Question

Let Q1, Q2, Q3 be constants so that (Q1, Q2) is the critical point of the function f(x, y) = (175)x 2 + (−150)xy + (175)y 2 + (−200)x + (400)y + (230), and Q3 = 1 if f has a local minimum at (Q1, Q2), Q3 = 2 if f has a local maximum at (Q1, Q2), Q3 = 3 if f has a saddle point at (Q1, Q2), and Q3 = 4 otherwise. Let Q = ln(3 + |Q1| + 2|Q2| + 3|Q3|). Then T = 5 sin2 (100Q)

satisfies:— (A) 0 ≤ T < 1. — (B) 1 ≤ T < 2. — (C) 2 ≤ T < 3. — (D) 3 ≤ T < 4. — (E) 4 ≤ T ≤ 5

Answer 1

The given two variable function is :

f(x, y) = (175)x2+ (−150)xy + (175)y2 + (−200)x + (400)y + (230)

At critical point :

fx=fy=0 ...(fx denotes partial derivative of f w.r.t x and so on)

Now,

fx = 175*2*x-150y-200=350x-150y-200 ...(1)

fy = -150x+175*2*y+400 =150x+350*y+400 ...(2)

Since ( Q1,Q2) is a critical point at ( Q1,Q2) (given) eqns. 1 and 2 equals 0 .So,

175*2*x-150y-200=350x-150y-200 =0 ...(3)

-150x+175*2*y+400 =-150x+350y+400=0 ...(4)

On solving (3) and (4)  we get

x=1/10

y=-11/10

So ( Q1,Q2) =(1/10,-11/10) .....(5)

To check weather  ( Q1,Q2) is a local maximum or minimum or saddle point we calculate discriminant

D=fxxfyy-(fxfy)2 ...(6)

from eqn (1)

fxx=350 ...(7)

and from eqn (2)

   fyy=350 ...(8)

from either (1) or (2)

fxy=-150 ...(9)

Therefore, from eqns (6), (7), (8) and (9)

D=350*350-(-150)2=107100 ...(10)

Now four cases arise:

1) If D > 0 and f xx ( Q1,Q2) > 0, then f has a local minimum at ( Q1,Q2),then Q3=1.
   2) If D > 0 and f xx ( Q1,Q2)< 0, then f has a localmaximum at (Q1,Q2),then Q3=2.
3) If D < 0, then f has a saddle point at (Q1,Q2),then Q3=3.
4) If D = 0, then no conclusion can be drawn,then Q3=4.

Since D>0 (eqn 10) and f xx ( Q1,Q2) > 0 ( eqn 7) so there is a local minimum at   (Q1,Q2) .And hence Q3=1 (for local minimum  Q3=1 is given).

Now,

Q = ln(3 + |Q1| + 2|Q2| + 3|Q3|) =ln(3+1/10+2*|(-11/10)|+3|1|)=ln((3*10+1+2*11+3*10)/10)=ln(8.3)

Now again,

T = 5 sin2 (100Q)= 5*sin2 (100*(ln(8.3))=0.6821

Hence T satisfies  0 ≤ T < 1

so (A) 0 ≤ T < 1 is the right option