+2sqrt(1-sin^(2)x))/(2sinx)}` `=tan^(-1){(2(1+cosx))/(2sinx)}` `=tan^(-1)((1+cosx)/(sinx))` `=tan^(-1)((2cos^(2)x/2)/(2sinx/2cosx/2))` `=tan^(-1)((cosx/2)/(sinx/2))` `=tan^(-1)(cotx/2)` `=tan^(-1)[tan(pi/2-x/2)]` `rArry=pi/2-x/2` `therefore(dy)/(dx)=0-1/2=-1/2` |
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| 175. |
यदि `y=sin{2tan^(-1)sqrt(((1-x)/(1+x)))}`, दिखाएँ कि `(dy)/(dx)=(-x)/(sqrt(1-x^(2)))` |
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Answer» Correct Answer - `-(1)/(2(1+x^(2)))` `x=costheta` |
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| 176. |
निम्नलिखित फलनों का x के सापेक्ष अवकलन कीजिए- `log{tan(pi/4+x/2)}` |
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Answer» माना `y=log{tan(pi/4+x/2)}` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=d/(dx)[log{tan(pi/4+x/2)}]` `=1/(tan(pi/4+x/2)).d/(dx){tan(pi/4+x/2)}` `=1/(tan(pi/4+x/2)).sec^(2)([pi/4+x/2).d/(dx)(pi/4+x/2)` `=(sec^(2)(pi/4+x/2))/(tan(pi/4+x/2))xx1/2` `=(cos(pi/4+x/2))/(2sin(pi/4+x/2)).1/(cos^(2)(pi/4+x/2))` `=1/(2sin(pi/4+x/2)cos(pi/4+x/2))` `=1/(sin{2(pi/4+x/2)}),[because2sinthetacostheta=sin2theta]` `=1/(sin(pi/2+x))` `=1/(cosx)=secx` |
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| 177. |
यदि (If) `xy=x^(3)+y^(3)`, (find) `(dy)/(dx)` निकालें । |
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Answer» दिया है ,`xy=x^(3)+y^(3)` दोनों तरफ x के सापेक्ष अवकलित (Differentiate) करने पर हमें मिलता है, `(d)/(dx)(xy)=(d)/(dx)(x^(3))+(d)/(dx)(y^(3))` `1*y+x*(dy)/(dx)=3x^(2)+3y^(2)(dy)/(dx)` या `(x-3y^(2))(dy)/(dx)=3x^(2)-y " " therefore(dy)/(dx)=(3x^(2)-y)/(x-3y^(2))` |
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| 178. |
यदि (If) `y=a^(u)`, सिद्ध करें कि (prove that) `(dy)/(dx)=a^(u)loga(du)/(dx)`. |
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Answer» दिया है, `y=a^(u) thereforelogy=u loga` दोनों तरफ x के सापेक्ष अवकलित करने पर हमें मिलता है, `(1)/(y)(dy)/(dx)=(loga)(du)/(dx) " " therefore(dy)/(dx)=yloga(du)/(dx)` अतः `(dy)/(dx)=a^(u)loga(du)/(dx) " " [becausey=a^(u)]` |
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| 179. |
यदि (If) `y=e^(x+e^(x+e^(x+..."to"infty)))` तो साबित करें कि (prove that) `(dy)/(dx)=(y)/(1-y)` |
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Answer» दिया है, `y=e^(x+e^(x+e^(x+..."to"infty)))` `thereforey=e^(x+y)` … (1) दोनों तरफ लघुगणक लेन पर हमें मिलता है , `logy=(x+y)log_(e)e` या ` logy=x+y " " [becauselog_(e)e=1]` दोनों तरफ x के सापेक्ष अवकलित करने पर हमें मिलता है, `(1)/(y)*(dy)/(dx)=1+(dy)/(dx),or((1)/(y)-1)(dy)/(dx)=1` या `((1-y)/(y))(dy)/(dx)=1 therefore(dy)/(dx)=(y)/(1-y)` |
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| 180. |
यदि (If) `y={x+sqrt(x^(2)+a^(2))}^(n)`, सिद्ध करें कि (prove that) `(dy)/(dx)=(ny)/(sqrt(x^(2)+a^(2)))`. |
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Answer» दिया है ,`y={x+sqrt(x^(2)+a^(2))}^(n)` `u=x+sqrt(x^(2)+a^(2))` रखें तो `y=u^(n)` `therefore(dy)/(du)=n u^(n-1)=n{x+sqrt(x^(2)+a^(2))}^(n-1)` तथा `(du)/(dx)=1+(1)/(2)(x^(2)+a^(2))^(-1/2)(2x)` `=1+(x)/(sqrt(x^(2)+a^(2)))=(x+sqrt(x^(2)+a^(2)))/(sqrt(x^(2)+a^(2)))` अब `(dy)/(dx)=(dy)/(du)*(du)/(dx)=n{x+sqrt(x^(2)+a^(2))}^(n-1)*(x+sqrt(x^(2)+a^(2)))/(sqrt(x^(2))+a^(2))` `=(n{x+sqrt(x^(2)+a^(2))}^(n))/(sqrt(x^(2)+a^(2)))=(ny)/(sqrt(x^(2)+a^(2)))` [ (1) से] |
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| 181. |
यदि (If) `y=sqrt((1-x)/(1+x))`, सिद्ध करें कि (prove that) `(1-x^(2))(dy)/(dx)+y=0` |
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Answer» `y=sqrt((1-x)/(1+x))=((1-x)/(1+x))^(1//2)` `rArrlogy=(1)/(2)[log(1-x)-log(1+x)]` ...(1) x के सापेक्ष अवकलन करने पर , हमें मिलता है `(1)/(y)(dy)/(dx)=(1)/(2)[(-1)/(1-x)-(1)/(1+x)]rArr(1)/(y)(dy)/(dx)+(1)/(2)*((2)/(1-x^(2)))=0` अतः `(1-x^(2))(dy)/(dx)+y=0` |
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| 182. |
`(dy)/(dx)` ज्ञात कीजिए, यदि- `y=cot^(-1){(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))}` |
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Answer» `y=cot^(-1){(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))}` `rArry=cot^(-1)(cotx/2),` [उपरोक्तानुसार हल करने पर] `rArry=x/2` `therefore(dy)/(dx)=1/2` |
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| 183. |
यदि (If) `y=e^(x)(sinx+cosx)`, सिद्ध करें कि (prove that) `(d^(2)y)/(dx^(2))-2(dy)/(dx)+2y=0` |
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Answer» `y=e^(x)(sinx+cosx)` `rArr(dy)/(dx)=e^(x)*(sinx+cosx)+e^(x)(cosx-sinx)=2e^(x)cosx` `rArr(d^(2)y)/(dx^(2))=2[e^(x)cosx+e^(x)(-sinx)]` `=2e^(x)(cosx-sinx)` L.H.S. `=(d^(2)y)/(dx^(2))-2(dy)/(dx)+2y` `=2e^(x)(cosx-sinx)-2xx2e^(x)cosx+2e^(x)(sinx+cosx)` `=2e^(x)[cosx-sinx-2cosx+sinx+cosx]` `=2e^(x)xx0=0=` R.H.S. |
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| 184. |
यदि `y=tan^(-1)(cotx)+cot^(-1)(tanx)`, सिद्ध करें कि `(dy)/(dx)= -2` |
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Answer» Correct Answer - `-(1)/(2(1+x^(2)))` `y=tan^(-1){tan((1)/(2)pi-x)}+cot^(-1){cot((1)/(2)pi-x)}` `=((1)/(2)pi-x)+((1)/(2)pi-x)=pi-2x` |
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| 185. |
निम्नलिखित फलनों का x के सापेक्ष अवकलन कीजिए- `log((x+sqrt(x^(2)-a^(2)))/(x-sqrt(x^(2)-a^(2))))` |
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Answer» माना `y=log((x+sqrt(x^(2)-a^(2)))/(x-sqrt(x^(2)-a^(2))))` `rArry=((x+sqrt(x^(2)-a^(2)))/(x-sqrt(x^(2)-a^(2)))xx(x+sqrt(x^(2)-a^(2)))/(x+sqrt(x^(2)-a^(2))))` `rArry=log{((x+sqrt(x^(2)-a^(2)))^(2))/(x^(2)-(x^(2)-a^(2)))}` `rArry=log{((x+sqrt(x^(2)-a^(2)))^(2))/(a^(2))}` `rArry=log(x+sqrt(x^(2)-a^(2)))^(2)-loga^(2)` `rArry=2log(x+sqrt(x^(2)-a^(2)))-loga^(2)` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=2d/(dx)[log(x+sqrt(x^(2)-a^(2)))]-d/(dx)loga^(2)` `=2.1/(x+sqrt(x^(2)-a^(2)))d/(dx)(x+sqrt(x^(2)-a^(2)))-0` `=2/(x+sqrt(x^(2)-a^(2))).{1+1/(2sqrt(x^(2)-a^(2))).(2x-0)}` `=2/(x+sqrt(x^(2)-a^(2))){1+x/(sqrt(x^(2)-a^(2)))}` `=2/(x+sqrt(x^(2)-a^(2))).(sqrt(x^(2)-a^(2))+x)/(sqrt(x^(2)-a^(2)))` `=2/(sqrt(x^(2)-a^(2)))` |
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| 186. |
निम्नलिखित फलनों का x के सापेक्ष अवकलन कीजिए- `sqrt(log{sin(x^(2)/3-1)})` |
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Answer» माना `y=sqrt(log{sin(x^(2)/3-1)})` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=d/(dx)[sqrt(log{sin(x^(2)/3-1)})]` `=d/(dx)[log{sin(x^(2)/3-1)}]^(1/2)` `=1/2(1)/(sqrt(log{sin(x^(2)/3-1)})).d/(dx)log{sin(x^(2)/3-1)}` `=1/(2sqrt(log{sin(x^(2)/3-1)})).(1)/(sin(x^(2)/3-1))d/(dx)sin(x^(2)/3-1)` `=1/(2sqrt(log{sin(x^(2)/3-1)}))` `.(1)/(sin(x^(2)/3-1))cos(x^(2)/3-1).d/(dx)(x^(2)/3-1)` `=(cot(x^(3)/3-1))/(2sqrt(log{sin(x^(2)/3-1)}))xx((2x)/3)` `=(xcot(x^(2)/3-1))/(3sqrt(log{sin(x^(2)/3-1)}))` |
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| 187. |
निम्नलिखित फलनों का x के सापेक्ष अवकलन कीजिए- `sqrt(3x+2)+1/(sqrt(2x^(2)+4))` |
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Answer» माना `y=sqrt(3x+2)+1/(sqrt(2x^(2)+4))` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=d/(dx)[sqrt(3x+2)+1/(sqrt(2x^(2)+4))]` `=d/(dx)(3x+2)^(1//2)+d/(dx)(2x^(2)+4)^(-1//2)` `=1/2(3x+2)^(1/2-1)d/(dx)(3x+2)+(-1/2)(2x^(2)+4)^(1/2-1)d/(dx)(2x^(2)+4)` `=1/(2sqrt(3x+2))xx3+(-1/2)1/((2x^(2)+4)^(3//2)).4x` `=3/(2sqrt(3x+2))-(2x)/((2x^(2)+4)^(3//2))` |
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| 188. |
यदि (If) `y=sin^(-1)[xsqrt(1-x)-sqrt(x)sqrt(1-x^(2))]`,(find) `(dy)/(dx)` निकालें । |
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Answer» `y=sin^(-1)[xsqrt(1-x)-sqrt(x)sqrt(1-x^(2))]=sin^(-1)[xsqrt(1-(sqrt(x))^(2))-sqrt(x)sqrt(1-x^(2))]` या `y=sin^(-1)x-sin^(-1)sqrt(x)." "[becausesin^(-1)x-sin^(-1)y=sin^(-1)[xsqrt(1-y^(2))-ysqrt(1-x^(2))]` x के सापेक्ष अवकलित (Differnitiate) करने पर हमें मिलता है, `(dy)/(dx)=(1)/(sqrt(1-x^(2)))-(1)/(sqrt(1-(sqrt(x))^(2)))(d)/(dx)(sqrt(x))=(1)/(sqrt(1-x^(2)))-(1)/(sqrt(1-x))*(1)/(2sqrt(x))` |
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| 189. |
यदि (If) `sqrt(1-x^(6))+sqrt(1-y^(6))=a(x^(3)-y^(3))`, सिद्ध करें कि (prove that) `(dy)/(dx)=(x^(2))/(y^(2))sqrt((1-y^(6))/(1-x^(6)))` |
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Answer» दिया है, `sqrt(1-x^(6))+sqrt(1-y^(6))=a(x^(3)-y^(3))` ... (1) दोनों तरफ x के सापेक्ष अवकलित करने पर हमें मिलता है, `(d)/(dx)sqrt(1-x^(6))+(d)/(dx)sqrt(1-y^(6))=a[(d)/(dx)(x^(3))-(d)/(dx)(y^(3))]` या `(d)/(d(1-x^(6)))sqrt(1-x^(6))*(d)/(dx)(1-x^(6))+(d)/(d(1-y^(6)))sqrt(1-y^(6))*(d)/(dy)(1-y^(6))*(dy)/(dx)=a[3x^(2)-(d)/(dy)(y^(3))*(dy)/(dx)]` या `(1)/(2sqrt(1-x^(6)))*(-6x^(5))+(1)/(2sqrt(1-y^(6)))*(-6y^(5))(dy)/(dx)=a*3x^(2)-3y^(2)*a(dy)/(dx)` या `(3ay^(2)-(3y^(5))/(sqrt(1-y^(6))))(dy)/(dx)=3ax^(2)+(3x^(5))/(sqrt(1-x^(6)))` या `[3y^(2)((sqrt(1-x^(6))+sqrt(1-y^(6)))/(x^(3)-y^(3)))-(3y^(5))/(sqrt(1-y^(6)))](dy)/(dx)` `=3x^(2)((sqrt(1-x^(6))+sqrt(1-y^(6)))/(x^(3)-y^(3)))+(3x^(5))/(sqrt(1-x^(6)))` [(1) से A का मान रखने पर क्योंकि जो संबंध लाना है उसमें A नहीं है ।] या `(3y^(2)(sqrt(1-x^(6))sqrt(1-y^(6)))+3y^(2)(1-y^(6))-3y^(5)x^(3)+3y^(8))/((x^(3)-y^(3))sqrt(1-y^(6)))(dy)/(dx)` या `=(3x^(2)(1-x^(6))+3x^(2)sqrt(1-x^(6))sqrt(1-y^(6))+3x^(8)-3x^(5)y^(3))/((x^(3)-y^(3))sqrt(1-x^(6)))` `(3y^(2)(sqrt(1-x^(6))sqrt(1-y^(6))+1-3xy^(3)))/(sqrt(1-y^(6)))(dy)/(dx)` `=(3x^(2)(1+sqrt(1-x^(6))sqrt(1-y^(6))-3x^(3)y^(3)))/(sqrt(1-x^(6)))` `therefore(dy)/(dx)=(x^(2))/(y^(2))*sqrt((1-y^(6))/(1-x^(6)))` Second method : दिया है, `sqrt(1-x^(6))+sqrt(1-y^(6))=a(x^(3)-y^(3))` (1) में, `x^(3)=sinA` तथा `y^(3)=sinB` रखने पर हमें मिलता है, `sqrt(1-sin^(2))A+sqrt(1-sin^(2)B)=a(sinA-sinB)` या `cosA+cosB=a(sinA-sinB)` या `2cos((A+B)/(2))cos((A-B)/(2))=2asin((A-B)/(2))cos((A+B)/(2))` या `cot((A-B)/(2))=a` या `(A-B)/(2)=cot^(-1)(a)` या `A-B=2cot^(-1)(a)` या `sin^(-1)x^(3)-sin^(-1)y^(3)=2cot^(-1)(a)` दोनों तरफ x के सापेक्ष अवकलित करने पर हमें मिलता है , `(1)/(sqrt(1-x^(6)))*(d)/(dx)(x^(3))-(1)/(sqrt(1-y^(6)))*(d)/(dx)(y^(3))=0` या `(1)/(sqrt(1-x^(6)))*3x^(2)-(1)/(sqrt(1-y^(6)))*3y^(2)(dy)/(dx)=0` `(dy)/(dx)=(x^(2))/(y^(2))*sqrt((1-y^(6))/(1-x^(6)))` |
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| 190. |
यदि `y=e^(x+e^(x+e^(x+...))` तब `(dy)/(dx)`=A. `(y)/(y+1)`B. `(y)/(y-1)`C. `(y)/(1-y)`D. none of these |
| Answer» Correct Answer - C | |
| 191. |
If `y=sqrt(sinxsqrt(sinx+sqrt(sinx+.."to"infty)))`,then `(dy)/(dx)` , is equal toA. `(cosx)/(2y+1)`B. `(cosx)/(2y-1)`C. `(sinx)/(2y+1)`D. `(sinx)/(2y-1)` |
| Answer» Correct Answer - B | |
| 192. |
निम्नलिखित फलनों का x के सापेक्ष अवकलन कीजिए- `sin^(-1)((sqrt(1+x)+sqrt(1-x))/2)` |
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Answer» माना `y=sin^(-1)((sqrt(1+x)+sqrt(1-x))/2)` `x=costhetarArrtheta=cos^(-1)x` रखने पर, `thereforey=sin^(-1)((sqrt(1+costheta)+sqrt(1-costheta))/2)` `rArry=sin^(-1)[(sqrt(2cos^(2)theta/2)+sqrt(2sin^(2)theta/2))/2]` `rArry=sin^(-1)[(sqrt2costheta/2+sqrtsintheta/2)/2)]` `rArry=sin^(-1)[1/(sqrt2)costheta/2+1/(sqrt2)sintheta/2],` `[becausesinpi/4=cospi/4=1/(sqrt2)]` `rArry=sin^(-1)[sin(pi/4+theta/2)]` `rArry=pi/4+theta/2` `rArry=pi/4+1/2cos^(-1)x` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=d/(dx)(pi/4)+1/2d/(dx)(cos^(-1)x)` `rArr(dy)/(dx)=0+1/2xx(-1)/(sqrt(1-x^(2)))` `rArr(dy)/(dx)=-1/(2(sqrt(1-x^(2))))` |
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| 193. |
निम्नलिखित फलनों का x के सापेक्ष अवकलन कीजिए- `sin^(-1)((3x+4sqrt(1-x^(2)))/5)` |
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Answer» माना `y=sin^(-1)((3x+4sqrt(1-x^(2)))/5)` `x=sinthetarArrtheta=sin^(-1)x` रखने पर, `thereforey=sin^(-1)((3sintheta+4sqrt(1-sin^(2)theta))/5)` `rArry=sin^(-1)(3/5sintheta+4/5costheta)` पुनः `3/5=cosalpha` और `4/5=sinalpha` जहाँ `alpha` अचर है,रखने पर, `y=sin^(-1)(cosalphasintheta+sinalphacostheta)` `rArry=sin^(-1)[sin(theta+alpha)]`, `[becausesinAcosB+cosAsinB=sin(A+B)]` `rArry=theta+alpha` `rArry=sin^(-1)x+alpha` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `therefore(dy)/(dx)=d/(dx)(sin^(-1)x)+d/(dx)(alpha)` `rArr(dy)/(dx)=1/(sqrt(1-x^(2)))+0` `rArr(dy)/(dx)=1/(sqrt(1-x^(2)))` |
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| 194. |
यदि `xsqrt(1+y)+ysqrt(1+x)=0` और `xney` हो,तो सिद्ध कीजिए कि `(dy)/(dx)=-1/(1+x^(2))` |
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Answer» `xsqrt(1+y)+ysqrt(1+x)=0` `rArrxsqrt(1+y)=-ysqrt(1+x)` दोनों पक्षों का वर्ग करने पर, `x^(2)(1+y)=y^(2)(1+x)` `rArrx^(2)+x^(2)y=y^(2)+xy^(2)` `rArrx^(2)-y^(2)=xy^(2)-x^(2)y` `rArr(x+y)(x-y)=xy(y-x)` `rArr(x+y)(x-y)=-xy(x-y)` `rArrx+y=-xy` `rArry+xy=-x` `rArry(1+x)=-x` `rArry=(-x)/(1+x)` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=-[((1+x).1-x.(0+1))/((1+x)^(2))]` `rArr(dy)/(dy)=(-1)/((1+x^(2))` यही सिद्ध करना था| |
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| 195. |
`(d^(2)x)/(dy^(2))` का मान होगा-A. `((d^(2)y)/(dx^(2)))^(-1)`B. `-((d^(2)y)/(dx^(2)))^(-1) ((dy)/(dx))^(3)`C. `((d^(2)y)/(dx^(2))) ((dy)/(dx))^(-2)`D. `-((d^(2)y)/(dx^(2))) ((dy)/(dx))^(-3)` |
| Answer» Correct Answer - B | |
| 196. |
यदि`xsqrt(1+y)+sqrt(1+x)=0`, `(dy)/(dx)` =A. `(x+1)/(x)`B. `(1)/(1+x)`C. `-(2+x)/(x)^(3)`D. `(x)/(1+x)` |
| Answer» Correct Answer - C | |
| 197. |
यदि (If) `xsqrt(1+y)+ysqrt(1+x)=0`, सिद्ध करें कि `(dy)/(dx)=-(1+x)^(-2)` |
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Answer» दिया है ,`xsqrt(1+y)=-sqrt(1+x)` या `x^(2)(1+y)=y^(2)(1+x)` या, `x^(2)-y^(2)=xy(y-x)` या `x+y=-xy` या `y(1+x)=-x` …(1) दोनों तरफ x के सापेक्ष अवकलित करें पर हमें मिलता है, `y*1+(1+x)(dy)/(dx)=-1` या, `(1+x)(dy)/(dx)=-(y+1)` या `(dy)/(dx)=-((y+1)/(1+x))=-(((-x)/(1+x)+1))/(1+x)` [(1) से] `=-(-x+1+x)/((1+x)^(2))=-(1+x)^(-2)` |
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| 198. |
निम्नलिखित फलनों का अवकलज ज्ञात कीजिए- `e^(cos^(-1)(sqrt(1-x^(2))))` |
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Answer» माना `y=e^(cos^(-1)(sqrt(1-x^(2))))` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=d/(dx)[e^(cos^(-1)(sqrt(1-x^(2))))]` `=e^(cos^(-1)(sqrt(1-x^(2))))(-1)/(sqrt(1-(sqrt(1-x^(2)))^(2))).d/(dx)(sqrt(1-x^(2)))` `=e^(cos^(-1)(sqrt(1-x^(2))))/x.1/2(1-x^(2))^(-1//2)d/(dx)(1-x^(2))` `=e^(cos^(-1)(sqrt(1-x^(2))))/x.1/(2sqrt(1-x^(2)))(-2x)` `=e^(cos^(-1)(sqrt(1-x^(2))))/(sqrt(1-x^(2)))` |
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| 199. |
निम्नलिखित फलनों का अवकलज ज्ञात कीजिए- `sin^(-1)((1-x^(2))/(1+x^(2)))` |
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Answer» `y=sin^(-1)((1-x^(2))/(1+x^(2)))` माना `x=tantheta`, तब `theta=tan^(-1)x` `thereforey=sin^(-1)((1-tan^(2)theta)/(1+tan^(2)theta))` `rArry=sin^(-1)(cos2theta)` `rArry=sin^(-1)[sin(pi/2-2theta)]` `rArry=pi/2-2theta` `rArry=pi/2-2tan^(-1)x` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=d/(dx)(pi/2-2tan^(-1)x)` `rArr(dy)/(dx)=0-2xx1/(1+x^(2))` `rArr(dy)/(dx)=(-2)/(1+x^(2))` |
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`sec^(-1)(1/(2x^(2)-1))` का अवकल गुणांक ज्ञात कीजिए जबकि `0ltxlt1/(sqrt2)` |
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Answer» माना `y=sec^(-1)(1/(2x^(2)-1))`, जहाँ `0ltxlt1/(sqrt2)` `x=costheta` रखने पर, `y=sec^(-1)(1/(2cos^(2)theta-1))` `rArry=cos^(-1)(2cos^(2)theta-1)`, `[becausesec^(-1)1/x=cos^(-1)x]` `rArry=cos^(-1)(cos2theta)` `rArry=2theta` `[because0ltxlt1/(sqrt2)rArr0ltcosthetalt1/(sqrt2)rArr0ltthetaltpi/4rArr0lt2thetaltpi/2]` `rArry=2cos^(-1)x`, `[becausex=costhetarArrtheta=cos^(-1)x]` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=(-2)/(sqrt(1-x^(2)))` |
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