Practice Test 2 — Detailed Solutions

GRE Math Subject · Axiom Academy

LESSON Practice Test 2 — Detailed Solutions Step-by-step solutions for 16 graduate-level problems covering multivariable calculus, group theory, real analysis, and more Evaluate ... where D is the disk x^2+y^2 . Convert to polar: x^2+y^2 = r^2 , dA = r dr dtheta . Find the directional derivative of f = x^2yz + xz^3 at (1,1,1) in the direction of v = (1,-1,2) . f_x = 2xyz + z^3 , at (1,1,1): 3 f_z = x^2y + 3xz^2 , at (1,1,1): 4 So grad f = (3,1,4) . Unit vector: v_hat = (1,-1,2)/sqrt(6) . The dot product gives 3-1+8 = 10 , so the directional derivative is 10/sqrt(6) , which is choice (A). Use Green's theorem to evaluate ... where C is the unit circle traversed counterclockwise. Here P=2y, Q=3x , so dQ/dx - dP/dy = 3-2 = 1 . Find the critical points of f = x^3 - 3xy + y^3 and classify them. Set grad f = 0 : f_x = 3x^2 - 3y = 0 and f_y = -3x + 3y^2 = 0 . From y = x^2 and x = y^2 : substituting gives x = x^4 , so x(x^3-1) = 0 . Critical points: (0,0) and (1,1). Second derivative test with D = f_xx*f_yy - (f_xy)^2 : At (0,0): D = -9 -- saddle point At (1,1): 0">D = 27 > 0 and 0">f_xx = 6 > 0 -- local minimum Let G be a group of order 15. Then G is: Since 15 = 3*5 and gcd(3,5) = 1, by the classification: if |G|=pq with p p does not divide q-1 , then G is cyclic. Here 3 does not divide 4 . So G must be cyclic: G = Z_15 . The center Z(G) consists of elements that commute with everything. In S_3 :

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