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Relations And Functions

Question
CBSEENMA12036073

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 false

Solution

B.

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

(i) A function f, such that f(x)= log |x| +bx2 +ax, x≠0
(ii) The function 'f' has extrema at x = -1 and x =2 i.e, f'(1) = f'(2) = 0 and f''(-1) ≠ 0≠f''(2)
Now, given function f is given by 
f(x) = log |x| +bx2 +ax
rightwards double arrow space straight f apostrophe left parenthesis straight x right parenthesis space equals space 1 over straight x space plus 2 bx space plus straight a
rightwards double arrow space straight f apostrophe apostrophe space left parenthesis straight x right parenthesis space equals space fraction numerator negative 1 over denominator straight x squared end fraction space plus 2 straight b
Since 'f' has extrema at x = - 1 and x =2
Hence, f'(-1) = 0 =f'(2)
f'(-1) = 0 
⇒ a-2b =1 ..... (i)
and f'(2) = 0 
⇒ a+ 4b = -1/2
solving eq. (i) and (ii), we get
a =1/2 and b = -1/4
straight f apostrophe apostrophe space left parenthesis straight x right parenthesis space equals space fraction numerator negative 1 over denominator straight x squared end fraction space plus fraction numerator negative 1 over denominator 2 end fraction space equals space minus space open parentheses fraction numerator straight x squared space plus 2 over denominator 2 straight x squared end fraction close parentheses
⇒ f'' has local maxima at both x = - 1 and x =2
Thus, a statement I is correct. Also, while solving for the statement I, we found values of a and b, which justify that statement 2 is also correct.

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.