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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. |
निम्न फलन का अवकल गुणांक ज्ञात कीजिये -` x^(y) +y^(x)=1 ` |
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Answer» Correct Answer - ` " " (-y^(x)log y+ yx^(y-1))/(x^(y) log x + xy^(x-1))` माना ` " " y=y_1+ y_2 ` जहाँ ` y_1 -x^(y) ` और ` y_2 =y^(x)` अब `" "y_ 1 =x^(y)" "rArr log y_1 =ylog x ` `rArr " "(dy_1)/(dx) =y_1 ((y)/(x)+ log x (dy)/(dx))` ` " "y_2 =y^(x)` `rArr " "log y_2 =x log y ` `rArr (dy_2)/(dx) =y_2 ((x)/(y) (dy)/(dx) +log y)` |
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| 2. |
निम्न फलन का अवकल गुणांक ज्ञात कीजिये -` x^(a) y^(b) =(x+y )^(a+b) ` |
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Answer» Correct Answer - ` (y)/(x)` दोनों पक्षों का log लेने पर ` " "alog x +blog y =(a+b )log (cx+y)` `rArr " "a*(1)/(x)+ b*(1)/(y)(dy)/(dx) =(a+b) (1)/(x+y) *(d)/(dx)(x+y)` अब सरल करने पर |
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| 3. |
यदि ` e^(x) =(sqrt(1+t) -sqrt(1-t ))/(sqrt(1+t)+sqrt(1-t) ) ` व ` tan ""(y)/(2) =sqrt((1-t)/(1+t) ), t =(1)/(2) ` पर ` (dy)/(dx) =`A. `1//2`B. ` -1//2`C. ` 0`D. इनमें से कोई नहीं |
| Answer» Correct Answer - B | |
| 4. |
फलन का अवकलन गुणांक ज्ञात कीजिए| ` tan 7x + e^(tan 4x)` |
| Answer» Correct Answer - ` 7sec ^(2)7x + 4sec ^(2) 4x* e^(tan 4x )` | |
| 5. |
निम्न फलन का अवकल गुणांक ज्ञात कीजिये -` (cos x )^(y) =(sin y )^(x)` |
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Answer» Correct Answer - ` (ytan x+ log sin y)/(log cos x - x cot y)` `(cos x)^(y) =(siny)^(x) " "rArr ylog cos x =x log sin y ` `rArr" "y*(1)/(cosx )(d)/(dx) (cos x ) +log cos x *(dy)/(dx) ` ` " "=x *(1)/(sin y) *(d)/(dx) (sin y) +log sin y*1` |
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| 6. |
यदि ` x=a cos theta + bsin theta ,y =a sin theta -b cos theta,` दिखाइए की `y^(2)(d^(2y))/(dx^(2))-x (dy)/(dx) +y=0` |
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Answer» `because " "x =acos theta +b sin theta ` `rArr (dx)/(d theta )=- a sin theta + b cos theta ` व् ` y= asin theta - bcos theta ` `rArr (dy)/(d theta ) =acos theta + b sin theta ` अब ` (dy)/(dx) =(dy//d theta )/(dx//d theta ) =(acos theta + b sin theta )/(-a sin theta +b cos theta )= (x)/(-y)` ` rArr " " y(dy)/(dx)+ x=0 " "...(1)` x के सापेक्ष पुनः अवकलन करने पर ` " "y (d^(2)y)/(dx^(2))+(dy)/(dx) *(dy)/(dx)+ 1=0` ` rArr " "y^(2)(d^(2)y)/(dx^(2) )+ (y(dy)/(dx)) (dy)/(dx)+y=0` ` rArr " "y^(2) (d^(2y))/(dx^(2))-x (dy)/(dx)+ y=0` |
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| 7. |
`(d)/(dx) [sqrt (1-x^(2) )sin ^(-1) x-x ]` का मान ज्ञात कीजिए| |
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Answer» `(d)/(dx) [sqrt (1-x^(2) )sin ^(-1) x-x ]` ` therefore (d)/(dx) [sqrt (1-x^(2) ) sin ^(-1) x] -(d)/(dx) [x]` ` =(sqrt (1-x^(2) )) (d)/(dx)sin ^(-1) x+sin^(-1)""x(d)/(dx) (sqrt(1-x^(2)))-(d)/(dx)(x) ` ` sqrt (1-x^(2))*(1)/(sqrt(1-x^(2)))+sin ^(-1) x*(1)/(2) (1-x^(2) )^(-1//2) *(-2x ) -1` ` " "{ 1-(xsin ^(-1) x)/(sqrt (1-x^(2)))-1} =(-xsin ^(-1) x)/(sqrt (1-x^(2)))` |
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| 8. |
यदि ` y= e^(msin ^(-1))x,` तब दिखाइए की ` " " (1-x^(2)) (d^(2)y)/(dx^(2))-x (dy)/(dx) -m^(2) y=0` |
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Answer» यहाँ ` " "y= e ^(msin ^(-1)x)` माना ` t= sin ^(-1) x " "rArr y= e^(mt ) ` `rArr " "(dy)/(dt)= me^(mt )=my (1)/(sqrt (1-x^(2)))` ` rArr " "sqrt (1-x ^(2))*(dy)/(dx) =my` x के सापेक्ष अवकलन करने पर ` (d^2y)/(dx^(2))*sqrt(1-x^(2))+(dy)/(dx)* (1)/(2) (1-x^(2))^(-1//2) (-2 x)= m (dy)/(dx)` ` rArr " "sqrt (1-x^(2))*(d^(2)y)/(dx^(2))-(x)/(sqrt(1-x^(2)))(dy)/(dx)=m (dy)/(dx)` ` rArr (1-x^(2))(d^2y) /(dx^(2))-x (dy)/(dx) =m (dy)/(dx ) sqrt(1-x^(2) )= m^*my` इसलिए ` " "(1-x^(2))(d^2y)/(dx^(2))-x (dy)/(dx)-m^(2)y=0 ` |
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| 9. |
`tan ^(-1) {(1+x)/(1-x)} ` का अवकलन गुणांक है-A. ` (2)/(1+x^(2))`B. ` (1)/(1+x^(2))`C. ` (1+x^(2))/(1-x^(2))`D. इनमें से कोई नहीं |
| Answer» Correct Answer - B | |
| 10. |
यदि ` y= sin (tan ^(-1) 2x )` तब सिद्ध कीजिए की `(dy)/(dx)= (2)/((1+4x^(2) )^(3//2) ) ` |
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Answer» `y= sin (tan^(-1) 2x ) ` माना ` tan ^(-1) 2x=t` ` therefore " "y= sin t ` ` rArr " "(dy)/(dt) =cos t ` व ` (dt)/(dx) ={ (1)/((1+ 4x^(2) ))*2} =(2)/((1+4x^(2) ) ) ` हम जानते है की ` " "(dy)/(dx) =((dy)/(dt)xx(dt)/(dx))` ` (dy)/(dt) ` व ` (dt)/(dx) ` के मान रखने पर ` " "(dy)/(dx) ={ cos txx (2)/((1+4x^(2) ))}" "...(1) ` ` t= tan ^(-1) 2xrArr tan t =2x ` ` rArr " "sec t = sqrt (1+tan ^(2) t )= sqrt ( 1+ 4x^(2) )` ` rArr " " cost = (1)/(sqrt (1+ 4x^(2) ) )` समीकरण (1 ) में cos t का मान रखने पर ` (dy)/(dx) ={ (1)/(sqrt (1+4x ^(2)))xx(2)/((1+ 4x^(2))) } =(2)/((1+4x^(2))^(3//2) ) ` |
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| 11. |
यदि ` y= (1)/(sqrt (a^(2) -x^(2) ) ),` तब ` (dy)/(dx)` का मान ज्ञात कीजिए| |
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Answer» यहाँ ` y= (1)/(sqrt( a^(2- x^(2)) ) )` माना ` (a^(2) -x^(2) )= t ` ` therefore " "y= (1)/(sqrt ( t))=t^(-1//2) ` ` rArr(dy)/(dx) =-(1)/(2) t^(-3//2) =(-1)/(2t^(3//2) ` और ` (dt)/(( dx))=-2x` हम जानते है की ` rArr " "(dy)/(dx) =((dy)/(dt) xx (dt)/(dx) ) =-(1)/(2t^(3//2) ) xx(-2x) =(x) /(t^(3//2) ) ` ` (dy)/(dx) =(x) /((a^(2) -x^(2) )^(3//2))` |
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| 12. |
`(dy)/(dx)` ज्ञात कीजिये यदि ` x= 2t^(2) +17t -1,y=3t^(4) -8t^(2) +9` |
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Answer» Correct Answer - ` (12t^(3) -6 t) /(4t+ 17)` ` x=2t^(2) +17t -1 " "rArr " "(dx)/(dt) =4t +17` ` y= 3t^(4) -8t^(2) +9" "rArr" "(dy)/(dt) =12 t^(3) -16t` ` (dy)/(dx) =(dy)/(dt) *(dt)/(dx) ` का प्रयोग करने पर| |
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| 13. |
दिखाइए की ` (d)/(dx) [ (x) /(2) sqrt ( a^(2)-x^(2) )+ (a^(2) )/(2) sin ^(-1) ""(x)/(a) ]= sqrt (a^(2) -x^(2) )` |
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Answer» यहाँ ` (d)/(dx) [(x)/(2) sqrt (a^(2) - x^(2) )+ (a^(2))/(2) sin ^(-1)"" (x)/(a) ]` ` rArr (d)/(dx) [(x)/(2) sqrt(a^(2) -x^(2) )] +(d)/(dx) [ (a^(2)/(2)sin ^(-1)"" (x)/(a) ) ` ` =(x)/(2) (d)/(x) (sqrt (a^(2) -x^(2)))+ sqrt ( a^(2) -x^(2) )(d)/(dx) ((x)/(2))` ` " "+ (a^2) /(a) (d)/(dx)(sin ^(-1)""(x)/(a)) ` ` =(x)/(2) *(1)/(2) (a^(2) -x^(2) )^(-1//2) *(-2x)+ sqrt (a^(2) -x^(2) ) *(1)/(2)` ` " "+ (a^(2) )/(2) *(1)/(sqrt (1-((x^(2) )/(a^(2)))))*(1)/(a) ` ` = (-x^(2) )/(2sqrt (a^(2)-x^(2) ))+ (sqrt (a^(2) -x^(2) ))/(2) +(a^(2) )/(2sqrt (a^(2)-x^(2)))` ` " "= (-x^(2)+(a^(2)-x^(2))+ a^(2))/(2sqrt (a^(2)-x^2)) =(a^(2)-x^(2))/(sqrt (a^(2) -x^(2) )= sqrt ((a^(2)-x^(2))) )` |
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| 14. |
`sin ^(-1) ""(1-x^(2))/(1+x^(2))` का अवकलन गुणांक है-A. ` -2 //(1+x^(2))`B. ` 2//(1+x^(2))`C. ` 1//(2+x^(2)) `D. इनमें से कोई नहीं |
| Answer» Correct Answer - A | |
| 15. |
यदि ` x^(y) =e^(x-y) ` तब `dy//dx=`A. ` (log x)/((1+log x )^(2))`B. ` (x-y)/((1+log x )^(2))`C. ` (x+y )/( (1+log x )^(2))`D. ` (1)/((1+log x ))` |
| Answer» Correct Answer - A | |
| 16. |
यदि ` y= sin [sqrt sin sqrt x 1,` तब ` (dy)/(dx) ` का मान ज्ञात कीजिए |
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Answer» यहाँ ` " " y= sin [ sqrt (sin sqrtx) ]` माना ` sqrt(x) =t, sin sqrt(x)= sin t=u ` व `sqrt (sin sqrt x) = sqrt(u) =v ` ` therefore " " y= sin v ` ` (dy)/(dx) =cos v, " "(du)/(dt)= cos t, " " (dv)/(du) =(1)/(2) u^(-1//2) ` व ` (dt)/(dx) =(1)/(2sqrt(x))` हम जानते है की ` (dy)/(dx) =((dy)/(dv)xx(dv)/(du)xx(du)/(dt)xx(dt)/(dx)) " "...(1)` सभी मान समीकरण (1 ) में रखने पर ` " "(dy)/(dx) =[ cos v*""(1)/(2sqrtu)*cos t*""(1)/(2sqrt ( x))]` ` " "= [cos sqrtu *(1)/(2sqrt(u))cost*""( 1)/(2sqrtx) ] " " जहाँ v= sqrt u ` `" "(1)/(4) cos ""( sqrt (sin t)) *(1)/(sqrt (sin sqrt x ) ) *cos sqrtx *(1)/(sqrt(x) ) ` ` " "` जहाँ ` u= sin t` ` =(1)/(4) cos (sqrt sin sqrt x ) * ( 1)/(sqrt sin sqrt x )*cos sqrtx *(1)/(sqrt(x) ) ` ` " " ` (जहाँ `t= sqrt x ` ) ` = (cos sqrt sin sqrtx ) /( 4 sqrtx sqrt sin sqrt x ) * cos sqrt x ` |
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| 17. |
यदि `y= sin (sqrt (sin x + cos x ) ),` तब `(dy)/(dx) ` का मान ज्ञात कीजिए| |
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Answer» `y= sin (sqrt (sin x + cos x ) ),` माना ` (sin x +cos x ) =t` और ` sqrt t =u` ` therefore y= sin u,u =sqrt (t),t =sin x + cos x ` ` " "=(dy)/(du )= cos u ,""(du)/(dt) =(1)/(2) t(-1//2) =(1)/(2sqrt (t))` तथा ` (dt)/(dx) = (cos x - sin x ) ` हम जानते है की ` (dy)/(dx) =((dy)/(du)xx(du)/(dt)xx(dt)/(dx))` ` =[ cos uxx(1)/(2sqrt (t))xx(cos x- sin x ) ]` ` =(cos sqrtt )/(2sqrt t ) (cos x - sin x ) ` ` =(cos sqrt (sinx +cos x) ) /(2 sqrt( sin x+ cos x ))*(cos x -sin x ) ` |
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| 18. |
`sqrt(x)` के सापेक्ष ` sin ^(-1)"" (1-x)/(1+x) ` का अवकलन है-A. ` 1//1+x`B. ` (-2)/(1+x)`C. ` (1)/(1+x^(2))`D. इनमें से कोई नहीं |
| Answer» Correct Answer - B | |
| 19. |
यदि ` sin y= xsin (a+y ),` तब `dy//dx=`A. ` (sin ^(2) (a+y))/(sin a )`B. ` sin (a+y )`C. `sin ^(2) (a+y)`D. इनमें से कोई नहीं |
| Answer» Correct Answer - A | |
| 20. |
यदि ` y= (5x) /(3sqrt 1-x^(2) )+ sin ^(2) (2x+ 3),` तब ` (dy)/(dx) ` का मान ज्ञात कीजिए | |
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Answer» यदि ` y= (5x) /(3sqrt 1-x^(2) )+ sin ^(2) (2x+ 3),` ` therefore (dy)/(dx) =(d)/(dx) {5x(1-x^(2) )^(-1//3) }+ (d)/(dx) {sin ^(@) (2x+ 3)} ` ` = {5x*(-(1)/(3) ) (1-x^(2) )^(-4//3) *(-2x ) + ( 1-x^(2) ) ^(-1//3)*5} ` ` " "+ {2sin (2x+3) cos (2x+ 3)*2}` ` =(10x^(2))/( 3(1-x^(2) )^(4//3) )+ (5)/((1-x^(2) )^(1//3))+2sin (4x+ 6) ` ` =((10x^(2) + 15(1-x^(2)))/(3(1-x^(2))^(4//3) )+2 sin (4x+6) ` ` = (15-5x^(2) ) /(3( 1-x^(2) ) ^(4//3))+2sin (4x+6 ) ` |
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| 21. |
यदि ` y= sin ^(-1) [x sqrt (1-x) -sqrt ( x)sqrt(1-x^(2) )]` तब `(dy)/(dx) ` का मान ज्ञात कीजिए| |
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Answer» `y=sin ^(-1) [x sqrt (1-x) -sqrt(x) sqrt(1-x^(2))]` माना ` x= sin theta, sqrt x =sin phi, ` तब ` y= sin ^(-1) [sin theta cos phi - sin phi cos theta ] =sin ^(-1) [sin (theta -phi )] ` ` y= theta-phi rArr " "sin ^(-1) x- sin ^(-1) sqrt (x) ` ` therefore (dy)/(dx) =(d)/(dx) { sin ^(-1) x-sin ^(-1) sqrtx) ` ` " "= (d)/(dx) (sin ^(-1) x) -(d)/(dx) (sin ^(-1) sqrt(x))` ` " "= [ (1)/(sqrt( 1-x^(2)) )-(1)/(2sqrt(x)*sqrt(1-x))]` |
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| 22. |
यदि ` y= sin ^(-1) { (sqrt (1+x)-sqrt (1-x))/(2)}` तब ` (dy)/(dx)` का मान ज्ञात कीजिए | |
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Answer» `y= sin ^(-1) ""{( sqrt(1+x)-sqrt(1-x))/(2)}` माना ` x= cos 2theta ,` तब ` " "y=sin ^(-1) { (sqrt (1+cos 2theta) -sqrt( 1-cos 2theta))/(2)}` ` sin ^(-1) {(sqrt( 2cos ^(2)theta )-sqrt (1-cos 2theta ))/(2)}` ` =sin ^(-1) {(sqrt ( 2)cos theta - sqrt (2) sin theta )/(2)}` ` = sin ^(-1) { (1)/(sqrt(2) )cos theta -(1)/(sqrt(2))sin theta }` ` " "=sin ^(-1) {sin ""(pi)/(4) cos theta - cos ""(pi)/(4) sin theta }` ` =sin ^(-1) {sin ((pi)/(4) -theta )}` `" "((pi)/(4)-theta ) =((pi)/(4) -(1)/(2) cos ^(-1) x ),` जहाँ `theta =(1)/(2) cos ^(-1)x ` ` y= (pi)/(4) -(1)/(2) cos ^(-1) x ` ` therefore " "(dy)/(dx) =(d)/(dx) ((pi)/(4) -(1)/(2)cos ^(-1)x )` ` =(d)/(dx) ((pi)/(4) )-(d)/(dx) ((1)/(2) cos ^(-1) x )` ` " "= (0-(1)/(2) *((-1))/(sqrt (1-x^(2))) ) =(1)/(2sqrt(1-x^(2)))` |
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| 23. |
यदि ` x=a sin 2theta (1+ cos 2theta ),y =bcos 2theta (1-cos 2theta ), ` तब ` dy//dx=`A. ` (btan theta )/(a) `B. ` (atan theta )/(b ) `C. ` (bcot theta )/(a) `D. इनमें से कोई नहीं |
| Answer» Correct Answer - A | |
| 24. |
यदि `y=cot ^(-1) (sqrt(1+x^(2))+x)` तब `(dy)/(dx) ` का मान ज्ञात कीजिए| |
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Answer» यहाँ ` y= cot ^(-1) (sqrt(1+x^(2) +x))` माना ` x=cot theta ` तब ` " "= cot ^(-1) (cosec theta+ cot theta ) =cot ^(-1) ((1)/(sin theta )+(cos theta )/(sin theta))` ` =cot ^(-1) ((1+ costheta )/(sin theta ))=cot^(-1) {(2cos ^(2)(theta //2))/(2sin (theta //2)cos (theta //2))}` ` =cot ^(-1) (cot ""(theta )/(2) )_=(theta)/(2)` ` y= (1)/(2) cot ^(-1) x ` ` therefore (dy)/(dx) =(d)/(dx) ((1)/(2)cot ^(-1) x )=(1)/(2) (d)/(dx) (cot ^(-1) x) = (-1)/(2(1+x^(2)))` |
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| 25. |
यदि ` y= cot ^(-1) ((1- x )/(1+x)) ` तब `(dy)/(dx) ` का मान ज्ञात कीजिए| |
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Answer» `y= cot ^(-1) ((1-x )/(1+x))` माना ` " "x= tan theta ` ` y= cot ^(_1) ((1-tan theta )/(1+tan theta )) =cot ^(-1) ""{tan ((pi )/(4) -theta )}` ` " "= cot ^(-1) [ cot {(pi)/(2) - ((pi)/(4)-theta )}]` ` " "= cot^(-1) [cot ""((pi)/(4) +theta )] =(pi)/(4) +theta ` ` y= (pi)/(4) +tan ^(-1) x,` जहाँ `x= tan thetarArrtheta tan^(-1) x (dy)/(dx) =(d)/(dx) ((pi)/(4) +tan ^(-1) x)` ` rArr (d)/(dx) ((pi)/(4) )+ (d)/(dx) (tan ^(-1) x ) =0+( 1)/((1+x^(2) )) =(1)/((1+x^(2))` |
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| 26. |
यदि ` y= tan ^(-1) ((x^(1//3)+a^(1//3))/(1-x^(1//3)a^(1//3)))` तब ` (dy)/(dx) ` का मान ज्ञात कीजिए| |
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Answer» `y= tan ^(-1) ((x^(1//3)+a^(1//3))/( a-x^(1//3)a^(1//3)))` माना `x^(1//3) =tan theta ` तथा ` a^(1//3) =tan phi ,` तब `y=tan ^(-1) ((tan theta + tanphi )/(1-tan theta tan phi ))= tan ^(-1) [tan (theta +phi ]` ` y= theta +phi=tan ^(-1)(x^(1//3) )+tan ^(-1) (a^(1//3))` ` therefore " "(dy)/(dx) =(d)/(dx) {tan ^(-1) (x^(1//3)) +tan ^(-1) (a^(1//3))}` ` " "= (d)/(dx) {tan ^(-1) (x^(1//3))} +(d)/(dx) {tan ^(-1) (a^(1//3))}` ` =(1)/(1+x^(2//3) )*(1)/(3)x ^(-2//3) +0 =(1)/(3x^(2//3)(1+x^(2//3)))` |
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| 27. |
यदि ` y= sin ^(-1) {(5x+ 12sqrt (1-x^(2)))/(13)}` तब `(dy)/(dx)` का मान ज्ञात कीजिए| |
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Answer» `" "y= sin ^(-1) {(5x+12sqrt(1-x^(2)))/(13)}` ` rArr y=sin ^(-1) {(5x)/(13)+ (12)/(13) sqrt(1-x^(2))}` माना ` (5)/(13)=sin alpha ` तथा ` x =cos theta ,` तब ` cos alpha =sqrt(1-(25)/(169))= sqrt((144)/(169))=(12)/(13)` व् ` sqrt(1-x^(2))=sqrt (1-cos ^(2))theta =sqrt (sin ^(2) theta =sin theta )` ` y=sin ^(-1) {sin alpha cos theta +cos alpha sin theta }` ` =sin ^(-1) {sin (alpha +theta ) } ` ` " "y= (alpha +theta )= sin ^(-1) ""(5)/(13) +cos ^(-1) x ` ` therefore " "(dy)/(dx) =(d)/(dx) {sin ^(-1) ""(5)/(13) +cos ^(-1) x}` ` =(d)/(dx) {sin ^(-1)"" (5)/(3)} +(d)/(dx)(cos ^(-1) x)` ` =0 -(1)/(sqrt(1-x^(2)))=-(1)/(sqrt(1-x^(2)))` |
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| 28. |
`(dy)/(dx)` ज्ञात कीजिये यदि ` x=a(cost+ sin t) ,y=a (sin t-cost)` |
| Answer» Correct Answer - ` (cost+sin t )/(cost-sin t )` | |
| 29. |
` (d)/(dx) {cos ^(-1)""(x-x^(-1))/(x+x^(-1))} =`A. ` (1)/(1+x^(2) ) `B. ` (-2)/(1+x^(2) ) `C. ` (2)/(1-x^(2))`D. इनमें से कोई नहीं |
| Answer» Correct Answer - B | |
| 30. |
यदि ` y= sec tan ^(-1) x,` तब ` (dy)/(dx) =`A. ` x//(1+x^(2))`B. ` xsqrt((1+x^(2)))`C. ` 1//sqrt(1+x^(2))`D. ` x//(1+x^(2))` |
| Answer» Correct Answer - D | |
| 31. |
` e^(sin x ) ` का sin x के सापेक्ष अवकल गुणांक ज्ञात कीजिये| |
| Answer» Correct Answer - ` e^(sinx )` | |
| 32. |
यदि `y= cot ^(-1) sqrt(( 1-sin x)/( 1+sin x ),)` तब `(dy)/(dx) ` का मान ज्ञात कीजिए| |
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Answer» प्रश्नानुसार `y= cot ^(-1) sqrt((1-sin x ) /( 1+sin x ) ) ` ` =cot ^(-1) sqrt ((1+cos ""((pi)/(2) +x))/( 1- cos""((pi )/(2)+x)) )` ` = cot ^(-1) sqrt((2cos ^(2) "((pi) /(4) + (x)/(2)))/( 2sin ^(2) ((pi )/( 4)+ (x)/(2) ) ))` ` " "= cot^(-1) { cot((pi )/(4) +(x)/(2) )} =((pi )/(4)+ (x)/(2) ) ` ` y= ((pi)/(4) + (x)/(2)) ` ` therefore " "(dy)/(dx) =(d)/(dx) ((pi)/(4) +(x)/(2) ) ` ` rArr" "(d)/(dx) ((pi)/(4) )+ (d)/(dx) ((x)/(2))=0 +(1)/(2) =(1)/(2)` |
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| 33. |
यदि `y= tan ^(-1) ((sqrt(1+ x^(2))+1)/(x))` तब ` (dy)/(dx)` का मान ज्ञात कीजिए | |
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Answer» `y=tan ^(-1) ((sqrt(1+x^(2) )+1) /(x))` माना ` x tan theta ,` तब ` y= tan ^(-1) ((sectheta +1)/(tan theta ))=tan ^(-1) ((1+costheta )/(sin theta ))` ` =tan ^(-1) { (2cos ^(2) (theta //2))/(2sin (theta //2)cos (theta //2))}` ` =tan ^(-1) {cot ""(theta)/(2) } =tan ^(-1) {tan ((pi)/(2)-(theta )/(2))}` ` =((pi)/(2)-(theta )/(2) )= (pi)/(2) -(1)/(2) tan ^(_1) x ` ` therefore " "(dy)/(dx) =-(1)/(2(1+x^(2)))` |
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| 34. |
यदि ` y= tan ^(-1)"" (sqrt( 1+sin x)+ sqrt( 1-sin x ))/( sqrt ( 1+sin x)- sqrt(1-sin x )) ` तब ` (dy)/(dx) ` का मान कीजिए| |
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Answer» माना ` y= tan ^(-1) ""(sqrt ( 1+sin x )+ sqrt (1-sin x ))/(sqrt(1+sin x )- sqrt(1- sin x ) )` ` = tan ^(-1)""( sqrt(1+ sin x + )sqrt ( 1-sin x ) )/( sqrt ( 1+sin x )- sqrt (1-sin x ))xx (sqrt (1+sin x )+ sqrt ( 1-sinx ))/( sqrt(1+sinx )+ sqrt( 1-sin x ))` ` " "= tan ^(-1)"" {( 1+sin x ) + (1-sin x)+ 2sqrt (1-sin ^(2)x ))/((1+sin x )-(1-sin x ))` ` =tan ^(-1)""((1+ cos x ) /( sin x )) =tan ^(-1)""{ (2cos ^(2) (x//2))/(2sin (x//2)cos (x//2))}` ` =tan ^(-1) {cot ""(x)/(2) }=tan ^(-1) {tan ""((pi)/(2)-(x)/(2))} ` ` " "y= ((pi)/(2) -(x)/(2))` ` therefore (dy)/(dx) =(d)/(dx) ((pi)/(2)-(x)/(2) ) =(d)/(dx) ((pi)/(2) )-(d)/(dx) ((x)/(2))` `" "= 0 -(1)/(2) =-(1)/(2) ` |
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| 35. |
`e^(tan ^(-1))sqrt (x) ` के लिए `(dy)/(dx)` का मान ज्ञात कीजिए| |
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Answer» `y=e ^(tan ^(-1)sqrt (x))` माना ` sqrt x=t , tan ^(-1) sqrt (x) =tan ^(-1)t=u` ` therefore " "y= e^(u) ` ` rArr (dy)/(du) =e^(u) ,(du)/(dt) =(1)/((1+ t^(2) ) ),(dt)/(dx) =(1)/(2) xx^(-1//2) =(1)/(2sqrt x)` हम जानते है की ` (dy)/(dx) =((dy)/(du) xx(du)/(dt)xx(dt)/(dx))=e^(u) (1)/((1+t^(2) ) )*(1)/(2sqrt (x) ) ` ` =e ^(tan ^(-1) t)*(1)/((1+t^(2) ))*(1)/(2sqrt (x) )` जहाँ `u= tan ^(-1) t ` ` =(e^(tan -1 sqrtx ) )/(2sqrt x(1+x) ) , " "` जहाँ ` t= sqrt x` ` (dy)/(dx) =(e^(tan -1sqrt x))/(2sqrt x (1+x))` |
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| 36. |
`(dy)/(dx)` ज्ञात कीजिये यदि` x= (3t )/(1+3t^(2)),y =(3t^(3))/(1+3t^(3))` |
| Answer» Correct Answer - ` (3t^(2) (1+3t^(2))^(2))/((1+3t^(3))^(2)(1-3t^(2)))` | |
| 37. |
`log tan x ` का अवकलन गुणांक है-A. ` 2sec 2x`B. ` 2cosec 2x `C. ` 2sec^(3)x `D. ` 2cosec ^(3)x ` |
| Answer» Correct Answer - B | |
| 38. |
`sin^(-1) x ` का `e^(sin ^(-1)) ` के सापेक्ष अवकल गुणांक ज्ञात कीजिये| |
| Answer» Correct Answer - ` (1)/(e^(sin^(-1))x ) ` | |
| 39. |
`(dy)/(dx)` ज्ञात कीजिये यदि` x=ct ,y =(c)/(t^(2))` |
| Answer» Correct Answer - ` (2)/(t^(3))` | |
| 40. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिये-` tan ^(-1) ((x)/(a)) ` |
| Answer» Correct Answer - ` (a)/(a^(2)+x^(2) )` | |
| 41. |
` x= (pi)/(6)` पर `cot ^(-1) [(cos 2x) ^*1//2) ]` का अवकलन गुणांक है-A. ` (2//3) ^(1//2)`B. ` (1//3)^(1//2)`C. `3^(1//2) `D. ` 6^(1//2) ` |
| Answer» Correct Answer - A | |
| 42. |
` x^(6) +3x^(2) -9` का `2x^(5) +7x^(3) +11x^(2) -17` के सापेक्ष अवकल गुणांक ज्ञात कीजिये| |
| Answer» Correct Answer - ` (6x^(5) +6x)/(10x^(4) +21x ^(2)+ 22x ) ` | |
| 43. |
यदि ` y= (sin x )^(x) +sin ^(-1) sqrt (x) ,` तब ` (dy)/(dx) ` का मान ज्ञात कीजिए| |
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Answer» यहाँ ` " "y= (sin x ) ^(x) +sin^(-1) sqrt(x) ` माना, ` " "y_1 =(sinx )^(x) ` व ` y_2 =sin ^(-1) sqrt x` तब ` " " y=y_1 +y_2 ` ` rArr " " (dy)/(dx) =(dy_1)/(dx) +(dy_2)/(dx) " "....(1)` ` " "y_1 =(sin x ) ^(x) ` ` " "logy_1 =x log (sinx )` x के सापेक्ष अवकलन करने पर ` (1)/(y_1) *(dy_1)/(dx) =x* (1)/(sin x ) *cos x+ log (sim x )*1 ` ` " "= (sin x )^(x) [x cot x +log (sin x)]" "....(2) ` अब, ` " "y_2 = sin ^(-1) sqrt x ` x के सापेक्ष अवकलन करने पर ` " "(dy_2)/(dx) =(1)/(sqrt( 1-x) )*(1)/(2) x^(1//2) =(1)/(sqrt(x-x^(2))) ` समीकरण (1 ),(2 ) व (3 ) से,` (dy)/(dx)=(sin x )^(x) {x cot x + log (sin x )} +(1)/(2sqrt (x-x^(2)) )` |
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| 44. |
यदि ` y= (sqrt(a+x ))/ sqrt(a-x)` तब ` (dy)/(dx)` का मान ज्ञात कीजिए| |
| Answer» Correct Answer - ` (a^(2)-asqrt(a^(2)-x^(2)))/(x^2sqrt(a^(2) -x^(2)))` | |
| 45. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिये- `sin ^(-1) (mx) ` |
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Answer» Correct Answer - ` (m)/(sqrt( (1-m^(2)x^(2))))` `(d)/(dx) sin ^(-1) (mx) =(1) /(sqrt(1-(mx)^(2)))*(mx ) =(m)/(sqrt( 1-m^(2)x^(2)))` |
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| 46. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` 10^(10^(x))` |
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Answer» Correct Answer - `10^(x) *10^(10^(x))(log _e10)^(2) ` माना ` " "10^(x) =t ` ` rArr " "10^(x) log _e 10* dx = dt ` ` rArr" "(dt)/(dx )= 10^(x) log _e 10` ` " "(d)/(dx) (10^(10x)) =(d)/(dt) (10^(t))(dt)/(dx) ` का प्रयोग करने पर ` " "= (d)/(dt) (10^(t))(10^(x) log_e 10 )= 10^(10x ) *10^(x) (log _e 10 )^(2)` ` " "= 10 ^(x) *10 ^(10x) (log _e 10)^(2)` |
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| 47. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए - ` tan ^(-1) ((sqrt(x)-x ) /(1+x^(3//2)))` |
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Answer» Correct Answer - ` (1)/(2sqrt(x) (1+x))- (1)/(1+x^(2))` ` x^(1//2 ) tan theta ` व ` x= tan alpha ` रखने पर |
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| 48. |
` (log sin ^(-1) x ^(2) )cos (cot ^(-1) x^(2))` |
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Answer» Correct Answer - ` (log sin ^(-1) x^(2))(2xsin (cot^(-1)x^(2)))/(1+x^(4))+ (2x cos (cot ^(-1)x^(2)))/(sin ^(-1) x^(2)sqrt(1-x^(4)))` ` (d)/(dx) [(log sin ^(-1) x^(2) ) cos(cot ^(-1) x^(2))]` ` =(log sin ^(-1) x^(2))(d)/(dx) [cos (cot ^(-1) x ^(2))] +cos (cot ^(-1) x^(2) )(d)/(dx) (log sin ^(-1)x^(2))` ` =[-(log sin ^(-1) x^(2) )] sin (cot ^(-1) x^(2) )(d)/(dx) (cot^(-1) x^(2))` ` " "+[cos (cot^(-1) x^(2) )] [(-1)/(sin ^(-1) x^(2) )* (d)/(dx) (sin ^(-1) x^(2))]` ` " "=[ -(log sin ^(-1) x^(2) )sin (cot ^(-1) x^(2) )* ((-1)/(1+(x^(2))^(2) ) ) (d)/(dx) (x^(2))]+` ` " "[ cos (cot ^(-1) x^(2) )* (1)/(sin ^(-1) x^(2) )* (1)/(sin ^(-1) x^(2)) *(1)/(sqrt(1-(x^(2))^(2) ))*(d)/(dx) (x^(2))]` |
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| 49. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए - ` sin ^(-1) ((2x )/(1+x^(2)))` |
| Answer» Correct Answer - ` (2)/(1+x^(2)) ` | |
| 50. |
`tan ^(2) ((pi x ^(2))/(2))` का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए| |
| Answer» Correct Answer - ` 2pixtan ((pix^(2))/(2) ) sec^(2) ((pix^(2))/(2))` | |