InterviewSolution
Saved Bookmarks
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.
| 51. |
यदि ` y= sqrt (sinx +sqrt(sin x +sqrt sin x +......))` तब सिद्ध कीजिए की `(dy)/(dx) =(cos x )/(2y-1) ` |
|
Answer» `because " "ysqrt (sinx +sqrt sin x +sqrt(sinx +sqrt...+infty))` ` " "= sqrt (sinx +y)` ` " "(because y= sqrt (sin x + sqrtsin x + sqrt(sin x +sqrt ...+infty ) ))` ` " "therefore y^(2) =sinx +y` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, ` " "2y(dy)/(dx) =(d)/(dx) sin x + (dy)/(dx) ` ` rArr" "(2y -1) (dy)/(dx)=cos x ` ` therefore " "(dy)/(dx) =(cos x )/(2y-1) ` |
|
| 52. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` tan ^(n) (px^(m))` |
|
Answer» Correct Answer - ` p mnx ^(m-1) tan ^(n-1) (px^(m)) sec^(2) (px^(m) ` माना ` px^(m) =t " "rArr (dt)/(dx) =pm x^(m-1)` |
|
| 53. |
(i ) यदि `y= log sqrt(((1+ sin x )/(1-sin x ))),` तब ` (dy)/(dx)` ज्ञात कीजिए (ii) यदि ` y= log sqrt(((1-cos x)/(1+cos x ))), `तब `(dy)/(dx)` ज्ञात कीजिये| |
|
Answer» Correct Answer - `(i) secx " "(ii) cosecx` (i ) माना ` sqrt((1+sin x )/(1-sinx ))= t" "rArr " "(1)/(2) log ((1+sin x )/(1-sin x ))=log t ` ` rArr " "[ (1)/(2) (1)/(((1+sin x)/(1-sinx)))*(d)/(dx) ((1+ sin x )/(1-sinx ))]dx =(1)/(t) dt ` ` rArr (1)/(t)(dt)/(dx) =((1-sinx ))/(2(1+sin x ) ) ((1-sinx ) (d)/(dx) (1+ sinx ) -(1+sinx )(d)/(dx) (1- sin x ))/((1-sin x)^(2))` सरल करने के बाद `(dt)/(dx) =sqrt((1+sin x)/(1-sinx ))sec x` तब ` " "(dy)/(dx) =(d)/(dt) (logt ) (dt)/(dx) ` का प्रयोग करने पर |
|
| 54. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए - ` sin ^(-1) ((1-x^(2))/(1+x^(2)))` |
|
Answer» Correct Answer - `(-2)/(1+x^(2))` `y= sin ^(-1) ((1-x^(2))/(1+x^(2)))" "rArr " "(dy)/(dx) =(d)/(dx) [sin ^(-1) ((1-x^(2))/(1+x^(2)))]` ab ` x= tan theta ` rakhne पर v ` (1-tan ^(2) theta )/(1+tan theta ) ` का प्रयोग करने पर |
|
| 55. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए - ` cos ^(-1) ((x-x^(-1))/(x+x^(-1)))` |
|
Answer» Correct Answer - ` (-2)/(1+x^(2))` माना ` y= cos ^(-1) ((x^(2) -1)/(x^(2)+1))` व ` x = tan theta ` रखने पर tb ` y= cos ^(-1) [-((1-tan ^(2)theta )/(1+ tan ^(2) theta ))]` ` " "= cos ^(-1) (-cos 2 theta )= cos ^(-1) [cos (pi -02theta )]` ` " "=pi - 2theta = pi -2tan ^(-1) x ` ` " "rArr " "(dy)/(dx) =(-2)/(1+x^(2))` |
|
| 56. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए - ` tan ^(-1) [cot x + cose cx ]` |
| Answer» Correct Answer - ` -(1)/(2)` | |
| 57. |
`tan ^(-1)x ` का ` sin ^(-1) x ` के सापेक्ष अवकल गुणांक ज्ञात कीजिये| |
| Answer» Correct Answer - ` (sqrt( 1-x^(2)))/( 1+x^(2))` | |
| 58. |
यदि ` y= log sec (ax+b )^(3)`तब `(dy)/(dx) ` का मान ज्ञात कीजिए| |
|
Answer» Correct Answer - `3(ax+b) ^(2) tan (ax+_b)^(3) ` माना ` " "sec (ax+ b )^(3) =t ` ` rArr 3a (ax+b ) ^(2) *sec (ax+b ) ^(3) *tan (ax+ b) ^(3) dx =dt ` ` rArr " "(dt)/(dx) =3a (ax+b ) ^(2) tan (ax+b) ^(2) sec (ax+b ) ^(3)` ` " "(dy)/(dx) =(d)/(dt)(logt ) *(dt)/(dx) ` का प्रयोग करने पर |
|
| 59. |
यदि ` y= e ^(x) (sin x +cos x ),` सिद्ध कीजिए की` (d^(2)y)/(dx^(2) )-2 (dy)/(dx) +2y =0` |
|
Answer» यहाँ ` y= e^(x) (sinx +cos x ) ` x के सापेक्ष अवकलन करने पर `(dy)/(dx) =e^(x) *(d)/(dx) (sin x +cos x )+ (sin x +cos x )(d)/(dx)e^(x)` ` " "e^(x)(cos x- sin x)+ (sin x +cos x )*e^(x)` पुनः अवकलन करने पर ` (d^(2)y) /(dx^(2))=2*(d)/(dx) (e^(x) cos x ) ` ` " =2 { e^(x) *(d)/(dx) (cos x )+ cos x (d)/(dx) (e^(x))}` ` " "=2 *(e^(x) (-sin x )+ (cos x ) e^(x) } ` ` " "= 2e^(x) (cos x -sin x )` अब ` ((d^(2)y)/(dx^(2))-2 (dy)/(dx) +2y )=2e^(x) (cos x -sin x ) -4e ^(x) cos x + 2e ^(x) (sin x +cos x )=0` |
|
| 60. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए - ` cos ^(-1) (4x^(3) -3x )` |
|
Answer» Correct Answer - ` (-3)/(sqrt(1-x^(2)))` `x= cos theta `rakhne पर tb,`(d)/(dx) [cos ^(-1) (4x^(3) -3 x)]= (d)/(dx) [cos ^(-1) (4cos ^(3) theta -3cos theta ) ` ` " "= (d)/(dx) [cos ^(-1) (cos 3theta)]` ` " "(d)/(dx) (3theta )=(d)/(dx) (3cos ^(-1)x )= (-3 )/(sqrt( 1-x^(2)))` |
|
| 61. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए - ` tan ^(-1) [(4sqrt (x))/(1-4x)]` |
| Answer» Correct Answer - ` (2)/((1+4x)sqrt(x))` | |
| 62. |
x के सापेक्ष `see (tan ^(-1)x ) ` का अवकल गुणांक ज्ञात कीजिए | |
|
Answer» माना ` " "y= sec (tan ^(-1)x)` माना ` tan ^(-1) x=t ` ` therefore " y= sec t ` या ` (dy)/(dx) =(d)/(dt) sect (dt)/(dx) =sec t tan t(d)/(dx) (tan ^(-1)x )` ` " " =(sec (tan ^(-1) x) tan (tan ^(-1)x ))/(1+ x^(2))` ` " "=(x)/( 1+x ^(2))sec(tan ^(-1) x )` |
|
| 63. |
यदि ` e^(y) (x+1)=1 ` सिद्ध कीजिए की ` (d^(2)y)/(dx^(2))=((dy)/(dx))^(2)` |
|
Answer» यहाँ ` " "e^(y) (x+1) =1 ` अथवा ` e^(y)=(1)/(x+1)` दोनों पक्षों का log लेने पर ` " "y= log {(1)/((x+1))} =log 1-log (x+1)` ` " "y= -log (x+1 )` अब, x के सापेक्ष अवकलन करने पर ` (dy)/(dx) =(-1)/((x+1))` व ` (d^(2)y)/(dx^(2))=(1)/((x+1)^(2))=((dy)/(dx))^(2)` |
|
| 64. |
`(i) (tan ^(3) x )/(ax^(2) +b)" "(ii) (log(cos x ))/(tan (log x ))" "(iii) (sqrt(sin x ))/(sin sqrt(x ))` |
| Answer» Correct Answer - `(i) (3tan ^(2) xsec ^(2)x (ax^(2)+b )- 2ax tan ^(3) x )/((ax^(2) +b)^(2) ) " "(ii) (-xtanx tan (logx )-log (cos x )sec^(2)(logx ))/(xtan ^(2) (log x )) (iii) (sqrt(x ) (sinx )^(-1//2) cos x sin sqrt (x) -sqrt sin x cos sqrt (x))/(2sqrt (x) (sin sqrt(x))^(2))` | |
| 65. |
x के सापेक्ष ` tan ^(-1)""(4x)/(4-x^(2)) ` का अवकल गुणांक ज्ञात कीजिए| |
|
Answer» माना ` y= tan ^(-1) ""(4x)/( 4-4x^(2)) ` ` rArr " "y= tan ^(-1) (x) /(1-(x//2) ^(2)) ` ` " "= tan ^(-1) [ (2(x//2))/(1-(x//2)^(2)) ] =2tan ^(-1) ((x)/(2))` माना ` x//2 =t " "therefore " "y= 2tan ^(-1)(t) ` ` therefore(dy)/(dx) =2( d)/(dt) tan ^(-1) t (d)/(dx) ((x)/(2))` `" "= 2 (1)/(1+t^(2) )*(1)/(2) =(1)/(1+((x)/(x))^(2) )= (4)/(4+x^(2))` |
|
| 66. |
`(dy)/(dx) ` का मान ज्ञात कीजिए|`(i) y=tan ^(-1) {sqrt( 1+cos x ) /( 1-cos x )}` ` y= tan ^(-1) { (1+ sin x )/(1-sin x ) } ` |
|
Answer» (i ) हम जानते है `y= tan ^(-1) { (1+cos x) /( 1-cos x ) } ` `= tan ^(-1) { sqrt(( 2cos ^(2) (x//2)) /( 2sin ^(2)(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)` (ii) हम जानते है ` y= tan^(-1) {sqrt((1+sin x )/( 1- sin x )}} ` ltbr gt` =tan ^(-1) {(1-cos ""((pi )/(2) +x) )/( 1+ cos""((pi)/(2) +x) ) } ^(1//2)` ` " "=tan ^(-1) { (2sin ^(2) ""((pi )/(4) + (x)/(2) ) )/( 2cos""(( pi)/(4) +(x)/(2) ))} ^(1//2)` ` " "=tan ^(-1) { tan ""( (pi)/(4) + (x)/(2))}= ((pi)/(4)+(x)/(2)) ` ` therefore " "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) ` |
|
| 67. |
` (i) e^(sqrt( sin x ) )" "(ii) e^(sqrt(tan x))" "(iii)e^(sqrt(sec x ))` |
| Answer» Correct Answer - `(i) (e^(sqrt(sinx )))/(2sqrtsin x)*cos x " "(ii) (sec^(2)x) /(2sqrttan x )e^(sqrt(tan x ))" "(iii) (tan xsqrtsecx)/(2) e^(sqrt(secx ))` | |
| 68. |
निम्न फलनों का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए| ` (i) tan ^(-1) (1+ x+x^(2))` ` (ii) )(sin ^(-1) x)^(2) -(cos ^(-1) x)^(2)` ` (iii) llog tan ^(-1) x^2)` |
|
Answer» माना ` y= tan ^(-1)(1+ x+x^(2) ),` माना ` (1+ x+x^(2) ) =t` ` therefore y= tan ^(-1) t ` ` therefore (dy)/(dx) =(d)/(dt ) tan ^(-1) t (d)/(dx)(1+ x+x^(2)) ` ` " "= (1)/(1+ t^(2) )(0 +1 +2x) ` ltbr gt` " "= ( 1+ 2x) /( 1+ (1+ x+x^(2)) ^(2))` माना `y=(sin ^(-1) x)^(2) -(cos ^(-1) x)^(2) ` माना `sin ^(-1) x=t_1` तथा `cos ^(-1) x=t_2 ` ` therefore " " y= t _1^(2) -t_2^(2) ` ` therefore (dy)/(dx)=(d)/(dt_1 )t_1^(2) (d)/(dx) t_1 -( d)/(dt_2 ) t_2^(2) (d)/(dx) t_2` ` " "=2t _1 (d)/(dx) sin ^(-1) x- 2t_2 (d)/(dx) cos ^(-1) x ` ` = (2sin ^(-1) x )/( sqrt ( 1-x^(2)) ) + (2cos ^(_1) x)/( sqrt (1-x^(2)))` ` " "= (2) /(sqrt ( (1-x^(2)) ) )[ sin ^(-1) x + cos ^(-1) x ] ` (iii) माना ` y= log tan ^(-1) x^(2) , `माना ` tan ^(-1) x^(2) =t ` ` therefore y= log t ` ` therefore *(dy)/(dx)= (d)/(dt )log t (dt)/(dx) =(1)/(t) (d)/(dx) tan ^(-1) x^(2) ` ` =(1)/(t) *( 1) /(1+(x^(2) )^(2) )*2x =(2x) /((1+ x^(4))tan ^(-1) x^(2))` |
|
| 69. |
निम्न फलनों का x का सापेक्ष अवकल गुणांक ज्ञात कीजिए| ` (i) tan ^(-1) [(sin x) /( 1+ cos x }]` ` (ii) tan ^(-1) [ sqrt ((1+ cos x )/( 1- cos x ))]` `(iii) tan ^(-1)[(sin x +cos x )/( cos x- sin x )] ` |
|
Answer» माना ` y= tan ^(-1) [(sin x)/( 1cos x ) ]` ltbr gt ` rArr " "y= tan ^(-1) [(2sin ( x)/(2) cos ""( x)/(2) ) /(1+2cos ^(2)""( x)/(2) -1)]` ` =tan ^(-1) [tan ((x)/( 2 ) )] =(x)/(2)` ` " " ( because tan ^(-1) (tan theta ) =theta ) ` ` therefore " "(dy)/(dx) =(1)/(2)` (ii) माना ` y=tan ^(-1) [ sqrt ((1+cos x ) /( 1- cos x ))]` ` " " y=tan ^(-1) [(1+2cos ^(2) (x//2)-1 ) /(1-1+2sin ^(2)(x//2))]` ` =tan ^(-1) cot (x//2) ` ` =tan ^(-1) tan ""((pi )/(2)- ( x)/(2) )= (pi)/(2) -(x)/(2) ` ` " "(because tan ^(-1) (tan theta )= theta ) ` ` therefore (dy)/(dx) =-(1)/(2) ` (iii) माना `y= tan ^(-1)[ ( sin x + cos x ) /(cos x - sin x )]` दाँयें पक्ष में अंश व् हर को cos x से भाग देने पर ` y= tan ^(-1) [(1+tan x) /( 1- tan x) ] ` ` = tan ^(-1) tan ""((pi)/(4) +x ) =(pi)/( 4) +x ` ` therefore" "(dy)/(dx) =(d)/(dx) ((pi) /(4) +x) =1 rArr (dy)/(dx) = 1 ` |
|
| 70. |
निम्न फलनों का x के सापेक्ष अवकल गुणांक ज्ञात कीजिये ` (i) (tan ^(-1) x ) /( x ) ` ` (ii) x sec ^(-1) x ^(3) `` (iii) (sin ^(-1) x )^(m)` |
|
Answer» (i ) माना ` y= (tan ^(-1) x ) /( x ) ` ` therefore (dy)/(dx) =( x(d)/(dx) tan ^(-1) x-tan ^(-1) x (d)/(dx) x)/( x^(2) ) ` ` " "( ( x)/( 1+x^(2) )- tan ^(-1) x)/( x^(2) )` या ` (dy)/(dx) =(x -(1+x^(2) )tan ^(-1) x)/( x^(2) (1+x^(2) ) ) ` (ii) माना ` y = sec ^(-1) x^(3) ` ` therefore " "(dy)/(dx) =x (d)/(dx) sec ^(-1) x^(3) + sec^(-1)x^(3) (d)/(dx) x ` ` = x(1xx3x^(2) )/( x^(3) sqrt (x^(6)) -1)+ sec ^(-1) x^(3) ` ` = (3)/(sqrt (x^(6)) -1 ) +sec ^(-1) x^(3)` माना ` y= (sin ^(-1) x)^(m) ` ` therefore " "(dy)/(dx) =(d)/(dx) (sin^(-1) x )^(m) ` ` " "=m(sin ^(-1) x ) ^(m-1) (d)/(dx) sin ^(-1) x ` ` rArr" "(dy)/(dx) =(m (sin ^(-1) x) ^(m-1))/(sqrt(1-x^(2)))` |
|
| 71. |
निम्न फलनों का x के सापेक्ष अवकल गुणांक ज्ञात कीजिये -` (i) cot ^(-1) ((x)/(a) ) ` `(ii) tan ^(-1) (2x+1) `` " (iii) log tan ^(-1) x ` `(iv) tan ^(-1) (mtan x ) ` |
|
Answer» (i ) माना ` y= cot ^(-1) ((x)/(a) ) ` माना ` (x)/(a) =t ` ` therefore " " y= cot^(-1) (t) ` ` " " therefore (dy)/(dx) =(d)/(dt) cot ^(-1) (t) (d)/(dx) ((x)/a) ` ` " "= (-1)/( a+t^(2) )*(1)/(a) =- ( a^(2))/( a^(2) + x^(2)) *(1)/(a ) =(-a)/( a^(2) + x^(2)) ` (ii) माना ` y= tan ^(-1) (2x+ 1) ` माना ` (2x+ )=t ` ` therefore y= tttan ^(-1) (t)` ` therefore ( dy)/(dx) =(d)/(dt) tan ^(-1) ""(d)/(dx) (2x+ 1) =(1.2) /(1+t^(2))` ` " "= (2) /(1+( 2x+ 1) ^(2) )` (iii) माना ` y= log tan ^(-1) x, tan ^(-1) x=t ` ` therefore " "(dy)/(dx)=(d)/(dt) log t (d)/(dx) tan ^(-1) x ` ` =(1)/(t)* (1) /(1+x^(2)) = (1)/(tan ^(-1)x(1+ x^(2))` (iv) माना ` y=tan ^(-1) (mtan x) ` ` therefore (dy)/(dx) =(d)/(dt) tan ^(-1) (m tan x ) ` ` " "= (1)/( 1+ ( m tan x )^(2))(d)/(dx) ( mtan x )` या ` (dy)/(dx) =( msec^(2)x ) /( 1+ (m tan x )^(2))` |
|
| 72. |
यदि ` y^(x) =x ^(sin y)` तो `(dy)/(dx) ` का मान ज्ञात कीजिए| |
|
Answer» दिया है ` " " y^(x) =x ^(sin y) ` `rArr " "xlog y= sin ylog x ` (दोनों पक्षों का लघुगणक लेने पर) x के सापेक्ष अवकलन करने पर ` x*(1)/(y) (dy)/(dx) +log y*1sin y((1)/(x))+log x cos y(dy)/(dx)` `rArr ((x)/(y)- cos y log x )(dy)/(dx) =(sin y)/(x ) -log y` ` rArr" "((x-ycos ylog x )/(y))(dy)/(dx) =(sin y- xlog y)/(x )` ` rArr " "(dy)/(dx)= (y(sin y-x log y))/(x(x-ycos ylog x ))` |
|
| 73. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` sin ^(n) (ax+ b)` |
| Answer» Correct Answer - ` ansin ^(n-1) (ax+b) cos (ax_b) ` | |
| 74. |
` e^(x^(m))` |
|
Answer» Correct Answer - ` mx^(m-1)e^(x^(m))` माना ` x^(m) =t " "rArr mx ^(m-1 ) dx =dt ` |
|
| 75. |
निम्न फलनों का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए| ` (i)" "(1)/((x+p)(x+q)(x+r)) ``(ii)" "((x-sin x )^(3//2))/(sqrt(x))` |
|
Answer» (i )माना ` " "y= (1)/((x+p)(x+q)(x+r))` ` rArr log y=log [(1)/((x+p)(x+q)(x+r))]` ` " "` (दोनों ओर का लघुगणक लेने पर) ` " "= log 1- log (x+p)-log (x+q) -log (x+r)` x के सापेक्ष दोनों पक्षों का अवकलन करने पर ` (1)/(y) (dy)/(dx) =0-(1)/(x+p)-(1)/(x+q) -(1)/(x+r)` ` rArr (dy)/(dx) =-y[(1)/(x+p) +(1)/(x+q)+(1)/(x+r) ]` ` " "=- (1)/((x+p)(x+q)(x+r) )[(1)/(x+p)+ (1)/(x+q)+(1)/(x+r)]` `(ii) y= (x-sin x )^(3//2)/(sqrt(x))` ` rArr " "log y= log [ ((x-sin x )^(3//2))/sqrt(x)]` ` " "=(3)/(2) log (x-sin x ) -(1)/(2)log x ` दोनों पक्षों का x के सापेक्ष अवकलन करने पर ` " "(1)/(y)(dy)/(dx) =(3)/(2) *(1)/(x-sin x )(1-cos x ) -(1)/(2)*(1)/(x)` `rArr " "(dy)/(dx) = y [(3)/(2)*(1-cos x )/((x-sin x ) )- (1)/(2x )]` ` rArr" "(dy)/(dx) =((x - sin x)^(3//2))/(2sqrt(x))[(3(1-cos x ))/(x-sin x ) -(1)/(x)]` |
|
| 76. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` cos(tan x ^(2))` |
| Answer» Correct Answer - ` -2x sin (tan x^(2) ) sec^(2) x^(2)` | |
| 77. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|`log tan (x//2)` |
|
Answer» Correct Answer - `cosecx` माना ` tan ""(x)/(2) =t " "rArr (dt)/(dx) =(1)/(2) sec ^(2) ""(x)/(2)` तब ` (d)/(dx) (log tan ""(x)/(2)) =(d)/(dt) (log t ) (dt)/(dx) =(d)/(dt) (logt )*(1)/(2) sec ^(2)""(x)/(2)` ` " "=(1)/(t)*( 1)/(2) sec ^(2) ""(x)/(2)=(1)/(2t) sec ^(2) ""(x)/(2) ` ` " "= (1)/(2) (cos x//2)/(sin x //2) *(1)/(cos ^(2)x//2) = (1)/(sin x )= cosec x ` |
|
| 78. |
यदि ` y= (1+(1)/(x) )^(x) + x^(1+(1)/(x))` तो `(dy)/(dx) ` का मान ज्ञात कीजिए| |
|
Answer» दिया है- ` " "y= (1+(1)/(x))^(x)+x ^(1+(1)/(x))` माना ` " "u =(1+(1)/(x) ) ^(x)` तथा ` " "v=x^(1+(1)/(x))` ` rArr " "y=u +v " "rArr(dy)/(dx)= (du)/(dx)+(dv)/(dx)" "...(1)` अब ` " "u= (1+(1)/(x))^(x) rArr log u =x log (1+(1)/(x))` दोनों पक्षों का x के सापेक्ष अवकलन करने पर ` " "(1)/(u) (du)/(dx) =x (d)/(dx) log (1+(1)/(x) )+log (1+(1)/(x))(d)/(dx) (x)` ` " "= x *(1)/(1+(1)/(x))*(d)/(dx) (1+(1)/(x)) +log(1+(1)/(x))*1` `rArr " "(du)/(dx)=u [(x^(2)/x+1)(0-(1)/(x^(2)))+ log (1+(1)/(x))]` ` " "= (1+(1)/(x))^(x) [log (1+(1)/(2))-(1)/(1+x)] " "...(2)` अब ` v=x^(1+(1)/(x))" "rArrlog v =(1+(1)/(x) ) log x ` दोनों पक्षों का x के सापेक्ष अवकलन करने पर ` rArr " "(1)/(v) (dv)/(dx) =(1+(1)/(x))*(1)/(x) +log x*(0-(1)/(x^(2)))` ` rArr " "(dv)/(dx)=v [(1)/(x) +(1)/(x^(2) )-(1)/(x^(2))log x ]` ` " "= x^(1+(1)/(2))[(1)/(x)+(1)/(x^(2))(1-log x )]` समीकरण (1 ),(2 ),व (3 ) से, ` (dy)/(dx) = (1+(1)/(x))^(x) [log (1+(1)/(x))-(1)/(1+x)]` ` " "+ x^(1+(1)/(x))[ (1)/(x)+ (1)/(x^(2))(1-log x )]` |
|
| 79. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` sin ^(2) (log x ^(2))` |
| Answer» Correct Answer - ` (4)/(x) sin (logx^(2)) cos (log x^(2))` | |
| 80. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` log (sec x+ tan x ) ` |
| Answer» Correct Answer - `secx ` | |
| 81. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` log tan ((pi)/(4)+(x)/(2))` |
| Answer» Correct Answer - ` sec x ` | |
| 82. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` (i)tan x^(@) " "(ii) sec x ^(@)` |
|
Answer» Correct Answer - ` (pi)/(180 ) sec^(2) ""(pix)/(180)(ii) (pi)/(180 ) tan ""(pix)/(180 ) ` `(i) (d)/(dx) (tan x^(@) )= (d)/(dx) (tan""(pix )/(180 ))` माना ` " "(pix)/(180 ) =t " "rArr (dt)/(dx) =(pi)/(180 ) ` ` rArr (d)/(dx) (tan ^(@) ) =(d)/(dx)(tan""(pix )/(180 ) )= (d)/(dt) (tan t ) (dt)/(dx)` ` " "= (d)/(dt) (tan t ) ""*(pi)/(180) =(pi)/(180 ) sec ^(2)t= (pi)/(180 ) sec ^(2) ((pix)/(180))` |
|
| 83. |
यदि `x= a cos theta , y= a sintheta ` तब ` (dy)/(dx)` का मान ज्ञात कीजिए| |
|
Answer» दिया है-` " "x= a cos theta , y=asin theta ` ` rArr" "(dx)/(d theta )=-asin theta ,(dy)/(d theta )= a cos theta ` `therefore " "(dy)/(dx) =(dy//d theta )/(dx//d theta )=(acos theta )/(-asin theta )=-cot theta ` |
|
| 84. |
निम्नलिखित प्रश्न में x के सापेक्ष अवकलन कीजिये- ` sin (tan ^(-1) e^(-x))` |
| Answer» Correct Answer - ` (-e^(-x) cos (tan ^(-1) e^(-x)))/(1+e^(-2x ))` | |
| 85. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` (log sinx )^(2)` |
|
Answer» Correct Answer - ` 2cot x log sin x ` माना ` log sin x=t` ` rArr (dt)/(dx) =cot x " "rArr (d)/(dt) (t^(2) ) (dt)/(dx)` ` " "= (d)/(dt) (t^(2))cot x =2t cot x= 2cot x (log sin x)` |
|
| 86. |
`sin 3x cos 5x` का अवकलन कीजिये| |
| Answer» Correct Answer - ` (4cos 8x -cos 2x )` | |
| 87. |
यदि ` x= (3at )/(1+t^(3) )` तथा ` y=(3at ^(2))/(1+t^(3))` तो ` (dy)/(dx)` का मान ज्ञात कीजिए| |
|
Answer» दिया है -` x= (3at )/(1+t^(3))` तथा ` y= (3at ^(2))/(1+t^(3))` ` rArr " "(dx)/(dt) =(3a*(1+t^(3) )*(d)/(dt) *t-3at *(d)/(dt)(1+t^(3)))/((1+t^(3) )^(2))` ` =(3a (1+t^(3))*1-3at *3t^(2))/((1+t^(3))^(2))` ` " "= 3a ((1-2t^(3))/((1+t^(3) )^(2)))" "...(1)` इसी प्रकार ` " "y= 3a *(t^(2))/(1+t^(3))` `rArr(dy)/(dt)=3a *((1+t^(3))(d)/(dt) t^(2) -t^(2)*(d)/(dt) (1+t^(3)))/((1+t^(3))^(2))` ` " "=3a ((1+t^(3) )*2t -t^(2) *3t^(2))/((1+t^(3))^(2) )= 3a (t*(2-t^(3)))/((1+t^(3))^(2))` अतः ` (dy)/(dx) =(dy//dt)/(dx//dt)=(3at (2-t^(3)))/((1+t^(3))^(2))xx((1+t^(3))^(2))/(3a(1-2t^(3)))` ` " "(dy)/(dx)=(t(2-t^(3)))/(1-2t^(3))` |
|
| 88. |
निम्नलिखित प्रश्न में x के सापेक्ष अवकलन कीजिये- ` log (log x ) ,xgt 1 ` |
| Answer» Correct Answer - ` (1)/(xlog x ); xgt 1 ` | |
| 89. |
`sin ^(-1) x ` का ` cos ^(-1) sqrt((1-x^(2))) ` के सापेक्ष अवकल गुणांक ज्ञात कीजिये| |
| Answer» Correct Answer - `1` | |
| 90. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` sin [(x)/(1+sqrt(x))]` |
|
Answer» Correct Answer - ` cos ((x)/(1+sqrt(x))) (2+sqrt(x))/(2(1+sqrt(x))^(2))` माना ` (x)/(1+sqrt(x))=t rArr ((1+sqrt(x))(d)/(dx) (x)-x (d)/(dx) (1+sqrt(x)))/((1+ sqrt(x))^(2))dx=dt ` सरल करने पर ` " "(2+sqrt(x))/(2(1+sqrt(x))^(2))=(dt)/(dx)` इन मानो का उपयोग करने पर ` " "(d)/(dx) [sin ((x)/(1+sqrt(x)) ) ]=(d)/(dt) (sint )*(dt)/(dx)` |
|
| 91. |
यदि ` y=log sin x +tan x` तब `(dy)/(dx) ` का मान ज्ञात कीजिए| यदि ` x=(pi)/(3)*` |
|
Answer» ` " " y= logsin x+ tan x ` ` therefore " "(dy)/(dx) =(d) /(dx) log sin x +(d) /(dx) tan x ` माना ` sin x =t` ` therefore " "(dy)/(dx) =(d)/(dt) log t (d)/(dx) sin x +sec ^(2) x= (1)/(t) cos x+sec ^(2) x ` ` " "= (cos x ) /(sinx ) +sec ^(2) x = cot x +sec ^(2)x ` ` therefore " "((dy)/(dx) ) _((x=pi //3))=cot ""(pi)/(3)+sec ^(2)"" (pi)/(3) ` ` " "= (1)/(sqrt( 3))+ 4 =4 +(1)/(sqrt(3)) ` |
|
| 92. |
` sqrt(ax^(2) +bx +c)` का अवकलन कीजिये| |
| Answer» Correct Answer - ` (ax+b )/(sqrt(ax^(2) +bx +c)) ` | |
| 93. |
`sin (log x) ,xgt 0` का x के सापेक्ष अवकलन कीजिये| |
| Answer» Correct Answer - `(cos (log x ))/(x) ` | |
| 94. |
यदि ` y= [log (x+ sqrt(x^(2)+1))]^(2)` तो सिद्ध कीजिये की` " "(x^(2)+1) (d^(3)y)/(dx^(3))+3x (d^(2)y)/(dx^(2))+(dy)/(dx)=0` |
|
Answer» दिया है- ` " "y= [log (x+ sqrt(x^(2)+1)]^(2)` दोनों पक्षों का x के सापेक्ष अवकलन करने पर ` (dy)/(dx) =2log (x+sqrt(x^(2)+1))(d)/(dx) log (x+sqrt(x^(2)+1))` ` =2 log (x+ sqrt(x^(2)+1)*(1)/(x+sqrt(x^(2)+1))*(d)/(dx) (x+sqrt(x^(2)+1))` ` =(2log (x+sqrt(x^(2 +1))))/(x+sqrt(x^(2)+1))*(1+(1)/(2sqrt(x^(2)+1))2x)` ` (2log (x+sqrt(x^(2)+_1)))/(x+sqrt(x^(2)+1))*(sqrt(x^(2)+1)+x)/(sqrt(x^(2) +1))` ` =(2log (x+sqrt(x^(2) +1)))/(sqrt(x^(2)+1))` `rArr" " (sqrt(x^(2)+1))(dy)/(dx) =2log (x+sqrt(x^(2)+1 ))` ` rArr" "(x^(2)+1) ((dy)/(dx) )^(2)=4 [log (x+ sqrt(x^(2)+1 ))]^(2) ` ` rArr " "(x^(2)+1) ((dy)/(dx) )^(2)=4y` अब दोनों पक्षों का x के सापेक्ष अवकलन करने पर `(x^(2)+1) *2((dy)/(dx) )((d^(2)y)/(dx^(2))) +((dy)/(dx))^(2)*2x =4 (dy)/(dx)` ` rArr " "2(x^(2)+1)(d^(2)y)/(dx^(2))+x (dy)/(dx) =2 ` पुनः दोनों पक्षों का x के सापेक्ष अवकलन करने पर ` [(x^(2)+1)(d^(3)y)/(dx^(3) )+2x (d^(2)y)/(dx^(2)) ]+[(dy)/(dx)+ x(d^(2)y)/(dx^(2))]=0` ` rArr " " (x^(2)+1) (d^(3)y)/(dx^(3))+3x (d^2y)/(dx^(2))+(dy)/(dx) =0` |
|
| 95. |
निम्न फलन का अवकल गुणांक ज्ञात कीजिये- x^(sin ^(-1))` का ` sin ^(-1)x` के सापेक्ष |
| Answer» Correct Answer - ` (i) x^(sin^(-1))((sqrt(1-x^(2))*sin ^(-1)x )/(x )+ log x )(ii) (x ^(sin x ) [(sin x )/(x ) + cos x log x ])/((sin x )^(x) [x cotx +log sin x ])(iii) (2)/(x) (iv) e^(sin ^(-1)x)(v) a^(sin ^(-1))log _e a ` | |
| 96. |
फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए| ` e^(sqrtsin x+sqrt cos x)` |
|
Answer» Correct Answer - ` e^(sqrt(sin x)+ sqrt cos x ) [(cos x)/(2sqrtsin x )- (sinx )/(2sqrtcos x)]` ` (d)/(dx) (e^(sqrt(sin x )+ sqrt(cos x )))= (d)/(dt) (e^(t) ) (dt)/(dx)" "` जहाँ ` t= sqrt (sin x )+ sqrt (cos x)` ` rArr " "(dt)/(dx) =(cos x )/(2sqrt sin x )-(sinx )/(2sqrt (cos x ))` ` therefore " "(d)/(dx) (e^(sqrt(sin x)+ sqrt(cos x ) ) ) =(d)/(dt) " " ` (सरल करने पर) |
|
| 97. |
फलन ` y= log (sqrt ( x) +(1)/(sqrt(x)) ) , ` तब ` (dy)/(dx) ` का मान ज्ञात कीजिए| |
|
Answer» `y=log (sqrt(x)+ (1)/(sqrt(3))),` माना ` sqrt(x) +(1)/(sqrt(x) ) =t ` ` " "therefore " "y= logt ` ` therefore " "(dy)/(dx) =(d)/(dt) log t (d)/(dx)(x^(1//2) + x^(-1//2) ` ` " "(1)/(t) [ (1)/(2)x^(-1//2)+(-(1)/(2) ) x^(-3//2) ] ` ` " "= (1)/(2t) [ (1)/(sqrt(x) ) -(1)/(xsqrt(x)) ] =(1)/(2((x+1)/(sqrt(x)))) [(x-1) /(xsqrt(x) )]` ` " "= ((x-1))/(2x(x+1)) ` |
|
| 98. |
`(ax+b )^(m)` का अवकलन कीजिये| |
| Answer» Correct Answer - `ma (ax+ b)^(m-1) ` | |
| 99. |
निम्न फलनों का अवकलन गुणांक ज्ञात कीजिए ` (i) sin 3x " "(ii) (4x+ 7)^(4) " "(iii) log (ax+ b ) ` |
|
Answer» माना `" "y=sin 3x,` माना 3x =t अतः ` " "y=sin t ` `" " therefore " "(dy)/(dx) =(d )/(dx) sin t=(d)/(dt) sin t (dt)/(dx) ` ` " "=cos t (d)/(dx) (3x) =3cos t ` ` " " therefore (dy)/(dx) =3cos 3x ` (ii) माना ` y= (4x+7)^(4) , ` माना `4x+ 7=t ` अतः ` " "y= t^(4) ` ` therefore (dy)/(dx) =(d)/(dx) t^(4) =(d)/(dt) t^(4)*(dt)/(dx) ` `" " =4t^(3) (d)/(dx) (4x+ 7) ` ` " "=4t^(3) (4(d)/(dx) x+ (d)/(dx) 7) ` ` " "=4t^(3) (4*1+0 )= 16t^(3)` `therefore " "(dy)/(dx) =16 (4x+7 ) ^(3) ` ,माना `y =log(ax+b) ` , माना `(ax+b) =t` ` "therefore " "(dy) /(dx) =(d)/(dx) log t =(d) /(dt) log t (d)/(dx) (ax+b ) ` ` " "=(1)/(t) (a (d)/(dx) x+ (d) /(dx) b) =(a)/(t)=(a) /((ax+b) ) ` ` therefore " "(dy)/(dx) =(a)/((ax+ b))` |
|
| 100. |
यदि ` (cos x)^(y)` =(sin y) ^(x) , ` तो `(dy)/(dx) ` का मान ज्ञात कीजिए| |
|
Answer» दिया है `" " (cosx )^(y) =(sin y)^x` ` rArr " "ylog cos x = x log sin y` ` rArr y(-sin x)/(cos x ) +log cosx *(dy)/(dx) =x*(1) (siny)cos y* (dy)/(dx)` ` " "+ log sin y*1 ` ` rArr " "-y tan x+log cos x (dy)/(dx) =x*cot y(dy)/(dx) +log sin y` ` rArr " "(dy)/(dx) (log cos x -x cot y) =log sin y+y tan x ` ` therefore " "(dy)/(dx) =(log sin y+ y tan x )/(log xcos x -x cot y ) ` |
|