Tiêu đề 19. Differentiation Medium Ans.pdf ✅

1. (b) Given f(2) = 4, f(2) = 1  = − − + − = − − → → 2 (2) 2 (2) 2 (2) 2 ( ) lim 2 (2) 2 ( ) lim 2 2 x xf f f f x x xf f x x x 2 2 ( ) 2 (2) lim 2 ( 2) (2) lim 2 2 − − − − − → → x f x f x x f x x = 2 ( ) (2) (2) 2 lim 2 − − − → x f x f f x = f(2) − 2 f(2) = 4 − 2(1) = 4 − 2 = 2 Trick : Applying L-Hospital rule, we get 2 1 (2) 2 (2) lim 2 = −  → f f x . 2. (c) Let x = 5,y = 0  f(5 + 0) = f(5).f(0)  f(5) = f(5)f(0)  f(0) = 1 Therefore, h f h f f h (5 ) (5) (5) lim 0 + −  = → =       − = − → → h f h h f f h f h h ( ) 1 lim 2 (5) ( ) (5) lim 0 0 {f(5) = 2} = 2 (0) 2 3 6 ( ) (0) 2 lim 0 =   =  =       − → f h f h f h . 3. (b) x a g x f a g a f x x a − − → ( ) ( ) ( ) ( ) lim . We add and subtract g(a)f(a) in numerator = x a g x f a g a f a g a f a g a f x x a − − + − → ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) lim =       − − −      − − → → x a f x f a g a x a g x g a f a x a x a ( ) ( ) lim ( ) ( ) ( ) lim ( ) =       − − −      − − → → x a f x f a g a x a g x g a f a x a x a ( ) ( ) ( ) lim ( ) ( ) ( ) lim = f(a)g'(a) − g(a)f'(a) [by using first principle formula] = 3.4 – (–1)(–2) = 12 – 2 = 10 Trick : x a g x f a g a f x x a − − → ( ) ( ) ( ) ( ) lim Using L–Hospital’s rule, Limit = 1 '( ) ( ) ( ) '( ) lim g x f a g a f x x a − → ; Limit = g'(a) f(a)− g(a)f'(a) = (4)(3) − (−1)(−2) = 12 – 2 = 10. 4. (b) 2 1 5 ( ) 3  = +      + x x  f x f ......(i) Replacing x by x 1 in (i), 2 1 3 ( ) 1 5  + = +      x f x x f ......(ii) On solving equation (i) and (ii), we get, 4 3 16 ( ) = 5 − + x f x x ,  2 3 16 ( ) 5 x f x = +  y = xf (x)  f(x) xf (x) dx dy = +  = ) 3 (5 16 1 4) . 3 (5 16 1 2 x x x x − + + + at x = 1, (5 3) 16 1 (5 3 4) 16 1 = − + + + dx dy = 8 7 . 5. (a)     −   = , 0 , 0 ( ) 3 3 x x x x f x and     −    = 3 , 0 3 , 0 ( ) 2 2 x x x x f x (0 ) = (0 ) = 0 + − f f 6. (b) 3 ( ) sin2 .cos2 .cos3 log2 2 + = + x f x x x x , sin4 cos3 ( 3)log 2 2 1 ( ) = + + 2 f x x x x , [sin7 sin ] 3 4 1 f(x) = x + x + x + Differentiate w.r.t. x, [7 cos7 cos ] 1 4 1 f(x) = x + x + , cos 1 4 1 7 cos7 4 1 f(x) = x + x + , f() = −2 + 1 = −1 . 7. (c) Around , 3 2 x = | cos x | = −cos x and | sin x | = sin x  y = − cos x + sin x  x x dx dy = sin + cos At 3 2 x = , 3 2 cos 3 2 sin   = + dx dy = ( 3 1) 2 1 2 1 2 3 − = − . 8. (b) x x f x x x log log(log ) ( ) = log (log ) =  2 (log ) log(log ) 1 1 ( ) x x x x f x −  =  e e f e 1 1 0 1 ( ) = −  = 9. (d)     −   = = log , if 1 log , if 0 1 ( ) | log | x x x x f x x         −    = , if 1 1 , if 0 1 1 ( ) x x x x f x . Clearly (1 ) = −1 − f and (1 ) = 1 + f ,  f(x) does not exist at x = 1 10. (a)                 − = − 2 2 1 exp tan x y x x          − = − 2 2 1 log tan x y x x  tan(log ) 2 2 x x y x = −  2 2 y = x tan(log x) + x  x x x x x x dx dy 2 sec (log ) 2 . tan(log ) . 2 2 = + +  x x x x x dx dy 2 tan(log ) sec (log ) 2 2 = + +  2 [1 tan(log )] sec (log ) 2 x x x x dx dy = + +
11. (a)         − = − − 2 ( ) cot 1 x x x x f x Put = tan x x ,          − = = −   2 tan tan 1 ( ) cot 2 1 y f x = cot ( cot 2 ) 1 − θ − = cot (cot 2 ) 1   − −  y =  − 2 = 2 tan ( ) 1 x x −  −  . (1 log ) 1 2 2 x x dx x dy x x + + − =  f (1) = −1 . 12. (a) x x x x x x x y n n − − = − − + + + = + 1 1 1 (1 )(1 )(1 ).....( 1 ) 1 2 2 2  2 1 2 1 2 (1 ) 2 . (1 ) 1 1 1 x x x x dx dy n n n − − − + − = + + + −  At x = 0 , 1 1 2 0 .1 1 0 2 1 = − + − = n+ dx dy . 13. (a) x x x x x x x x x f x 2 sin sin 32 2 sin 2 sin . cos . cos 2 . cos 4 . cos 8 . cos16 ( ) 5 = =  x x x x x f x 2 sin 32 cos 32 .sin cos .sin 32 . 32 1 ( ) −  =  2 2 1 32 . . 0 2 1 2 1 32. 4 2 =         −  =        f . 14. (c) We are given that xe y x xy 2 = + sin When x = 0 , we get y = 0 Differentiating both sides w.r.t. x, we get, x x dx dy y dx dy e xe x xy xy = + 2 sin cos       + + Putting, x = 0 , y = 0 , we get = 1 dx dy . 15. (d) sin( x + y) = log( x + y) Differentiating with respect to x,       + + =      + + dx dy dx x y dy x y 1 1 cos( ) 1 1 0 1 cos( ) =       +      + + − dx dy x y x y  x y x y + +  1 cos( ) for any x and y. So, 1 + = 0 dx dy , = −1 dx dy . Trick: It is an implicit function, so 1 1 cos( ) 1 cos( ) / / = − + + − + + − = −     = − x y x y x y x y f y f x dx dy . 16. (a) ln(x + y) = 2xy        = + + + y dx dy x x y dy dx 2 ( ) (1 / )  2 2 1 1 2 2 2 2 + − − − = x xy xy y dx dy  1 1 1 2 (0) = − − y  = , at x = 0, y = 1 . 17. (c) m n m n x y x y + = 2( + )  m log x + n log y = log 2 + (m + n)log( x + y) Differentiating w.r.t. x both sides       + + + + = dx dy x y m n dx dy y n x m 1  x y dx dy = . 18. (b)   dx d dy d dx dy / / = =              tan cos sin [ sin cos sin ] [cos ( sin ) cos ] = = − + + − − − a a . 19. (d) Obviously 2 1 1 1 cos t x + = − and 2 1 1 sin t t y + = −  x t 1 tan − = and y t 1 tan − =  y = x  = 1 dx dy . 20. (c) 2 2 1 1 t t x + − = and 2 1 2 t t y + = Put t = tan in both the equations, we get    cos 2 1 tan 1 tan 2 2 = + − x = and    sin 2 1 tan 2 tan 2 = + y = . Differentiating both the equations, we get   = −2 sin 2 d dx and 2 cos 2.  = d dy Therefore y x dx dy = − = −   sin 2 cos 2 . 21. (a) x x x f x +                 + = − − 2 1 2 ( ) cos cos 1  x x x + + = − 2 1 2   (1 log ) 2 1 1 . 2 1 ( ) x x x f x x + + +  = −  4 3 1 4 1 f(1) = − + = 22. (a) n y (x 1 x ) 2 = + +          + = + + + − 2 2 1 1 ( 1 ) 1 x x n x x dx dy n  2 2 1 ( 1 ) x n x x dx dy n + + + =  n n x x dx dy x       + = + + 2 2 ( 1 ) 1          + + + 2 2 2 2 1 . 1 x x dx dy x dx d y         +  +      = + + − 2 1 2 2 1 1 1 x x n x x n
 n n x x dx dy x dx d y (1 x ). . ( 1 ) 2 2 2 2 2 + + = + +  n y dx dy x dx d y x 2 2 2 2 (1 + ) + . = 23. (c) f(x) = x  f(1) = 1 n , f x nx f n n  =   = − ( ) (1) 1 ( ) ( 1) (1) ( 1) 2  = −   = − − f x n n x f n n n ..... f (x) n! f (1) n! n n =  = , ! ( 1) (1) ...... 2! (1) 1! (1) (1) n f f f f n n − +  +   − ! ! ..... ( 1) 3! ( 1)( 2) 2! ( 1) 1! 1 n n n n n n n n n + + − − − − − = − + = 0 − 1 + 2 − 3 + ...... + (−1) n = 0 n n n n n n C C C C C . 24. (b) We have         − + +        + − =         − + +        + − = − − − − x x x x x x e x e x y 1 6log 3 2log tan 1 2log 1 2log tan 1 6log 3 2log tan log log log log tan 1 1 1 2 2 1 tan 1 tan (2 log ) tan 3 tan (2 log ) 1 1 1 1 x x − − − − = − + +  tan 1 tan 3 0 0 1 1 = +  =  = − − n n dx d y dx dy y . 25. (b) We have, n x n x i n x f x x x n x i x x x 2 2 .... (2 1) ) cos sin ( ) cos( 3 .... (2 1) ) sin( 3 5 + + − = + = + + + − + + +  ( ) (sin ) ( cos ) 2 2 2 2 f  x = −n n x + n i n x  f x n n x n i n x 4 2 4 2 ( ) = − cos − sin  ( ) (cos sin ) 4 2 2 f  x = −n n x + i n x  ( ) ( ) 4 f x = −n f x 26. (a) ( 1)( 2) 15 14 3 7 3 2 2 2 4 − − − = + + + − + = x x x x x x x x y = ( 2) 16 ( 1) 1 3 7 2 − + − + + − x x x x  ( ) (3 ) (7) [( 1) ] 16 [( 2) ] 2 −1 −1 y = D x + D x + D − D x − + D x − n n n n n n ( 1) ![ ( 1) 16( 2) ] − −1 − −1 = − − − + − n n n n x x = ( 1) ![16( 2) ( 1) ] − −1 − −1 − − − − n n n n x x . 27. (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) '( ) 1 2 3 ' 3 ' 2 ' 1 1 2 3 1 2 3 1 2 3 ' 3 ' 2 ' 1 h x h x h x g x g x g x f x f x f x h x h x h x g x g x g x f x f x f x F x = + + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ' 3 ' 1 2 1 2 3 1 2 3 h x h x h x g x g x g x f x f x f x   ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 3 1 2 3 ' 3 ' 2 ' 1 h a h a h a g a g a g a f a f a f a F a = + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 3 ' 2 3 ' 1 1 2 3 h a h a h a g a g a g a f a f a f a  + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ' 3 ' 2 ' 1 1 2 3 1 2 3 h a h a h a g a g a g a f a f a f a =. 0 + 0 + 0 = 0 [ f (a) = g (a) = h (a),r = 1, 2, 3]  r r r 28. (d) Given 2 3 3 1 6 1 0 sin cos ( ) p p x x x f x = − , 2nd and 3rd rows are constant, so only 1st row will take part in differentiation  2 3 3 3 3 3 3 3 3 3 3 1 6 1 0 sin cos ( ) p p x dx d x dx d x dx d f x dx d − = We know that ) 2 !, sin sin( n x x dx d x n dx d n n n n n = = + and ) 2 cos cos( n x x dx d n n = + Using these results, 2 3 3 3 1 6 1 0 2 3 cos 2 3 3! sin ( ) p p x x f x dx d −        +      + =   2 3 at 0 3 3 1 6 1 0 6 1 0 ( ) p p f x dx d x − − = = = 0 i.e., independent of p. 29. (a) Applying formula we get F'(x) (log x )3x (log x )2x 3 2 2 = − = (3 log )3 2 (2 log ) 2 x x − x x = 9x log x 4 x log x 2 − = (9x 4 x)log x 2 − . 30. (b)  + = y dt t x 0 2 1 4 1  2 1 4 1 y dy dx + =  2 1 4y dx dy = +  dx dy y y dx d y 2 2 2 1 4 4 + =  y y y y dx d y 1 4 4 1 4 4 2 2 2 2  + = + = 31. (b) Applying Leibnitz’s theorem by taking 2 x as second function. We get , ( . ) 2 D y D e x n n x = = ( ) ( ). ( ) ( ). ( ) .......... . 2 2 2 2 1 2 1 2 0 + + + − − C D e x C D e D x C D e D x n n x n n x n n x = .2 0 0 .......... 2! ( 1) . .2 2 + + + − + + x x x e n n e x ne x x n y {x 2nx n(n 1)}e 2 = + + − .