A wire of length 2 units is cut into two parts which are bent re

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

1.

On the ellipse 4x2 + 9y2 = 1, the points at which the tangents are parallel to the line 8x = 9y, are

  • 25, 15

  • - 25, 15

  • - 25, - 15

  • 25, - 15


2.

A container s the shape of an inverted cone. Its height is 6 m and radius is 4m at the top. If it is filled with water at the rate of 3m/min then the rate of change of height of water(in mt/min) when the water level is 3 m is

  • 34π

  • 29π

  • 16π

  • 2π


3.

 If α, β, γ are the lengths of the tangents from the vertices of a triangle to its incircle. Then

  • α + β + γ = 1r2αβγ

  • α + β + γ = 1rαβγ

  • 1α + 1β + 1γ = rαβγ

  • α2 + β2 + γ2 = 2rαβγ


4.

If a cylindrical vessel of given volume V with no lid on the top is to be made from a sheet of metal, then the radius (r) and height(h) of the vessel so that the metal sheet used is minimum is

  • r = πV3, h =  πV3

  • r = πV, h = πV

  • r = Vπ3, h = Vπ3

  • r = Vπ, h = Vπ


Advertisement
Advertisement

5.

A wire of length 2 units is cut into two parts which are bent respectively to form a square of side=x units and a circle of radius=r units. If the sum of the areas of the square and the circle so formed is minimum, then:

  • 2x=(π+4)r

  • (4−π)x=πr

  • x=2r

  • x=2r


C.

x=2r

According to give information, we have
Perimeter of a square + perimeter of a circle
= 2 units

⇒ 4 x + 2πr = 2
 Now, let A be the sum of the areas of the square and the circle.
Then, A = x22r

space equals space straight x squared space plus space straight pi space fraction numerator left parenthesis 1 minus 2 straight x right parenthesis squared over denominator straight pi squared end fraction
rightwards double arrow space straight A space left parenthesis straight x right parenthesis space equals space straight x squared plus space fraction numerator left parenthesis 1 minus 2 straight x right parenthesis squared over denominator straight pi end fraction
Now space minimum space value space ofA space left parenthesis straight x right parenthesis comma space dA over dx space equals space 0
rightwards double arrow space 2 straight x space plus fraction numerator 2 left parenthesis 1 minus 2 straight x right parenthesis over denominator straight pi end fraction. left parenthesis negative 2 right parenthesis space equals space 0
space straight x space equals space fraction numerator 2 left parenthesis 1 minus 2 straight x right parenthesis over denominator straight pi end fraction
rightwards double arrow space straight pi space straight x plus space 4 straight x space equals space 2
straight x space equals space fraction numerator 2 over denominator straight pi space plus 4 end fraction space left parenthesis ii right parenthesis

Now space from space left parenthesis straight i right parenthesis space we space get
straight r space equals space fraction numerator 1.2. begin display style fraction numerator 2 over denominator straight pi plus 4 end fraction end style over denominator straight pi end fraction
space equals space fraction numerator straight pi space plus space 4 minus 4 over denominator straight pi left parenthesis straight pi plus 4 right parenthesis end fraction space equals space fraction numerator 1 over denominator straight pi plus 4 end fraction space... space left parenthesis iii right parenthesis
from space eqs space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis comma space we space get
space straight x space equals space 2 straight r

295 Views

Advertisement
6.

Let f (x) be a polynomial of degree four having extreme values at x =1 an x =2. If limit as straight x rightwards arrow 0 of open square brackets 1 plus fraction numerator straight f left parenthesis straight x right parenthesis over denominator straight x squared end fraction close square brackets space equals space 3 comma then f(2) is equal to 

  • -8

  • -4

  • 0

  • 0

204 Views

7.

A spherical balloon is filled with 4500π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72π cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is

  • 9/7

  • 7/9

  • 2/9

  • 2/9

1221 Views

8.

Let a, b ∈ R be such that the function f given by f(x) = ln |x| + bx
2+ ax, x ≠ 0 has extreme values at x = –1 and x = 2.
Statement 1: f has local maximum at x = –1 and at x = 2.
Statement 2: straight a space equals space 1 half space and space straight b space equals space fraction numerator negative 1 over denominator 4 end fraction

  • Statement 1 is false, statement 2 is true

  • Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

  • Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

  • Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

143 Views

Advertisement
9.

If dy/dx = y + 3 > 0 and y(0) = 2, then y(ln2) is equal to:

  • 7

  • 5

  • 13

  • 13

158 Views

10.

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point (-3, 1) and has eccentricity square root of 2 over 5 end root is

  • 3x2 + 5y2 -32 = 0

  • 5x2 + 3y2 - 48 = 0

  • 3x2 + 5y2 - 15 = 0 

  • 3x2 + 5y2 - 15 = 0 

276 Views

Advertisement