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 Multiple Choice QuestionsMultiple Choice Questions

1.

limx01 - cos2xsin5xx2sin3x equals

  • 103

  • 310

  • 65

  • 56


A.

103

We have, limx01 - cos2xsin5xx2sin3x= limx01 - cos2xx2 limx0sin5xsin3x= limx02sin2xx2limx0sin5x5x 3xsin3x . 5x3x= limx02sinxx . 53= 2 × 53= 103


2.

limx0ax - bxex - 1 is equal to 

  • logeab

  • logeba

  • logeab

  • logea + b


A.

logeab

limx0ax - bxex - 1 = limx0ax - bxx . xex - 1= limx0ax - 1x - bx - 1x . xex - 1= limx0ax - 1x - limx0bx - 1x . limx0xex - 1= logea - logeb . limx01ex - 1x= logeab . 1= logeab 


3.

The value of limxa2x2 + ax + 1 - a2x2 + 1 is

  • 12

  • 1

  • 2

  • None of these


A.

12

Given, limxa2x2 + ax + 1 - a2x2 + 1= limxa2x2 + ax + 1 - a2x2 - 1a2x2 + ax + 1 + a2x2 + 1= limxaxa2 + ax + 1x2a2 + 1x2= aa2 + a2 = a2a = 12


4.

limx1 - 4x - 13x - 1 is equal to

  • e12

  • e- 12

  • e4

  • e3


B.

e- 12

Given, limx1 - 4x - 13x - 1= limx1 - 4x - 1- x - 14- 43x - 1x - 1= e- 4limx3 - 1x/1 - 1x= e- 4 × 3= e- 12


5.

If limit as straight x space rightwards arrow infinity of space open parentheses 1 plus straight a over straight x plus straight b over straight x squared close parentheses to the power of 2 straight x end exponent space equals space straight e squared then the values of a and b, are

  • straight a element of space straight R with equals below space straight b element of space straight R with equals below
  • straight a space equals 1 comma space straight b element of space straight R with equals below
  • straight a element of space straight R with equals below space comma space straight b space equals 2
  • straight a element of space straight R with equals below space comma space straight b space equals 2

D.

straight a element of space straight R with equals below space comma space straight b space equals 2
limit as straight x space rightwards arrow infinity of space open parentheses 1 plus straight a over straight x plus straight b over straight x squared close parentheses to the power of 2 straight x end exponent
space equals space limit as straight x space rightwards arrow infinity of space open parentheses 1 plus fraction numerator begin display style straight a end style over denominator begin display style straight x end style end fraction plus fraction numerator begin display style straight b end style over denominator begin display style straight x squared end style end fraction close parentheses to the power of open parentheses fraction numerator 1 over denominator begin display style straight a over straight x plus straight b over straight x squared end style end fraction close parentheses space straight x space 2 straight x space straight x space open parentheses straight a over straight x plus straight b over straight x squared close parentheses end exponent
space equals space straight e to the power of 2 straight a end exponent
rightwards double arrow space straight a space equals 1
straight b element of straight R
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6.

limxπ2acotx - acosxcotx - cosx

  • logeπ2

  • loge2

  • logea

  • a


C.

logea

limxπ2acotx - acosxcotx - cosx= limxπ21 + cotxlogea + cot2x2!logea2 + ... - 1 - cosxlogea - cos2x2!logea2 - ...cotx - cosx= limxπ2logea + cotx + cosx2!logea2 + ...cotx - cosxcotx - cosx= limxπ2logea + cotx + cosx2!logea2 + ...= logea


7.

If the normal to the curve y = f(x) at (3, 4) makes an angle 3π4 with the positive x-axis, then f'(3) is equal to :

  • - 1

  • 34

  • 1

  • 34


A.

- 1

 α = 3π4and we know that f' (3) = tanα                                    = tan3π4                                    = - 1


8.

The solution of the differential equation dy over dx space equals space fraction numerator straight x plus straight y over denominator straight x end fraction   satisfying the condition y (1) = 1 is  

  • y = ln x + x

  • y = x ln x + x2

  •  y = xe(x−1)

  •  y = xe(x−1)


D.

 y = xe(x−1)

y = vx

dy over dx space equals space straight v space plus space straight x dv over dx
straight v space plus space straight x dv over dx space equals space 1 space plus space straight v
rightwards double arrow space dv space equals space dx over straight x
therefore space straight v space equals space log space straight x space plus space straight c
rightwards double arrow space straight y over straight x space equals space log space straight x space space plus straight c

Since, y (1) = 1, we have y = x log x + x

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9.

dndxnlogx is equal to

  • n - 1!Xn

  • n !Xn

  • n - 2!Xn

  • - 1n - 1n - 1!Xn


D.

- 1n - 1n - 1!Xn

Let y = log(x)

On differentiating w.r.t. x from 1 to n times, we get

y1 = 1x, y2 = - 1x2y3 = 2x3, y4 = - 6x4yn = - 1n - 1n - 1!xn


10.

The value of limn1n + 1 + 1n + 2 + ... + 16n is

  • log2

  • log6

  • 1

  • log3


B.

log6

We know that,

limn1na + 1 + 1na + 2 + ... + 1nb = logba limn1n + 1 + 1n + 2 + ... + 16n = log61                                                               = log6