Engineering Mathematics Question Bank focuses on “Euler’s Theorem – 2”.
1. In euler theorem x ∂z⁄∂x + y ∂z⁄∂y = nz, here ‘n’ indicates?
a) order of z
b) degree of z
c) neither order nor degree
d) constant of z
Answer: a
Explanation: Statement of euler theorem is “if z is an homogeneous function of x and y of order ‘n’ then x ∂z⁄∂x + y ∂z⁄∂y = nz”.
2. If z = xn f(y⁄x) then?
a) y ∂z⁄∂x + x ∂z⁄∂y = nz
b) 1/y ∂z⁄∂x + 1/x ∂z⁄∂y = nz
c) x ∂z⁄∂x + y ∂z⁄∂y = nz
d) 1/x ∂z⁄∂x + 1/y ∂z⁄∂y = nz
Answer: c
Explanation: Since the given function is homogeneous of order n, hence by euler’s theorem
x ∂z⁄∂x + y ∂z⁄∂y = nz.
3. Necessary condition of euler’s theorem is?
a) z should be homogeneous and of order n
b) z should not be homogeneous but of order n
c) z should be implicit
d) z should be the function of x and y only
Answer: a
Explanation:
Answer ‘z should be homogeneous and of order n’ is correct as statement of euler’s theorem is “if z is an homogeneous function of x and y of order ‘n’ then x ∂z⁄∂x + y ∂z⁄∂y = nz”
Answer ‘z should not be homogeneous but of order n’ is incorrect as z should be homogeneous.
Answer ‘z should be implicit’ is incorrect as z should not be implicit.
Answer ‘z should be the function of x and y only’ is incorrect as z should be the homogeneous function of x and y not non-homogeneous functions.
4. If (z=e^{frac{x^2+y^2}{x+y}}) then, (x frac{∂z}{∂x} + y frac{∂z}{∂y}) is?
a) 0
b) zln(z)
c) z2 ln(z)
d) z
Answer: b
Explanation:
Given (z=e^{frac{x^2+y^2}{x+y}}),let u=ln(z)=(frac{x^2+y^2}{x+y}=frac{x(1+(frac{y}{x})^2)}{(1+frac{y}{x})}) = x f(y/x)
Hence u is homogeneous of order 1,
Hence,
(x frac{∂u}{∂x}+y frac{∂u}{∂y})=u
Putting, u = ln(z) we get,
(x frac{∂z}{∂x}+y frac{∂z}{∂y}) = zln(z)
5. If (z=sin^{-1}frac{x^3+y^3+z^3}{x+y+z}) then, (xfrac{∂z}{∂x}+yfrac{∂z}{∂y}).
a) 2 tan(z)
b) 2 cot(z)
c) tan(z)
d) cot(z)
Answer: a
Explanation:
Given (z=sin^{-1}frac{x^3+y^3+z^3}{x+y+z}), put u=sin(z)=(frac{x^3+y^3+z^3}{x+y+z}=x^2 f(frac{y}{x},frac{z}{x}))
Hence, (x frac{∂u}{∂x}+y frac{∂u}{∂y}=2u)
Putting u = sin(z), we get
(x frac{∂u}{∂x}+y frac{∂u}{∂y}=frac{2Sin(z)}{Cos(z)}=2Tan(z))
6. Value of (x frac{∂u}{∂x}+y frac{∂u}{∂y}) if (u=frac{Sin^{-1} (frac{y}{x})(sqrt{x}+sqrt{y})}{x^3+y^3}) is?
a) -2.5 u
b) -1.5 u
c) 0
d) -0.5 u
Answer: a
Explanation: Since the function can be written as,
u=(x^{frac{-5}{2}} frac{Sin^{-1} (frac{y}{x})(1+sqrt{frac{y}{x}})}{1+(frac{y}{x})^3}=x^n f(frac{y}{x})), by euler’s theorem,
(x frac{∂u}{∂x}+y frac{∂u}{∂y} = -frac{5}{2} u)
7. If f1(x,y) and f2(x,y) are homogeneous and of order ‘n’then the function f3(x,y) = f1(x,y) + f2(x,y) satisfies euler’s theorem.
a) True
b) False
Answer: a
Explanation: Since f1(x,y) and f2(x,y) are homogeneous and of order n hence,
(x frac{∂f_1}{∂x}+y frac{∂f_1}{∂y} = nf_1 (x,y))
(x frac{∂f_2}{∂x}+y frac{∂f_2}{∂y} = nf_2 (x,y))
Hence adding these two equations,
We get
(x frac{∂f_1+f_2}{∂x}+y frac{∂f_1+f_2}{∂y} = nf_2 (x,y)+nf_1 (x,y))
(x frac{∂f_3}{∂x}+y frac{∂f_3}{∂y} = nf_3 (x,y))
Hence f3 satisfies euler’s theorem.
8. If (z=ln(frac{x^2+y^2}{x+y})-e^{frac{x^2+y^2}{x+y}}) then find (x frac{∂z}{∂x}+y frac{∂z}{∂y}).
a) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
b) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
c) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1+frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
d) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
Answer: b
Explanation:
Given (z=ln(frac{x^2+y^2}{x+y})-e^frac{x^2+y^2}{x+y})
Let, (u = ln(frac{x^2+y^2}{x+y})) and (v=e^(frac{x^2+y^2}{x+y})) hence z=u-v
Now, let (u’ = e^u = frac{x^2+y^2}{x+y}=xf(frac{y}{x})) hence u’ satisfies euler’s theorem,
Hence,
(x frac{∂u’}{∂x}+y frac{∂u’}{∂y}=u’)
Hence, by putting u’=eu, we get
(x frac{∂u}{∂x}+y frac{∂u}{∂y}=frac{e^u}{e^u} = 1), ……(1)
Now, let v’ = ln(v)= (frac{x^2+y^2}{x+y}=xf(frac{y}{x})) hence v’ satisfies euler’s theorem,
Hence,
(x frac{∂v’}{∂x}+y frac{∂v’}{∂y}=v’)
Hence, by putting v’=ln(v),we get
(x frac{∂v}{∂x}+y frac{∂v}{∂y}=vln(v)), …… (2)
By subtracting eq(1) and eq(2), we get
(x frac{∂z}{∂x}-y frac{∂z}{∂y}=1-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
9. If z = Sin-1 (x⁄y) + Tan-1 (y⁄x) then x ∂z⁄∂x + y ∂z⁄∂y is?
a) 0
b) y
c) 1 + x⁄y Sin-1 (x⁄y)
d) 1 + y⁄x Tan-1 (y⁄x)
Answer: a
Explanation: Given z = Sin-1 (x⁄y) + Tan-1 (y⁄x)
Let, u = Sin-1 (x⁄y) and v = Tan-1 (y⁄x) hence z = u + v
Now, let u’ = Sin(u) = x⁄y = f(x⁄y) hence u’ satisfies euler’s theorem,
Hence,
(x frac{∂u’}{∂x}+y frac{∂u’}{∂y}=0)
Hence, by putting u’=eu, we get
(x frac{∂u}{∂x}+y frac{∂u}{∂y}=0/e^u = 0) ,……(1)
Now, let v’= Tan(v)=y/x=f(y/x) hence v’ satisfies euler’s theorem,
Hence,
(x frac{∂v’}{∂x}+y frac{∂v’}{∂y}=0)
Hence, by putting v’=ln(v), we get
(x frac{∂v}{∂x}+y frac{∂v}{∂y}=0 ),……(2)
By adding eq(1) and eq(2), we get
(x frac{∂z}{∂x}+y frac{∂z}{∂y}=1+frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
10. If f(x,y)is a function satisfying euler’ s theorem then?
a) (x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
b) (frac{1}{x}^2 frac{∂^2 f}{∂x^2}+2/xy frac{∂^2 f}{∂x∂y}+frac{1}{y}^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
c) (x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}=nf)
d) (y^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+x^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
Answer: a
Explanation: Since f satisfies euler’s theorem,
(x frac{∂z}{∂x}+y frac{∂z}{∂y}=nz)
Differentiating it w.r.t x and y respectively we get,
(x frac{∂^2 u}{∂x^2}+frac{∂u}{∂x}+y frac{∂^2 u}{∂x∂y}=n frac{∂u}{∂x}),
and
(x frac{∂^2 u}{∂y}∂x+frac{∂u}{∂y}+y frac{∂^2 u}{∂y^2}=n frac{∂u}{∂y})
Multiplying with x and y respectively,
(x^2 frac{∂^2 u}{∂x^2}+x frac{∂u}{∂x}+xy frac{∂^2 u}{∂x∂y}=nx frac{∂u}{∂x}),
and
(xy frac{∂^2 u}{∂y}∂x+y frac{∂u}{∂y}+y^2 frac{∂^2 u}{∂y^2}=ny frac{∂u}{∂y})
Adding above equations we get
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}=n(n-1)u)
11. If (u = Tan^{-1} (frac{x^3+y^3}{x+y})) then, (x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) is?
a) Sin(4u) – Cos(2u)
b) Sin(4u) – Sin(2u)
c) Cos(4u) – Sin(2u)
d) Cos(4u) – Cos(2u)
Answer: b
Explanation:
Let, v = Tan(u) = x2 f(y/x)
By euler’s theorem,
g(u) = (x frac{∂u}{∂x}+y frac{∂u}{∂y} = 2frac{Tan(u)}{Sec^2 (u)} = Sin(2u))
Hence,
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y} = g(u)[g’(u)-1])
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) = Sin(2u)[2Cos(2u)-1] = Sin(4u)-Sin(2u)
12. If (u = e^{frac{(x^2+y^2)}{x+y}}) Then, (x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y})=?
a) u ln(u)
b) u ln(u)2
c) u [1+ln(u)]
d) 0
Answer: b
Explanation: Let, v = ln(u) = (frac{x^2+y^2}{x+y} = x f(frac{y}{x}))
Hence by applying euler theorem,
(x frac{∂v}{∂x}+y frac{∂v}{∂y}=v)
Hence,
g(u) = (x frac{∂u}{∂x}+y frac{∂u}{∂y}=u ln(u))
Hence,
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) = g(u)[g’(u)-1]
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) = u ln(u)[1+ln(u)-1] = u ln(u)2
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