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 Multiple Choice QuestionsMultiple Choice Questions

1.

let straight f left parenthesis straight x right parenthesis space equals space fraction numerator 1 minus space tan space straight x over denominator 4 straight x minus straight pi end fraction space comma space straight x not equal to space straight pi over 4 space straight x space element of space open square brackets 0 comma space straight pi over 2 close square brackets . If f(x) is continuous in open square brackets 0 space comma space straight pi over 2 close square brackets comma then f (π/4) is

  • 1

  • -1/2

  • -1

  • 1/2


B.

-1/2

straight f left parenthesis straight x right parenthesis space space equals space fraction numerator 1 minus space tan space straight x over denominator 4 straight x space minus space straight pi end fraction
limit as straight x rightwards arrow space straight pi divided by 4 of space straight f left parenthesis straight x right parenthesis space equals stack space lim with straight x rightwards arrow straight pi divided by 4 below space open parentheses fraction numerator 1 minus tan space straight x over denominator 4 space straight x minus straight pi end fraction close parentheses
By space straight L apostrophe space Hospital space rule
limit as straight x rightwards arrow space straight pi divided by 4 of space open parentheses fraction numerator negative sec squared space straight x over denominator 4 end fraction close parentheses space equals space fraction numerator negative sec squared space straight pi divided by 4 over denominator 4 end fraction space equals space minus 2 over 4
rightwards double arrow space limit as straight x rightwards arrow straight pi divided by 4 of space straight f left parenthesis straight x right parenthesis space equals space minus space 1 divided by 2
Also f(x) is continuous in [0, π/ 2] ,
so f(x) will be continuous at π / 4 .
∴ Value of function = Value of limit ⇒ f(π/ 4) = −1/ 2
118 Views

2.

The differential equation representing the family of curves straight y squared space equals space left parenthesis straight x plus square root of straight c right parenthesis , where c > 0, is a parameter, is of order and degree as follows:

  • order 1, degree 2

  • order 1, degree 1

  • order 1, degree 3

  • order 2, degree 2


C.

order 1, degree 3

straight y squared space equals 2 straight c space left parenthesis straight x space plus square root of straight c right parenthesis space..... space left parenthesis straight i right parenthesis
2 yy apostrophe space equals space 2 straight c.1 space or space yy apostrophe space equals space straight c space.... space left parenthesis ii right parenthesis
rightwards double arrow space straight y squared space equals space 2 yy apostrophe space left parenthesis straight x plus square root of yy apostrophe end root right parenthesis space left square bracket on space putting space value space of space cfrom space left parenthesis ii right parenthesis space and space left parenthesis straight i right parenthesis right square bracket
On space simplifying comma space we space get space
left parenthesis straight y minus 2 xy apostrophe right parenthesis squared space equals space 4 yy apostrophe cubed space left parenthesis iii right parenthesis
Hence space equation space left parenthesis iii right parenthesis space is space of space order space 1 space adn space degree space 3.
112 Views

3.

The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants is of

  • second order and second degree

  • first order and second degree

  • first order and first degree

  • second order and first degree


D.

second order and first degree

Ax squared space plus By squared space equals space 1 space space space space..... space left parenthesis 1 right parenthesis
Ax space plus space By dy over dx space equals 0 space.... space left parenthesis 2 right parenthesis
straight A space plus By space fraction numerator straight d squared straight y over denominator dx squared end fraction space plus straight B open parentheses dy over dx close parentheses squared space equals 0 space.... space left parenthesis 3 space right parenthesis
straight x open curly brackets negative By fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight B open parentheses dy over dx close parentheses squared close curly brackets plus By dy over dx space equals space 0
rightwards double arrow xy fraction numerator straight d squared straight y over denominator dx squared end fraction space plus straight x space open parentheses dy over dx close parentheses squared minus straight y dy over dx space equals space 0
281 Views

4.

The differential equation for the family of curves,x2 +y2 -2ay = 0  where a is an arbitrary constant is

  • 2(x2-y2)y' = xy

  • (x2+y2)y' = xy

  • 2(x2+y2)y' = xy

  • (x2- y2)y' = xy


A.

2(x2-y2)y' = xy

2x + 2yy′ - 2ay′ = 0

straight a space equals space fraction numerator straight x space plus space yy apostrophe over denominator straight y apostrophe end fraction space left parenthesis eliminating space straight a right parenthesis
rightwards double arrow space left parenthesis straight x squared minus straight y squared right parenthesis straight y apostrophe space equals space 2 xy space

144 Views

5.

If f is a real-valued differentiable function satisfying |f(x) – f(y)| ≤ (x – y)2 , x, y ∈ R and f(0) = 0, then f(1) equals

  • -1

  • 0

  • 2

  • 1


B.

0

straight f apostrophe left parenthesis straight x right parenthesis space equals space limit as straight h rightwards arrow 0 of space fraction numerator straight f left parenthesis straight x plus straight h right parenthesis minus straight f left parenthesis straight x right parenthesis over denominator straight h end fraction
vertical line straight f apostrophe space left parenthesis straight x right parenthesis vertical line space equals space limit as straight h rightwards arrow 0 of open vertical bar fraction numerator straight f left parenthesis straight x plus straight h right parenthesis minus straight f left parenthesis straight x right parenthesis over denominator straight h end fraction space close vertical bar less or equal than space limit as straight h rightwards arrow 0 of open vertical bar fraction numerator left parenthesis straight h right parenthesis squared over denominator straight h end fraction close vertical bar
⇒ |f′(x)| ≤ 0
⇒ f′(x) = 0
⇒ f(x) = constant As
f(0) = 0 ⇒ f(1) = 0
129 Views

6.

The solution of the differential equation ydx + (x + x2y) dy = 0 is

  • -1/ XY  =C

  • -1/XY + log y = C

  • 1/XY + log y = C

  • log y =Cx


B.

-1/XY + log y = C

y dx + x dy + x2y dy = 0

fraction numerator straight d space left parenthesis xy right parenthesis over denominator straight x squared straight y squared end fraction space plus space 1 over straight y space dy space equals space 0 space
rightwards double arrow space minus 1 over xy space plus space log space straight y space equals space straight C

168 Views

7.

A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness than melts at a rate of 50 cm3 /min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is

  • fraction numerator 1 over denominator 36 space straight pi end fraction space cm divided by min
  • fraction numerator 1 over denominator 18 space straight pi end fraction space cm divided by min
  • fraction numerator 1 over denominator 54 space straight pi end fraction space cm divided by min
  • fraction numerator 5 over denominator 6 space straight pi end fraction space cm divided by min

B.

fraction numerator 1 over denominator 18 space straight pi end fraction space cm divided by min
dv over dt space equals space 50
4 πr squared space dr over dt space equals space 50
rightwards double arrow space dr over dt space equals space fraction numerator 50 over denominator 4 straight pi space left parenthesis 15 right parenthesis squared end fraction space where space straight r space equals space 15
space equals space fraction numerator 1 over denominator 16 space straight pi end fraction
161 Views

8.

If straight x dy over dx space equals space straight y space left parenthesis log space straight y space minus space log space straight x space plus 1 right parenthesis commathen the solution of the equation is

  • straight y space log space open parentheses straight x over straight y close parentheses space equals cx
  • space straight x space log space open parentheses straight y over straight x close parentheses space equals space cy
  • log space open parentheses straight y over straight x close parentheses space equals space cx
  • log space open parentheses x over y close parentheses space equals space cx

C.

log space open parentheses straight y over straight x close parentheses space equals space cx
fraction numerator straight x space dy over denominator dx end fraction space equals space straight y space left parenthesis log space straight y space minus space log space straight x plus 1 right parenthesis
dy over dx space equals space straight y over straight x open parentheses log open parentheses straight y over straight x close parentheses plus 1 close parentheses
put space straight y space equals vx
dy over dx space equals space straight v plus space xdv over dx
rightwards double arrow space straight v space plus xdv over dx space equals space straight v space left parenthesis log space straight v plus 1 right parenthesis
xdv over dx space equals space straight v space log space straight v
rightwards double arrow space fraction numerator dv over denominator straight v space log space straight v end fraction space equals space dx over straight x
put space log space straight v equals space straight z
1 over straight v space dv space equals space dz
rightwards double arrow dz over straight z space equals space dx over straight x
In space straight z space equals space In space straight x space plus In space straight c
straight z space equals cx
log space straight v space equals space cx
log space open parentheses straight y over straight x close parentheses space equals cx
118 Views

9.

A function y = f(x) has a second order derivative f″(x) = 6(x – 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x – 5, then the function is

  • (x-1)2

  • (x-1)3

  • (x+1)3

  • (x+1)2


B.

(x-1)3

f''(x) = 6(x - 1) f'(x) = 3(x - 1)2 + c ........ (i)
At the point (2, 1) the tangent to graph is y = 3x - 5 Slope of tangent = 3
∴ f'(2) = 3(2 - 1)2 + c = 3
3 + c = 3
⇒ c = 0
∴ From equation (i) f'(x) = 3(x - 1)2
f'(x) = 3(x - 1)2 f(x) = (x - 1)3 + k ...... (ii)
Since graph passes through (2, 1)
∴ 1 = (2 - 1)2 + k k = 0
∴ Equation of function is f(x) = (x - 1)3

166 Views

10.

Let f be differentiable for all x. If f(1) = - 2 and f′(x) ≥ 2 for x ∈ [1, 6] , then

  • f(6) ≥ 8

  • f(6) < 8

  • f(6) < 5

  • f(6) = 5


A.

f(6) ≥ 8

As f(1) = - 2 & f′(x) ≥ 2 ∀ x ∈ [1, 6]
Applying Lagrange’s mean value theorem

fraction numerator straight f left parenthesis 6 right parenthesis space minus straight f left parenthesis 1 right parenthesis over denominator 5 end fraction space equals space straight f apostrophe space left parenthesis straight c right parenthesis space greater or equal than 2
rightwards double arrow space straight f left parenthesis 6 right parenthesis space greater or equal than space 10 plus space straight f left parenthesis 1 right parenthesis
rightwards double arrow space straight f left parenthesis 6 right parenthesis space greater or equal than space 10 minus 2
rightwards double arrow space straight f left parenthesis 6 right parenthesis space greater or equal than space 8

118 Views