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Differential Calculus

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 Joined: 24 Mar 2009 Post: 2
27 Mar 2009 20:05:04 IST
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differentiation
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Differential Calculus

1)             If y = sin-1  √(1+x) + √(1-x)

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then find dy/dx =

a)                 1/√1-x^2

b)                 -1/√1-x^2

c)                  -1/2√1-x^2

d)                 none

2)     If x=2logcot t and y=tan t + cot t, then (dy/dx)sin 2t +1 =

a)     cos^2 t            b)         sin^2 t             (c) cos 2t        (d)       2 cos^2 t

3)     If sin y = x sin (z+y)  and dy/dx=a/x^2+2xb+1 then

a)    a=b                 b)         a-b =1             (c) a+b =1      (d)       a^2 + b^2 = 1

4)     If y = f (2x -1/x^2 +1) and f’(x) = sin x^2. then dy/dx at x=0 equals

a)         ½ sin 1            b)         sin 1                c)         2 sin 1             d)         none

5)     If g is inverse of f and f’(x)= 1/1+x^n then g’(x) equals

a)         1 + x^n            b) 1+ {f(x)}^n               c)         1+{g(x)}^n       d)         none

6)     If y= (1+x)(1+x^2)(1+x^4)………………….(1+x^2n), then at x=0, dy/dx =

a)         -1         b)         0          c)         1          d)         none

Cool goIITian

Joined: 8 Sep 2008
Posts: 66
27 Mar 2009 20:59:58 IST
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ans. to 1st one is (c)

Hot goIITian

Joined: 15 Feb 2009
Posts: 186
28 Mar 2009 10:13:08 IST
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d ans to the second one is a). solve it by first differentiating x w.r.t  t  n  then y w.r.t  t.divide dy/dt from dx/dt to get the required ans.

hav a  nice day.

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Joined: 30 Mar 2009
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20 Apr 2009 11:13:16 IST
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1)            If y = sin-1  √(1+x) + √(1-x)

__________________

2

then find dy/dx =

a)                 1/√1-x^2

b)                 -1/√1-x^2

c)                  -1/2√1-x^2

d)                 none

1. put x = sin 2a then 1 + x = 1 + sin 2a = (sin a + cos a)2 or (1 + x )1/2 = (1 + sin 2a)1/2 = sina + cos a and similarly (1 - x)1/2 = sin a - cos a

so y = sin-1(sin a) = a = 1/2 sin -1 x so dy/dx = 1/2(1-x2)1/2

2)     If x=2logcot t and y=tan t + cot t, then (dy/dx)sin 2t +1 =

a)     cos^2 t            b)         sin^2 t             (c) cos 2t        (d)       2 cos^2 t

dy/dx = (dy/dt)/(dx/dt) = (sec2 t-cosec2 t)cot t/2(-cosec2 t) = cos 2t/sin 2t

(dy/dx)sin 2t + 1 = cos 2t + 1 = 2cos2 t

3)     If sin y = x sin (z+y)  and dy/dx=a/x^2+2xb+1 then

a)    a=b                 b)         a-b =1             (c) a+b =1      (d)       a^2 + b^2 = 1

sin y = x sin(z+y) = x (sin z cos y + cos z sin y) or 1 = x(sinz cot y + cos z) or 1/x = sin z cot y + cos z or

(1/x – cos z)/sin z = cot y now differentiate w.r.t x both side

4)     If y = f (2x -1/x^2 +1) and f’(x) = sin x^2. then dy/dx at x=0 equals

Take 2x – 1/x2 + 1 = v now dy/dx = df(v)/dx = df/dv*dv/dx = sin2v*(2+2/x3)

When x = 0, v = 1, dy/dx = sin21*(2 + 2/03) so answer in none

a)         ½ sin 1           b)         sin 1               c)         2 sin 1                        d)         none

5)     If g is inverse of f and f’(x)= 1/1+x^n then g’(x) equals

f(g(x)) = x or df/dx = df/dg*dg/dx = 1 or dg/dx = 1/(df/dg) = 1 + (g(x))n

a)         1 + x^n           b) 1+ {f(x)}^n              c)         1+{g(x)}^n      d)         none

6)     If y= (1+x)(1+x^2)(1+x^4)………………….(1+x^2n), then at x=0, dy/dx =

(1+x)(1+x2)(1+x4)…..(1+xp) where p = 2n, multiplying and dividing by (1-x) we get (1-x2p)/(1-x)

Differentiate this w.r.t x and putting x = 0, we get ans 1

a)         -1         b)         0          c)         1          d)         none

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