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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.
| 101. |
निम्न फलन का x के सापेक्ष अवकलन कीजिए| ` (i) y= tan ^(-1) ((sqrt(1+x^(2) -1))/x)` ` (ii) y=sin [2tan ^(-1) sqrt(((1-x)) /((1+x))` |
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Answer» (i ) ` y= tan ^(-1) [(sqrt(1+x^(2) )-1)/(x) ]` माना ` x= tan theta rArr theta tan ^(-1) x ` ` therefore " "y= tan ^(-1) [(sqrt((1+ tan ^(2) theta ))-1)/(tan theta )] =tan ^(-1) [(sectheta -1)/(tan theta )]` ` rArr " "y= tan ^(-1) [(1-cos theta )/(sin theta ) ] =tan ^(-1) tan((theta )/(2))=(theta )/(2)` ` rArr " "y= (1)/(2) tan ^(-1) x ` ` rArr " "(dy)/(dx) =(1)/(2) (d)/(dx) tan ^(-1) x=(1)/(2) ((1)/(a+x^(2)))` ` therefore (dy)/(dx) =(1)/(2(1+x^(2)))` (ii) ` " "y= sin [2tan ^(-1) sqrt(((1-x))/((1+x)))]` माना ` x= cos theta rArr theta cos ^(-1) x ` ` therefore " "=sin [2tan ^(-1)sqrt(((1-cos theta )/(1+costheta )))] ` ` =sin [ 2tan ^(-1) sqrt(((1-1+2sin ^(2)""(theta)/(2))/(1+2cos ^(2) ""(theta )/(2) -1)))]` ` =sin [2tan ^(-1) (tan ""(theta )/(2))] =sin theta ` ` rArr y=sin theta =sin (cos ^(-1) x )` ` therefore (dy)/(dx) =(d)/(dx) sin (cos ^(-1) x ) =cos (cos ^(-1) x ) ((-1)/(sqrt(1-x^(2))))` ` therefore (dy)/(dx) =(-x) /(sqrt((1-x^(2) ) ))` |
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| 102. |
निम्न फलन का x के सापेक्ष अवकलन कीजिए| ` y= sin ^(-1) ((x)/(sqrt(1+x^(2))))` |
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Answer» `" "y=sin ^(-1) ((x)/(sqrt(1+x^(2))))` माना `x=tan theta rArr theta =tan ^(-1) x ` ` y= sin ^(_1) [ (tan theta )/(sqrt(1+tan ^(2)theta )]]` ` therefore " "y=sin ^(-1) [(tan theta )/(sectheta )] =sin ^(-1) (sin theta ) ` ` rArry=theta " " (becausetheta =tan ^(-1) x )` ` therefore " "y= tan ^(-1) x ` ` rArr" "(dy)/(dx) =(d)/(dx) tan ^(-1) x= (1)/(1+x^(2))` अतः ` (dy)/(dx) =(1)/(a+x^(2))` |
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| 103. |
निम्नांकित फलन का द्वितीय अवकलज ज्ञात कीजिये| ` x^(3) e^(4x )` |
| Answer» Correct Answer - ` e^(4) [16 x^(3) +24x^(2) +6x]` | |
| 104. |
यदि ` y= (1-cos 2x) /( 1+cos 2x) ` तब ` (dy)/(dx) ` का मान ज्ञात कीजिए| |
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Answer» `y= (1-cos 2x) /(1+cos 2x) =(1-1 +2sin ^(2) x )/(1+2cos ^(2) x-1) ` ` =(sin ^(2)x )/( cos ^(2) x )= tan ^(2) x =(tanx )^(2) ` माना ` tan x=t ` ` therefore " "y= t^(2) ` ` therefore " "(dy)/(dx) =(d)/(dt) t^(2) (d)/(dx) tan x =2tsec^(2) x ` ` therefore " "(dy)/(dx) =2tan x sec^(2) x ` |
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| 105. |
यदि ` y= log x^(x) ` तो सिद्ध कीजिए की ` " "(dy)/(dx) =(1+log x )` |
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Answer» माना ` x^(x) =t " "rArrx log x =log t ` ` x*(d)/(dx) (log x )+ log x (d)/(dx) (x) =log t ` ` " "x *(1)/(x)+ log x =(1)/(t)(dt)/(dx)` ` rArr " "(dt)/(dx) =x^(x) (1+log x) rArr (dy)/(dx)= (d)/(dx) (log x^(x))` ` rArr " "(dy)/(dx)=(d)/(dt) (log t )(dt)/(dx)= (d)/(dt) (log t ) [x^(x) (1+ log x ) ]` ` " "= (1) /(t) [x ^(x) (1+ log x ) ] =91) /(x^(x)) [ x^(x) (1+log x) ] =1 + log x ` |
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| 106. |
` log log sin x ` का x के सापेक्ष अवकलन कीजिए| |
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Answer» माना ` y= log log sin x ` ` therefore " "(dy)/(dx) =(d)/(dx) (log log sin x ) =(1)/( log sin x )*(d)/(dx) (sin x ) ` ` = ( 1)/( log sin x ) *( 1)/( sinx ) * cos x =(cotx) /(log sin x ) ` |
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| 107. |
`cos ^(3)x ` का अवकल कीजिये| |
| Answer» Correct Answer - ` 3x^(2) cos x^(3) ` | |
| 108. |
यदि ` y= (sin ^(-1) x)/(sqrt((1-x^(2)))).` तब सिद्ध कीजिए की, ` (1-x^(2) )(dy)/(dx) =xy+1` |
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Answer» `" "y=(sin^(-1) x)/(sqrt(1-x^(2)))` ` therefore (dy)/(dx) (sqrt((1-x^(2)))(d)/(dx) sin ^(-1) x-sin ^(-1) x(d)/(dx) (1-x^(2))^(1//2))/((1-x^(2)))` ` rArr " "(1-x^(2))(dy)/(dx) =1-sin ^(-1) x*(1(-2x))/(2sqrt(1-x^(2)))` ` " "= 1+x(sin ^(-1) x)/(sqrt(1-x^(2)))=1+xy" "` [समी०(1 )से ] |
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| 109. |
निम्न फलन का x के सापेक्ष अवकलन कीजिए | ` log log x^(2)` ` |
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Answer» Correct Answer - `(i) (1)/(xlog x ) " "(ii) secx " "(iii) 3e^(3x) " "(iv) 2x sin (cot x^(2) ) cosec^(2)x^(2)" "(v) -4(e^(x) -e^(-x))" "(vi) (2sin x )/((1+cos x)^(2))` माना ` log x ^(2) =t " "rArr (1)/(x^(2))* 2x dx = dt rArr (2)/(x ) dx =dt (d)/(dx) (log log x ^(2)) =(d)/(dt)(log t ) (dt)/(dx) =(d)/(dt) =(d)/(dt) (log t ) (2)/(x) ` ` " "= (1)/(t)* (2)/(x) =(2) /(x log x ) =(2) /(x*log x^(2))=(1)/(xlog x )` |
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| 110. |
यदि ` y= sin x * cos 2x ` तो सिद्ध कीजिए की` " " (dy)/(dx) =y (cot x - 2tan 2x)` |
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Answer» ` (dy)/(dx )= (d)/(dx) (sin x cos 2x) =sin x (d)/(dx) (cos 2x )+ cos 2x (d)/(dx) (sin x ) " " =-sin x * 2sin 2 x+cos 2x cos x ` (सरल करने पर ) ` " "= y( (-2sin x sin 2x+ cosx cos2x )/(sin x cos 2x ))` |
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| 111. |
निम्न फलनों का अवकलन गुणांक ज्ञात कीजिए| ` (i) ((x-sin x )^(3//2))/( sqrt(x) ) " "(ii) sin (log x ) " "(iii) cos x^(3) " "(iv) e^(log_ax) ` |
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Answer» माना ` y= ((x-sin x )^(3//2)) /( sqrt(3)) ` ` therefore " "(dy)/(dx) =(sqrt(x) (d)/(dx) (x-sin x ) ^(3/2) -(x-sin x)^(3/2) (d)/(dx) (x) ^(1/2))/(x)` माना `x-sin x=t ` अतः `(dy)/(dx) =(sqrt(x) (d)/(dt) t^(3/2) (d)/(dx)(x-sin x) -(x-sin x )^(3/2)(d)/(dx) x^(1/2)/(x) ` ` = (sqrt (x) *(3)/(2) t^(1//2) (1-cos x ) -(x-sin x)^(3//2) (1)/(2sqrt(x)) )/(x) ` ` " "= (3xsqrt (x-sin )(1-cos x)-(x-sin x)^(3//2))/(2xsqrt x))` ` sqrt ((x-sin x) [3x (1-cos x-(x- sin x ) ]))/(2xsqrt (x))` ` =(sqrt ((x-sin x))[ 2x -3cos x+ sin x] )/(2xsqrt (x))` माना `y= sin (log x ),`माना `log x =t` ` therefore " "y= sin t ` ` therefore (dy)/(dx) =(d)/(dt) sin t (d)/(dx) log x=(cost )/(x) =(cos (log x ) )/(x) ` (iii) माना ` y cos x^(3),` माना ` x^(3) =t ` ` " " y= cos t ` ` therefore " "(dy)/(dx) =(d)/(dt) cos t (d)/(dx) x^(3) =-sin t*3x^(2) ` ` therefore (dy)/(dt) =- 3x ^(2) sin x^(3) ` (iv) माना ` y= e ^(log _ax) , ` माना ` log _a x=t ` ` therefore " "y= e^(t) ` ` therefore (dy)/(dx) =(d)/(dt) e^(t) (d)/(dx) log_a x ` ` " "= e ^(t) (1)/(x) log_a e= e^(log _a x) (1)/(x) log_ae ` |
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| 112. |
यदि ` y= tan ^(-1) ""(cos x)/(1+sin x ),` तो `(dy)/(dx) ` का मान ज्ञात कीजिए| |
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Answer» हम जानते है की ` cos x =sin ((pi)/(2)-x)` तथा ` " "sin x =cos ""((pi)/(2)-x)` ` therefore " "y=tan ^(-1)"" (cos x )/(1+sin x )=tan ^(-1)"" (sin ""((pi)/(2)-x))/(1+cos ((pi )/(2)-x))` ` " "= tan ^(-1) ""(2sin ((pi)/(4) -(x)/(2) )cos ((pi)/(4)-(x)/(2)))/(2cos ^(2) ((pi)/(4)-(x)/(2)))` `" "= tan ^(-1) [tan ((pi)/4-(x)/(2) )] =(pi)/(4)-(x)/(2) ` ` therefore " "(dy)/(dx) =(d)/(dx) ((pi)/(4)-(x)/(2))=(d)/(dx) ((pi)/(4))-(d)/(dx) ((x)/(2))=-(1)/(2)` |
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| 113. |
निम्न फलन का x के सापेक्ष अवकलन कीजिये|` (i) x^(x) +(sin x )^(sin x)` |
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Answer» माना ` " "y= x^(x) +(sin x )^(sin x )` माना ` " " y_1 =x^(x) ` तथा ` y_2 =(sin x)^(sin x ) ` ` therefore " " y=y_1 +y_2 " "...(1)` punha ` " "y_1 =x^(x) ` ` therefore " "(dy_1)/(dx) =x^(x) [1+log x ] ` [udahran (1 )(i ) से ] तथा ` " "y_2 =(sin x)^(sin x ) ` दोनों पक्षों का लघुगणक लेने पर, ` log y_2 =sin x log sin x ` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, ` (d)/(dy_2) log y_2""(dy_2)/(dx) =sin x(d)/(dx) log sin x+ log sin x (d)/(dx) sin x ` ` rArr " "(1)/(y_2) (dy_2) /(dx) =cos x+ cos x log sin x ` ` " " =cos x (1+log sin x ) ` ` rArr " "(dy_2)/(dx) =y_2 cosx (1+log sin ) ` ` " "= cos x (sinx ) ^(sin x) [ 1+ log sin x ] " "...(3)` समी० (1 ) ,(2 ) व् (3 ) से, ` (dy)/(dx) =x^(3) (1+log x ) + cos x (sin x ) ^(sinx) [1+ log sin x ]` |
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| 114. |
निम्नांकित फलन का द्वितीय अवकलज ज्ञात कीजिये| ` (ax+ b )^(p//q)` |
| Answer» Correct Answer - ` (a^(2)p(p-q))/(q^(2))(ax+b) ^((p)/(q) -2) ` | |
| 115. |
निम्न फलनों का x के सापेक्ष अवकलन ज्ञात कीजिए| ` (i)" " tan ^(3) x ``(ii) " "(logx )^(3)` ` (iii) " " log (x+(1)/(2))` ` (iv)" "log (sec x +tan x )` ` (v) " " sin log (x ^(3) +1)`` (vi)" "log tan"" (x)/(2)` ` (vii)" "tan x^(3)` ` (viii)" " (x+ sqrt (x^(2) +1))^(3)` |
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Answer» Correct Answer - ` (i) 3tan ^(2) x sec ^(2) x " "(ii) (3)/(x) (log x ) ^(2) " "(iii) (x^(2)-1)/(x(x^(2)+1))" "(iv ) sec x " "(v) (3x^(2) )/(x^(3)+1)cos log(x^(3) +1)(vi) cosecx" "(3x^(2)sec^(2)x^(3))" "(viii) (3[x+sqrt(x^(2) +1)]^(3))/(sqrtx^(2)+1)` (i ) माना ` tanx = t " "rArr sec^(2) xdx =dt ` |
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| 116. |
निम्न फलनों का x के सापेक्ष अवकलन कीजिए| `(i) y= sec ^(-1) ((sqrt x+1)/(sqrtx-1))+sin^(1) ((sqrt(x)-1)/(sqrtx+1))` `(iv ) ` यदि ` y= sec ^(-1) ((sqrt (x) )/(sqrtx-1)+sin ^(-1) ((sqrt(x)-1)/(sqrt(x)))` to ` (dy)/(dx)` का मान ज्ञात कीजिए| |
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Answer» `(i) y= sec ^(-1) ((sqrt(x)+1)/(sqrt(x)-1))+sin ^(-1)((sqrt(x)-1)/(sqrt(x)+1))` ` =cos ^(-1) ((sqrt (x)-1)/(sqrt(x)+1))+sin ^(-1) ((sqrt(x-1))/(sqrt(x)+1)) ` ` " "[ because sec^(-1) x=cos ^(-1) (1//3)]` ` " "y= (pi)/(2) " "therefore (dy)/(dx) =0` `(iv) ` माना ` " "y=sec ^(-1) ((sqrt(x) )/(sqrtx-1))+sin^(-1) ((sqrt(x)-1)/(sqrt(x)))` ` " "=cos ^(-1) ((sqrt(x) -1)/(sqrt(x)))+sin ^(-1) ((sqrt(x)-1)/(sqrt(x))) = (pi)/(2) ` `therefore (dy)/(dx) =(d)/(dx)((pi)/(2) )=0` |
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| 117. |
निम्न फलनों का अवकलन गुणांक ज्ञात कीजिए- ` (i)log [ x + sqrt (x^(2) -a ^(2)) ]` ` (ii) log sin (ax+ b) ` `(iii) sin [ cos (2x+ 3) ]` (iv) log [ sin mx + cos nx ]` |
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Answer» माना ` y= log [ x + sqrt (x^(2) -a^(2) ) ] ` माना ` x+ sqrt (x^(2) -a^(2) ) =t ` ` " " y = log t ` ` therefore " "(dy)/(dx) =(d)/(dt) log t* (d)/(dx) [ x+ sqrt( x^(2) -a^(2)]` ` " "= ( 1) /(t) [ (d)/(dx)x+ (d)/(dx) (x^(2) -a^(2) ) ^(1//2) ] ` पुनः ` x^(2)-a^(2) =t_1 ` ` therefore (dy)/(dx) =(1)/(t) [1+ (d)/(dt_1) t_1^(1//2) (d)/(dx) (x^(2) -a^(2) ) ]` ` " "= (1)/(t) [ 1+ (1)/( 2sqrt ( t_1) )(2x-0 ) ] =(1)/(t) [ 1+ (x)/(sqrt( x^(2) -a^(2)))] ` ` " "= [ (1) /(x+ sqrt (x^(2) -a^(2) ) )] [ (x+ sqrt(x^(2) -a^(2) )) /( sqrt( ( x^(2) -a^(2)) ) ]] ` ` " "= (1) /(sqrt( x^(2) -a^(2) ) ) ` माना ` y= log sin ( ax+ b) ` माना ` sin (ax+b) =t ` ` therefore " " y= log t ` ` therefore " "(dy)/(dx) =(d)/(dt) log t (d)/( dx) sin (ax+b) ` ` " "= (1)/(t) (d)/(dt) sin ""t_ 1 (d)/(dx) (ax+ b) ` जबकि ` ax+ b =t_1 =(1)/(t) cos t_1 *(a+0) ` ` " "(acos (ax+ b))/(sin (ax+ b) ) =a cot (ax+b) ` (iii) माना ` y= sin [cos (2x+ 3) ]` माना ` cos (2x+ 3) =t ` ` therefore " "y= sin t ` ` therefore " "(dy)/(dx) =(d)/(dt ) sin t (d)/(dx) cos (2x+ 3) ` ` " "=cost (d)/(dt_1) cos t_1"" (d)/(dx) (2x+ 3) ` जबकि ` (2x+3) =t_1` ` " "=- sin t_1 cost*2` ` =- 2cos [ cos (2x+ 3) sin (2x+ 3) ]` (iv) माना ` y= log [ sin mx + cos nx ]` माना ` sin mx + cos nx =t ` ` therefore " " y= log t ` ` therefore " "(dy)/(dx) =(d)/(dt) log t (d)/(dx ) [ sin mx+ cos nx]` ` " "= ( 1)/(t) (mcos mx -n sin nx ) ` ` " "= (mcos mx - n sin nx ) /( sin mx +cos nx ) ` |
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| 118. |
निम्न फलनों का x के सापेक्ष अवकलन कीजिए| ` (i) y= tan ^(-1) ((3a^(2)x-x^(3))/(a(a^(2)-3x^(2))))` ` (ii) y=sin ^(-1) [x sqrt(1-x)-sqrtxsqrt((1-x^(2)))` |
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Answer» ` (i) y= tan ^(-1) [(3a^(2) x-x^(3) )/(a(a^(2) -3x^(2))]]` माना ` x=a tan theta " "rArr theta =tan ^(-1) ((x)/(a))` ` therefore " "y=tan ^(-1) [(a^(3) (3tan theta -tan ^(3) theta ))/(a^(3) (1-3 tan ^(2) theta ))]` ` =tan ^(-1) tan 3 theta =3 theta ` `rArr y= 3 tan ^(-1) ((x)/(a)) ` ` therefore (dy)/(dx) =3(1)/(1+((x)/(a))^(2))*(1)/(a) rArr (dy)/(dx) =(3a)/(a^(2)+x^(2))` `" " y=sin ^(1) [x sqrt (1-x) -sqrt( x)sqrt (1-x^(2))` माना ` x= sin alpha rArr alpha = sin ^(-1) x ` तथा ` sqrt x =sin beta " "rArr beta =sin ^(-1) sqrt(x) ` ` therefore y= sin ^(-1) [sin alpha sqrt(1-sin ^(2) beta ) -sin beta sqrt (1-sin ^(2) alpha ]` `" "= sin ^(-1) [sin alpha cos beta - cos alpha sin beta ]` ` " "= sin ^(-1) sin (alpha -beta )= alpha - beta ` ` " "y= sin ^(-1) x-sin ^(-1) sqrt(x)` ` therefore (dy)/(dx) =(d)/(dx)sin ^(-1) x-(d)/(dx) sin ^(-1) (x)^(1//2) ` ` (dy)/(dx) =(1) /sqrt(1-x^(2))-(1)/(2sqrt xsqrt( 1-x) )` |
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| 119. |
निम्नांकित फलन का द्वितीय अवकलज ज्ञात कीजिये| ` sin (cos x )` |
| Answer» Correct Answer - `-[sin ^(2) xsin (cos x )+ cos x cos (cos x)]` | |
| 120. |
निम्नांकित फलन का द्वितीय अवकलज ज्ञात कीजिये| ` x^(4) log x ` |
| Answer» Correct Answer - ` x^(2) [12log x +7]` | |
| 121. |
निम्न फलनों का अवकलन कीजिये| `(i) x^(x)` ` (ii) x^(sin-1)x` `(iii) (sin x )^(tanx )` |
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Answer» (i ) माना `" "y =x^(x)` दोनों पक्षों का लघुगणक लेने पर ` " "log y= log (x^(x))=x log x ` दोनों पक्षों का x के सापेक्ष अवकलन करने पर ` " "(d)/(dy) logy(dy)/(dx) =(d)/(dx) [x log x]` ` rArr " "(1)/(y)(dy)/(dx) =x(d)/(dx) log x +log x (d)/(dx) x ` ` rArr " "(1)/(y)(dy)/(dx) =x*(1)/(x) +log x` `rArr" "(dy)/(dx) =y (1+log x)` अतः ` " "(dy)/(dx) =x^(3) (1+log x ) " "(because y= x^(x))` `(ii)` माना ` " "y=x ^(sin -1)x ` दोनों पक्षों का लघुगुणक लेने पर, ` " "log y =log x ^(sin-1x)` ` " "log y =sin ^(-1) xlog x ` ` therefore (d)/(dy) log y (dy)/(dx)=sin ^(-1) x(d)/(dx) log x +log x(d)/(dx) sin ^(-1)x` ` rArr " "(1)/(y)(dy)/(dx) =(sin ^(-1)x)/(x) +(log x)/(sqrt(1-x^(2)))` ` rArr " "(dy)/(dx)= y[(sin ^(-1) x)/(x)+ (log x)/(sqrt(1-x^(2))]]` ` therefore" "(dy)/(dx) =x^(sin -1_x)[(sin ^(-1)x)/(x)+(log x)/(sqrt(1-x^(2)))]` ` " " (because y= x^(sin ^(-1_x)))` `(iii)` माना ` " " y= (sinx )^(tan x)` दोनों पक्षों का लघुगुणक लेने पर `" "log y=log (sinx )^(tan x ) ` ` " "=tan x log (sinx ) ` दोनों पक्षों का x के सापेक्ष अवकलन करने पर ` (d)/(dy) log y (dy)/(dx) =tan x (d)/(dx) log sin x+ log sin x (d)/(dx) tan x ` ` rArr " "(1)/(y) (dy)/(dx) =tan x cot x +sec ^(2) x log sin x ` ` " "= 1+sec ^(2) x log sin x ` ` rArr " "(dy)/(dx) =y [1+sec^(2) x log sin x ]` ` " "= (sin x ) ^(tan x ) [1+sec^(2)x log sin x ]` |
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| 122. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` (sin ^(-1)x)/(x^(2))` |
| Answer» Correct Answer - ` (x-2sin ^(-1) (x)sqrt(1-x^(2)))/(x^(3)sqrt(1-x^(2)))` | |
| 123. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए -` (xcos ^(-1) x )/(sqrt(1-x^(2)))` |
| Answer» Correct Answer - ` (-x)/((1-x^(2)))+(cos^(-1)x)/((1-x^(2))^(3//2))` | |
| 124. |
निम्न फलनों के x के सापेक्ष अवकलन कीजिये| ` (i) ((x+1)^(2)sqrt(x-1))/((x+4)^(e)e^x)` ` (ii) (xlog x ) ^(log log x )` |
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Answer» (i ) माना ` y= ((x+1)^(2) (x+1)^(1//2))/((x+4)^(3)e^(x))` दोनों पक्षों का लघुगणक लेने पर ` log y= log [((x+1)^(2) (x-1)^(1//2))/((x+4)^(3)e^(x))]` ` =log (x+1)^(2) +log (x-1)^(1//2) -log (x+4) ^(3) -log e^(x)` या ` log y =2 log (x+1) +(1)/(2) log (x-1) -3 log (x+4) -x ` ` " "(because log _e e=1)` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, ` (d)/(dx) log y (dy)/(dx) =2 (d)/(dx) log (x+1) +(1)/(2) (d)/(dx) log (x-1)` ` " "-3(d)/(dx) log (x+4)-1` ` rArr (1)/(y) (dy)/((dx) )=(2)/(x+1) +(1)/(2(x-1))-(3)/(x+4)-1` ` rArr (dy)/(dx) =y [(2)/(x+1)+ (1)/(2(x-1))-(3)/(x+4) -1]` ` rArr (dy)/(dx)=((x+1)^(2) sqrt((x-1) ))/((x+4 )^(3) e^(x) )[ (2) /(x+1)+(1)/(2(x-1))-(3)/(x+4)-1` `(ii)` माना ` y= (xlog x ) ^(log log x ) ` दोनों पक्षों का लघुगणक लेने पर, `log y= log [(xlog x ) ^(log log x )]` `" " =(log log x )log (xlog x )` दोनों पक्षों का x के सापेक्ष अवकलन करने पर ` (d)/(dy) log y (dy)/(dx) =[log log (x) ] (d)/(dx) log (xlog x ) ` ` " "+ log (xlog x ) (d)/(dx) log log (x) ` ` rArr " "(1)/(y) (dy)/(dx) =log log(x) [(1) /(xlog x ) {x(d) /(dx) log x +log x (d)/(dx) x }]` ` " "= + log (x log x ) [ (1)/(x log x )]` ` rArr (1)/(y)(dy)/(dx) =log log (x) [(1) /(xlog x ) (1+ log x ) ]` ` " "+ log (xlog x ) [(1) /(xlog x )]` ` rArr " "(dy)/(dx) =(xlog x )^(log log x ) [(log log (x))/(x log x )](1+log x ) ` ` " " + log (x log x ) /( xlog x )` |
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| 125. |
निम्नांकित फलन का द्वितीय अवकलज ज्ञात कीजिये| ` tan ^(-1) x ^(3)` |
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Answer» Correct Answer - ` (6x(1-2x^(6)))/( (1+x^(6))^(2))` ` y= tan ^(-1) x ^(3) rArr (dy)/(dx) =(1)/( { 1+(x^(3))^(2)} ) (3x^(2))=(3x ^(2) )/(1+ x^(6))` ` rArr(d)/(dx) ((dy)/(dx) )= (d^(2)y)/( dx^(2) )=3 *(d)/(dx) ((x^(2))/(1+x^(6)))` ` =3 [ ((1+x^(6) ) (d)/(dx) (x^(2))-x^(2) (d)/(dx) (1+ x^(6)))/((1+ x ^(6) )^(2))]` |
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| 126. |
निम्नलिखित प्रश्न में x के सापेक्ष अवकलन कीजिये-` (sin x )^(x) +sin ^(-1) sqrt(x)` |
| Answer» Correct Answer - ` (sin x)^(x) (xcot x+ log sin x )+ (1)/(2) (1)/(sqrt( x-x^(2)))` | |
| 127. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` x^(4)cos ^(-1) x ` |
| Answer» Correct Answer - ` 4x^(3) cos ^(-1) x - (x^(4))/(sqrt(1-x^(2)))` | |
| 128. |
निम्न फलनों के लिए `theta =(pi)/(2) ` पर ` (d^(2)y)/(dx^(2))` का मान ज्ञात कीजिये| `(i) x=a (theta -sin theta ) ,y=a (1+cos theta )` `(ii) x=a (1-cos theta ), y=a (theta + sin theta )` |
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Answer» Correct Answer - ` (i)(1)/(a) " "(ii) -(1)/(a) ` `(i) " "x=a (theta -sin theta ) " "rArr " " (dx)/(d theta )=a (1-cos theta )` ` " "y=a (1+cos theta )" "rArr " "(dy)/(dx) =-asin theta ` अब ` (dy)/(dx) =(dy//d theta)/(dx//d theta ) ` व ` theta = (pi)/(2)` का प्रयोग करने पर| |
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| 129. |
`(dy)/(dx)` का मान ज्ञात कीजिये यदि `(i) x= sqrt(1+t ), y= sqrt(1-t)` `(ii) x=a cos theta ,y= asin theta ` |
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Answer» Correct Answer - ` (i) -sqrt((1+t)/(1-t))" "(ii) -cot theta ` ` (i) x=sqrt(1+t)`ltbr gt` rArr (dx)/(dt)=2 (1)/(sqrt(1+t))*(d)/(dt) (1+t) =(1)/(2sqrt(1+t))` और ` y=sqrt(1-t )" "rArr (dy)/(dx) =(-1)/(2sqrt(1-t))` |
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| 130. |
`cot (cos ^(-1) x ) ` का x के सापेक्ष अवकलन कीजिये| |
| Answer» Correct Answer - ` (1)/((1-x^(2))^(3//2))` | |
| 131. |
`e^(x) ` का `sqrt(x) ` के सापेक्ष अवकलन कीजिये| |
| Answer» Correct Answer - ` 2e^(x) sqrt(x)` | |
| 132. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` (i)sin ^(-1)""(1)/(sqrt(1+x^(2)))` ` (ii) tan ^(-1) ""(x)/(sqrt( 1+x^(2)))` ` (iii) tan ^(-1) ((x)/(1+x^(2)))` |
| Answer» Correct Answer - ` (-1)/(sqrt(1-x^(2)cos ^(-1) x ))" "(ii) (1)/((1+ 2x^(2))sqrt(1+x^(2)))" "(iii) (1-x^(2))/(1+3x^(2)+x^(4))` | |
| 133. |
` cos ^(-1) (cot x) ` का x के सापेक्ष अवकलन कीजिये| |
| Answer» Correct Answer - ` (cosec ^(2)x)/(sqrt(1-cot^(2)x))` | |
| 134. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए -` cos ^(-1) [(1-x^(2n))/(1+x^(2n))]` |
| Answer» Correct Answer - ` (2nx ^(n-1))/(1+x^(2n))` | |
| 135. |
फलन का अवकलन गुणांक ज्ञात कीजिए| ` (1)/(sqrt( x+1)+sqrt(x ))" " (1) /(sqrt (x+1)-sqrt(x))` |
| Answer» Correct Answer - ` (i) (1)/(2) [(1)/(sqrt( x+1)) -(1)/(sqrt( x )) ] " "(ii) (1)/(2) [(1)/(sqrt( x+1))+ (1)/(sqrt(x))l]` | |
| 136. |
फलन का अवकलन गुणांक ज्ञात कीजिए| ` (i)sqrt(log x ) " "(ii)sqrt(sec x) " "(iii) sqrt(log _a x )` |
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Answer» Correct Answer - ` (i) (1)/(2xsqrt(log x ))" "(ii) (tan sqrt(sec x))/(2) " "(iii) (1) /(2x sqrt(log _a x ) )log _ae ` (ii ) माना ` sec x =t rArr " "(dt)/(dx) = sec x tan x ` |
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| 137. |
फलन का अवकलन गुणांक ज्ञात कीजिए| ` log ""(x^(2) +x+1 ) /(x^(2)- x+1)` |
| Answer» Correct Answer - ` (2 ( 1-x^(2)))/(x^(4 +x^(2)) +1 )` | |
| 138. |
`(dy)/(dx)` ज्ञात कीजिये यदि ` x=a sec theta ,y =b tan theta ` |
| Answer» Correct Answer - ` (b)/(a) cosec theta ` | |
| 139. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिये-` x ^(cos ax )` |
| Answer» Correct Answer - ` x ^(cos ax ) [(cos ax)/(x ) -a sin ax log x ]` | |
| 140. |
`ax^(2) +2hxy +by^(2) +2gx +2fy +x=0` तब ` (dy)/(dx) ` ज्ञात कीजिये | |
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Answer» `ax^(2) +2hxy +by^(2) +2gx +2fy +x=0` दोनों पक्षों का x के सापेक्ष अवकलन करने पर ` a(d)/(dx) x^(2)+ 2h (d)/(dx) (xy)+b(d)/(dx) y^(2) +2g""(d)/(dx) x+ 2f (dy)/(dx) +0=0 ` ` rArr " "2ax +2h [x (dy)/(dx) +y*1] +2by (dy)/(dx) +2g ""(dy)/(dx) +2g +2f"" (dy)/(dx)=0` `rArr " "(ax + hy +g) +(dy)/(dx) (hx +by+f)=0` ` rArr " "(dy)/(dx) =-((ax+hy+g)/(hx +by +f))` |
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| 141. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए -` y= tan ^(-1) [(sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt (1-x^(2)))]` |
| Answer» Correct Answer - ` (-x)/(sqrt((1-x^(4))))` | |
| 142. |
` cot ^(-1) ((1-x)/(1+x))` का x के सापेक्ष अवकलन कीजिये| |
| Answer» Correct Answer - ` (1)/(1+x^(2))` | |
| 143. |
फलन का अवकलन गुणांक ज्ञात कीजिए| ` (1) sqrt((ax+b)^(m)) " " (ii) (a-bx ) ^(7//3) " "(iii) sqrt(ax^(2) +bx +c)` |
| Answer» Correct Answer - ` (i) (am)/(2) (ax+ b ) ^((m)/(2) -1) " "(ii) -(7b)/(3)(a-bx) ^(4//3)" "(iii) (2ax+b)/(2sqrt( ax^(2)+bx+ c )) ` | |
| 144. |
`(dy)/(dx)` ज्ञात कीजिये यदि ` x=at^(2) ,y =2at ` |
| Answer» Correct Answer - ` (1)/(t)` | |
| 145. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिए|` e^(tan ^(-1_x) )log (tan x ) ` |
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Answer» Correct Answer - ` e^(tan ^(-1_x))[(sec^(2)x)/(tan x )+ (log tan x )/(1+x^(2))]` ` (d)/(dx) { e^(tan ^(-1_x))*log (tan x )}` ` " "= e^(tan ^(-1_x))*(d)/(dx) (log tan x )+ log (tan x ) (d)/(dx) (e^(tan ^(-1_x))) ` ` " "= e^(tan ^(-1_x))*(1)/(tan x ) *sec ^(2) x+ log tan x*e ^(tan -1_x) (1)/((1+x^(2)))` |
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| 146. |
निम्न फलन का x के सापेक्ष अवकल गुणांक ज्ञात कीजिये-` (sin ^(-1) x )^(x)` |
| Answer» Correct Answer - ` (sin ^(-1) x )^(x) [log sin ^(-1)x + (x)/(sqrt(1-x^(2)sin ^(-1) x ))]` | |
| 147. |
निम्न फलन का अवकल गुणांक ज्ञात कीजिये - ` (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 ` |
| Answer» Correct Answer - ` -(b^(2))/(a^(2))-(x)/(y);` | |
| 148. |
यदि ` ax^(2) +2hxy +by^(2) +2gx +2fy +c=0` तब ` (dy)/(dx) ` का मान ज्ञात कीजिये| |
| Answer» Correct Answer - ` -((ax+ hy+ g)/(hx + by + f))` | |
| 149. |
यदि `x^(3) +y^(3)=3axy ` तब ` (dy)/(dx) ` का मान ज्ञात कीजिये| |
| Answer» Correct Answer - ` (ay-x^(2))/(y^(2) -ax) ` | |
| 150. |
फलन ` (x) /(sqrt (a^(2) +x^(2) )) ` का अवकल गुणांक ज्ञात कीजिए| |
| Answer» Correct Answer - ` (i) (a^(2))/((a^(2)-x^(2) )^(3//2))" "(ii) (3a^(2)x-x^(3)) /((a^(2)-x^(2) )^(3//2))" "(iii) (-x (3a^(2) +x^(2)))/((a^(2)+x^(2) )^(3//2))" "(iv)(1)/(2) sqrt((x^(2)-x+_1 )/(x^(2)+x+1))[(2(1-x^(2)))/(x^(2)-x+1)^(2)]` | |