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Relations And Functions

Question
CBSEENMA12036013

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 = 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

Some More Questions From Relations and Functions Chapter

If a matrix A has 12 elements, what arc the possible orders it can have 7 What if it has 7 elements ?

Let A be the set of all students of a boys school. Show that the relation R in A given by R = {(a, b) : a is sister of b} is the empty relation and R’ = {(a, b) : the difference between heights of a and b is less than 3 meters} is the universal relation.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Give an example of a relation which is

(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1, L2) : L1 is perpendicular to L2}. Show that R is symmetric but neither reflexive nor transitive.

 Determine whether each of the following relations are reflexive, symmetric and transitive :

(i) Relation R in the set A = {1, 2, 3,....., 13, 14} defined as

R = {(x, y) : 3 x – y = 0}

(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x,y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x,y) : x – y is an integer}

(v) Relation R in the set A of human beings in a town at a particular time given by
(a)    R = {(x, y) : x and y work at the same place}
(b)    R = {(x,y) : x and y live in the same locality}
(c)    R = {(x, y) : x is exactly 7 cm taller than y}
(d)    R = {(x, y) : x is wife of y}
(e)    R = {(x,y) : x is father of y}

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.