Math - 2019 Past Year Paper - IIT JAM MCQ

 Correct Answer :- A

Explanation : f(x,y,z) = g(x² + y² - 2z²).

df'/dx = g'(x² + y² - 2z²) (2x)

df"/dx” = g"(x² + y² - 2z²) (4x²) + g'(x² + y² - 2z²)*2.........(1)

df/dy = g'(x² + y² - 2z²) (2y)

df"/dy” = g"(x² + y² - 2z²) (4y²) + g'(x² + y² - 2z²)*2.........(2)

df'/dz = g'(x² + y² - 2z²) (2y)

df"/dz” = g"(x² + y² - 2z²) (4z²) + g'(x² + y² - 2z²)*2.........(3)

Adding (1), (2) and (3)

g"(x² + y² - 2z²)(4x² + 4y² + 16z²) + g'(x²+ y² - 2z²) (2 + 2 - 4)

= 4(x² + y² + 4z²) g"(x² + y² - 2z²)

If the radius of convergenceis 4, Then

If x^h y^k is an I.F. of differential equation, Then given equation become exact differential equation.
x^h u^k+1 (1 + xy)dx + x^{h +1} yk (1 – xy) dy = 0
So

=> (k + 1)y^kx^h + (k + 2)x^{h + 1}y^{k + 1}
= (h + 1)x^hy^k - (h + 1)x^{k + 1}y^{k + 1}
Comparing coefficients of both the sides, we have
h – k = 0
h + k – 4
⇒ h = –2, k = – 2

equation of the tangent plane at point (2, 0, 1) is (x – 2) fx_{(2, 0, 1)} + (y – 0) fy_{(2, 0, 1)} + (z – 1) fz_{(0, 0, 1)} = 0

Here fx_(2,0,1) = 3, fy_(2,0,1) =0, fz_(2,0,1) = 4
so, we have
(x – 2).3 + (y – 0).0 + (z – 1)4 = 0
⇒ 3x + 4z = 10
which is required tangent plane.

By the change of order of integration



Let t = (1 – x)²
dt = –2(1 – x) dx

Given P is a 4 × 4 matrix with rational entries.
Let characteristic polynomial of P is
hp(x) = x⁴ + ax³ + bx² + cx + d,   ...(i)
where a, b, c and d are rational.
Since √2 + i is a root of (1), Then √2 − i is also a root of (1)
⇒ (x² − 22x + 3) is a factor of (1)
Since x² + 3 − 2√2x is factor of (1), Then
x² + 3 + 2√2x is also a factor of (1)
⇒ (x² + 3 − 2√2x) (x² + 3 + 2√2x) is a factor of (1)
x⁴ + 6x² + 9 – 8x².
= x⁴ – 2x² + 9 ...(ii)
From (1) & (2), we have
P⁴ – 2P² + 9I = 0
⇒ P⁴ = 2P² – 9I

Let be defined as
f(x) = e^ax, a > 1 x ← R
f′(x) = ae^ax.
f(o) = ae^ao = 1 & f′(x) = ae^ax > e^ax = f(x) ∀ x ∈ R.
hence f(1) = e^a
.
e < ea < ∞
⇒ f(1). lies in the interval (e, ∞)

For option (b);

Let vn = 1 / n3
Then,

Asis convergent, sois convergent. 
For option (c);

Let vn = 1 / n3
Then
Thenis also convergent.
For option (d)





Sois also convergent.

Hence option (a) is divergent
(By Cauchy-condensation test)

For the orthogonal trajectories of the family of curves,

integrating, we have
log_e y = log_e√x + log c₁ ⇒ y = c₁√x
The curve passes through the point (1, 2), then c₁ = 2
Now, orthogonal trajectories in y = 2√x
At point (2, 2√2) satisfy the given condition.
 

Here x, x + e^x are two linear independent
solution. so general solution can be written as
y = c₁x + c₂ (x + e^x)

= (c₁+ c₂) x + c₂e^x

y′= (c₁+ c₂) + c₂e^x
.
y(0) = c₂ = 3
y′ (0) = 4 = c₁+ 2c₂

⇒ c₁ = 4 – 6 = –2
so, y(x) (–2 + 3) x + 3e^x = 3e^x+ x

y(1) = 3e + 1

Here
f_x = 3x² + 2y ⇒ f_xx = 6x, f_xy = 2
f_y = 2x + 3y² ⇒  f_yy = 6y
then f_xx f_yy – (f_xy)² = 36xy – 4
so at the point (0, 0) f_xx f_yy – (f_xy)² < 0
⇒ (0, 0) is a saddle point.

By Green's Theorem,




Now by using polar x = rcosθ, y = r sinθ. r = 0 to 2 cosθ, θ = 0 to 2θ.
we obtain the solution.

By stoke theorem,

Here








 

Here, {an} is a sequence of a positive real number.
The series converges if the series converges.