Wiredfaculty Logo

For Daily Free Study Material Join wiredfaculty Whatsapp Group | Download Android App | Ncert Book Download

Application of Derivatives

More Topic from Mathematics

Question 1
Wired Faculty App

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

  • 9/2

Solution

C.

2/9

Volume of balloon is,space straight V space equals space 4 over 3 πr cubed
After 49 minutes volume = 4500π – 49 (72π) = 972π
4 over 3 space πr cubed space equals space 972 space straight pi rightwards double arrow space straight r cubed space space equals space 729 rightwards double arrow space straight r space equals space 9 Now comma straight v equals 4 over 3 πr cubed differentiating space both space the space sides dv over dt space equals space 4 over 3 space straight pi 3 straight r squared dr over dt 72 space straight pi space equals space 4 πr squared space dr over dt dr over dt space equals space fraction numerator 72 over denominator 4.9.9 end fraction space equals space 2 over 9

Question 2
Wired Faculty App

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

  • 2x=r

Solution

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 = x2 +π2r
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

Question 3
Wired Faculty App

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 

  • 5x2 + 3y2 - 32 = 0

Solution

A.

3x2 + 5y2 -32 = 0

B.

5x2 + 3y2 - 48 = 0

straight x squared over straight a squared space plus straight y squared over straight b squared space equals space 1 9 over straight a squared space plus 1 over straight b squared space equals 1...... space left parenthesis 1 right parenthesis

case - 1 when a > b
b2 = a2 (1 - e2)
b2 = a2 (1 - 2/5)
5b2 = 3a2......... (2)
from (1) & (2)
fraction numerator 9 space straight x space 3 over denominator 5 straight b squared end fraction space plus space 1 over straight b squared space equals space 1 rightwards double arrow space straight b squared space space equals 32 over 5 therefore space straight a squared space equals space 32 over 3 therefore space fraction numerator 3 straight x squared over denominator 32 end fraction space plus fraction numerator 5 straight y squared over denominator 32 end fraction space equals space 1 rightwards double arrow space 3 straight x squared space plus space 5 straight y squared minus 32 Case space minus 2 When space straight b greater than straight a straight a squared space equals space straight b squared minus space left parenthesis 1 minus straight e squared right parenthesis space equals space 3 over 5 space straight b squared space........ space left parenthesis 3 right parenthesis from space left parenthesis 1 right parenthesis space & space left parenthesis 3 right parenthesis straight a squared space equals space 48 over 5 comma space straight b squared space equals space 16 therefore space fraction numerator 5 straight x squared over denominator 48 end fraction space plus straight y squared over 16 space equals space 1 rightwards double arrow space 5 straight x squared space plus 3 straight y squared minus 48 space equals space 0

Question 4
Wired Faculty App

Given P(x) = x4+ ax3 + cx + d such that x = 0 is the only real root of P′ (x) = 0. If P(–1) < P(1),then in the interval [–1, 1].

  • P(–1) is the minimum and P(1) is the maximum of P

  • P(–1) is not minimum but P(1) is the maximum of P

  • P(–1) is the minimum but P(1) is not the maximum of P

  • neither P(–1) is the minimum nor P(1) is the maximum of P

Solution

B.

P(–1) is not minimum but P(1) is the maximum of P

P(x) = x4+ ax3+ bx2+ cx + d
P′(x) = 4x3+ 3ax2+ 2bx + c
As P′(x) = 0 has only root x = 0
⇒ c = 0
P′(x) = x(4x2+ 3ax + 2b)
⇒ 4x3+ 3ax + 2b = 0 has non real root.
and 4x2+ 3ax + 2b > 0 ∀ x ∈ [−1, 1].

As P(−1) < P(1) ⇒ P(1) is the max. of P(x) in [−1, 1]

Question 5
Wired Faculty App

If straight x space equals space straight e to the power of straight y plus straight e to the power of straight y plus.... to space infinity end exponent space comma space straight x space greater than thin space 0 comma space then space dy over dx space is

  • x /x+1

  • x-1/x

  • 1/x

  • x+1/x

Solution

B.

x-1/x

straight x space equals space straight e to the power of straight y plus straight e to the power of straight y plus.... infinity end exponent end exponent straight x space equals straight e to the power of straight y plus straight x end exponent Taking space log space on space both space side log space straight x space equals space straight y plus straight x Diff. space straight w. straight r. straight t space to space straight x 1 over straight x space equals dy over dx plus 1 rightwards double arrow dy over dx space equals space fraction numerator 1 minus straight x over denominator straight x end fraction