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 Multiple Choice QuestionsMultiple Choice Questions

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11.

limxx + 5x + 2x + 3 equals

  • e

  • e2

  • e3

  • e5


C.

e3

limxx + 5x + 2x + 3 = limx1 + 3x + 2x + 3                                   = limx1 + 3x + 2x + 233x + 3x + 2                                  = elimx1 + 3x1 + 2x                                  = e3


12.

The differential equation which represents the family of curves y=c1ec2xe, where c1 and c2 are arbitrary constants, is

  • y' =y2

  •  y″ = y′ y

  • yy″ = y′

  • yy″ = y′


D.

yy″ = y′

y c1ec2x = …..(i)
y' = c1c2ec2x
y' = c2y.....(from (i) ....(ii)
y" = c2y' ....... (iii)
from (ii) & (iii)
fraction numerator straight y apostrophe over denominator straight y end fraction space equals space fraction numerator straight y " over denominator straight y apostrophe end fraction
rightwards double arrow space yy " space equals space left parenthesis straight y apostrophe right parenthesis squared

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13.

The value of limx252 - x is

  • 102

  • does not exist


D.

does not exist

limx 252 - xLHL = limx  2-52 - x       = limh  052 - 2 - h × 2 + 2 - h2 + 2 - h       = limh  052 + 2 + h2 - 2 + h = 

RHL = limx  2+52 - x       = limh  052 - 2 - h × 2 + 2 + h2 + 2 + h       = limh  052 + 2 + h2 - 2 - h = -         LHL  RHL

Thus, limit does not exist. 


14.

If y = tan-11 - sinx1 + sinx, then the value of dydx at x = π6 is

  • 12

  • 12

  • 1

  • - 1


A.

12

Given, y = tan-11 - sinx1 + sinx              = 1 - cosπ2 - x1 + cosπ2 - x             = tan-1tanπ4 - x2             = π4 - x2           x = π6  dydx = - 12


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15.

The value of limx2e3x - 6 - 1sin2 - x

  • 32

  • 3

  • - 3

  • - 1


C.

- 3

limx2e3x - 6 - 1sin2 - x= lim -x2e3x - 63cos2 - x            using L'Hospital's rule)= - 3e0cos0 = - 3


16.

The value of the limit limx1sinex - 1 - 1logx

  • 0

  • e

  • 1e

  • 1


D.

1

limx1sinex - 1 - 1logx= limh0sineh - 1log1 + h= limh0sineh - 1eh - 1 × eh - 1log1 + h= 1 × limh0h + h22! + ...h - h22! + ... = 1 × 1 = 1


17.

The value of limnnn2 + 12 + nn2 + 22 + ... + nn2 + n2 is

  • π4

  • log2

  • 0

  • 1


A.

π4

limnnn2 + 12 + nn2 + 22 + ... + nn2 + n2= limnr = 1nnn2 + r2 = limn1nr = 1n11 + rn2

01dx1 + x2 = tan-1x01 = π4


18.

Let f(x) = x + 3x + 1, then the value of limx- 3 - 0fx is

  • 0

  • does not exist

  • 12

  • - 12


B.

does not exist

limx- 3 - 0 x + 3x + 1

But x + 3 is not defined on left hand side of - 3.

Hence, function is not defined.


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19.

The equation of the tangent to the curve y = x +4/x2, that is parallel to the x-axis, is

  • y= 0

  • y= 1

  • y= 2 

  • y= 2 


D.

y= 2 

We have, 
straight y space equals space straight x space plus space 4 over straight x squared
On differentiating w.r.t x, we get

dy over dx space equals space 1 space minus space 8 over straight x cubed
since the tangent is parallel to X- axis, therefore
dy/dx = 0
⇒ x3 = 8

⇒ x = 2 abd y =3

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20.

If g(x) is a polynomial satisfying g(x) g(y) = g(x) + g(y) + g(xy) - 2 for all real x and y and g(2) = 5, then limx3 g(x) is

  • 9

  • 10

  • 25

  • 20


B.

10

Since, g(x) g(y) = g(x) + g(y) + g(xy) - 2

Now, at x = 0, y = 2, we get

g(0) g(2) = g(0) + g(2) + g(0) - 2

                       g2 = 5

 5g0 = 5 + 2g0 - 2 3g0 = 3   g0 = 1

g(x) is given in a polynomial and by the relation given g(x) cannot be linear.

Let g(x) = x2 + k       g0 = 1

 From Eq. (i)  x2 + 1y2 + 1 = x2 + 1 + y2 + 1 + x2y2 + 1 - 2 limx3gx = g3 = 32 + 1 = 10


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