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Relations And Functions

Question
CBSEENMA12032506

Prove that:  integral subscript 0 superscript straight pi over 2 end superscript left parenthesis square root of tanx space plus space square root of cotx right parenthesis space dx space equals space straight pi square root of 2.

Solution

Let I = integral subscript 0 superscript straight pi over 2 end superscript space left parenthesis square root of tan space straight x end root space plus space square root of cot space straight x end root right parenthesis space dx
Put square root of tanx space equals space straight y space space space space space space or space space tan squared straight x space equals space straight y
rightwards double arrow space space space space left parenthesis 1 plus tan squared straight x right parenthesis space dx space equals space 2 straight y space dy space space rightwards double arrow left parenthesis 1 plus straight y to the power of 4 right parenthesis space dx space equals space 2 straight y space dy
therefore space space space space space space space space space space space dx space space equals fraction numerator 2 straight y over denominator 1 plus straight y to the power of 4 end fraction dy
When    straight x space equals space 0 comma space space space space straight y space equals space square root of tan space 0 end root space equals space 0
When    straight x equals straight pi over 2 comma space space space space straight y space equals space square root of tan space straight pi over 2 end root space rightwards arrow space infinity
therefore       straight I space equals space integral subscript 0 superscript infinity. open parentheses straight y plus 1 over straight y close parentheses. space fraction numerator 2 straight y squared over denominator 1 plus straight y to the power of 4 end fraction dy space equals space 2 integral subscript 0 superscript infinity fraction numerator straight y squared plus 1 over denominator straight y to the power of 4 plus 1 end fraction dy space equals space 2 integral subscript 0 superscript infinity fraction numerator 1 plus begin display style 1 over straight y squared end style over denominator straight y squared plus begin display style 1 over straight y squared end style end fraction dy
Put    straight y minus 1 over straight y space equals space straight t comma space space space space space space space space space space space space therefore space space space open parentheses 1 plus 1 over straight y squared close parentheses dy space equals space dt
Also   straight y squared plus 1 over straight y squared minus 2 space equals space straight t squared               rightwards double arrow space space space space space straight y squared plus 1 over straight y squared space equals space straight t squared plus 2
When y = 0, t → – ∞. When y → ∞, t → ∞
therefore                 straight I space equals space 2 integral subscript negative infinity end subscript superscript infinity fraction numerator 1 over denominator straight t squared plus 2 end fraction dt space equals space 2. space 2 integral subscript 0 superscript infinity fraction numerator 1 over denominator straight t squared plus 2 end fraction dt
                        equals space 4 integral subscript 0 superscript infinity fraction numerator 1 over denominator straight t squared plus left parenthesis square root of 2 right parenthesis squared end fraction dt space equals space 4. space space fraction numerator 1 over denominator square root of 2 end fraction open square brackets tan to the power of negative 1 end exponent space open parentheses fraction numerator straight t over denominator square root of 2 end fraction close parentheses close square brackets subscript 0 superscript infinity
                         equals space 2 square root of 2 space left square bracket space tan to the power of negative 1 end exponent infinity space minus space tan to the power of negative 1 end exponent 0 right square bracket space space equals 2 square root of 2 open square brackets straight pi over 2 minus 0 close square brackets space equals space straight pi square root of 2

Some More Questions From Relations and Functions Chapter

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.

Check whether the relation R in R defined by R = {(a,b) : a ≤ b3} is refleive, symmetric or transitive.

If R and R’ arc reflexive relations on a set then so are R ∪ R’ and R ∩ R’.