Maths-
General
Easy

Question

Assertion left parenthesis A right parenthesis colon If u equals log space open parentheses fraction numerator x squared plus y squared over denominator x minus y end fraction close parentheses then x squared u subscript x x end subscript plus 2 x y u subscript x y end subscript plus y squared u subscript y y end subscript equals negative 1

Reason left parenthesis bold R right parenthesis colon If u equals f left parenthesis x comma y right parenthesis  is a homogeneous function in x comma y of degree ' n ' then x squared u subscript x x end subscript plus 2 x y u subscript x y end subscript plus y squared u subscript y y end subscript equals n left parenthesis n minus 1 right parenthesis u

  1. Both A and R true, but R is the correct explanation of A
  2. Both A and R are true, but R is not the correct explanation of A
  3. A is true, but R is false
  4. A is false, but space R space is true

hintHint:

We are given two statements. First is an assertion and second is the reason. We have to find which of the state is true. And if both are true, we have to find if second statement is correct explanation of first.

The correct answer is: Both A and R are true, but R is not the correct explanation of A


    We will check both the statements one by one.
    A: The given function is u equals log open parentheses fraction numerator x squared plus y squared over denominator x minus y end fraction close parentheses
    We have to find the value of x squared u subscript x x end subscript plus 2 x y u subscript x y end subscript plus y squared u subscript y y end subscript
    We will first simplify the given function and find it's degree. The function should be homogeneous for further operations. Hence, we need to simplify it.
    L e t space z space equals e to the power of u
s o comma space z space equals space fraction numerator x squared plus y squared over denominator x minus y end fraction
space space space space space space z space equals space fraction numerator x squared open parentheses 1 space plus space begin display style y squared over x squared end style close parentheses over denominator x open parentheses 1 space minus begin display style y over x end style close parentheses end fraction
space space space space space space space space space equals x f open parentheses y over x close parentheses
S o comma space t h e space d e g r e e space o f space t h e space f u n c t i o n space i s space 1.
n space equals space 1
    z= f(u)
    So, xuxx + 2xyuxy + yuyy = g(u)[g'(u) - 1]    ...(1)
    g left parenthesis u right parenthesis space equals space n fraction numerator f left parenthesis u right parenthesis over denominator f apostrophe left parenthesis u right parenthesis end fraction
    We will find all the values.
    z space equals space f left parenthesis u right parenthesis
f left parenthesis u right parenthesis space equals space e to the power of u
f apostrophe left parenthesis u right parenthesis space equals space e to the power of u
n space equals space 1
g left parenthesis u right parenthesis space equals left parenthesis 1 right parenthesis space e to the power of u over e to the power of u
g left parenthesis u right parenthesis equals 1
S o comma space t h e space d e r i v a t i v e space o f space t h e space a b o v e space f u n c t i o n space w i l l space b e space
g apostrophe left parenthesis u right parenthesis space equals space 0
    Substituting the values in equation (1) we get,
    x2uxx + 2xyuxy + y2uyy = 1(0 - 1)
    = -1
    So, the first statement is true.
    R: The second statement is directly the property of homogeneous equation of degree n and function of x and y. So, the second statement is true.
    Now, if we see the first statement has z as function of f(u). It cannot be directly found from the second statement. We have to make the changes to find the value for the first statement.
    Both the statements are true but, R is not correct explanation of A.

    For such questions, we should know the formulas related to homogeneous equation. Before solving the equation, we have to see if the given function is a homegenous function. A homegenous function has same degree for all its terms.

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