IIT JAM Mathematics Mock Test- 3 - Mathematics MCQ

T(x₀ , x₁ , x₂ ) = (x₀ , x₀ + x₁, x₀ + x₁ + x₂ )

Let basis are (1, 0, 0), (0, x, 0), and (0, 0, x² ) 

 Then 

T(1, 0, 0) = (1, 1, 1) 

 T (0, x, 0) = (0, x, x) 

 T(0, 0, x²) = (0, 0, x²) 

Cofactors of T

T₁₁ = 1       T₁₂ = 0         T₁₃ = 0

T₂₁ = – 1     T₂₂ = 1        T₂₃ = 0 

T₃₁ = 0        T₃₂ = – 1     T₃₃ = 1 

∴  adj. T = Transpose of co-factors matrix = 

Hence T^-1 = 

Given that 

and 

By a well known th. we know that 

Given ∈ > 0, it follows from (2), (3), and (1) that there exists N ∈ I such that 

and such that

Hence

For any n ≥ N, choose x such that 1 – 1/n < x < 1 – 1/(n + 1). Then 1 – 1/N ≤ 1 – 1/n < x < 1, and so, by (6),

Now I – x^k = (1 – x) (1 + x + x² + .... + x^{k – 1}) ≤ k(1 – x), for any k ∈ I. 

Hence, since 1 – x < 1/n, we have 

 1 – x^k ≤ k(1 – x) < k/n . .....(8) 

By (8) and (5) we then have (since n ≥ N)

To estimate I3 we have, using (4),

But x < 1 – 1/(n + 1) and so 1 – x > 1/(n + 1). Thus (n + 1) (1 – x) > 1 and so

From (7) we then have

which proves that   converges to L.

= 

= 

Since here a_n = p(n) 

If let 

Radius of convergence of power series is given by

Given differential equation 

now on observing we can see that. The differential equation is neither Linear and homogeneous nor separable now putting y² = v - x

The given equation reduces to 

which is a homogeneous equation 

Let Y = (y_n) be the sequence of real numbers given by 

Clearly, Y is not a monotone sequence. However, if m > n, then 

Since ,  2^r-1≤ r! it follows that if m > n, then 

Therefore, it follows that (y_n) is a Cauchy sequence. Hence it converges to a limit y. At the present moment we cannot evaluate y directly; however, passing to the limit (with respect to m) in the above inequality, we obtain 

Hence we can calculate y to any desired accuracy by calculating the terms yn for sufficiently large n. The reader should to this and show that y is approximately equal to 0.632 120559. 

we find that 

"foe all real x. Thus, since W(sin x , cos x) ≠ 0 for all real x, we conclude that sin x and cos x are indeed linearly independent solutions of the given differential equation on every real interval.

Let f(x) = Sin x ( 1 + cos x) 

⇒ f'(x) = cosx (1 + cosx) + sinx (–sinx) = cosx + cos²x – sin²x 

= cosx + cos²x 

f"(x) = – sin x – 2sin2x = – (sinx + 2sin2x) 

for maximum or minimum value of f(x), f'(x) = 0 

therefore cos x + cos2 x = 0 

⇒ cos x = – cos2x

⇒ cosx = –cos (π ± 2x) 

⇒ x = π ± 2x 

⇒ x = ��3 , –π 

Now 

Therefore f(x) is maximum at x =π/3. 

Let iof T : R² → R³ be a Linear transformation such that matrix a with respect to standard basis of T is 

here clearly the columns of A are linearly independent b ⇒ T is one one mapping since matrix is 3 × 2 the column of A span R³ if A has 3 pivot positions but it is contradiction as A has 2 columns only

⇒ Associated Linear transformation is not onto 

Rank of matrix = Rank of Linear transformation = 2 

Given differential equation 

multiply both sides of the given equation by e^y we get 

Integrating factor = 

solution is 

where c is an arbitrary constant 

When origin O is inside S. In this case, divergence theorem cannot be applied to the region V enclosed by S, since  has a point to discontinuity at the origin. To remove this  difficulty, let us enclose the origin by a small sphare Σ of radius ε. 

The function F is continuously differentiable at the points of the region v´ enclosed between S and Σ. Therefore applying divergence theorem for this region V´, we have 

Now on the sphere Σ, the outward drawn normal n is directed towards the centre. Therefore on Σ, we have 

Let T : R³ + R³

such that the matrix for T is given by 

First we check out the eigenvalues of A is given by 

The Eigen vectors for l =1 is u1 = 

as 

and Eigen vectors for λ = –2 are v2 = 

and if P = 

Since Given system of linear equation 

It can be represented as AX = B 

here [A] can have rank 3 or less than 3 

i.e. ρ (A) ≤ 3

and [Ab] have rank 3 as we have given 

case(I) if ρ(A) = 3 

Then ρ(A) = ρ(Ab) 

⇒ system have exactly one solution 

Case(II) if 

ρ(A) < 3 

⇒ system have no solution. 

⇒ system can have at most one solution. 

2 and – 2

We want extreme values of f(x,y) = xy subject to the constraints 

To do so, we first find the value of x,y and λ for which 

∇f = λ∇g 

and g(x,y) = 0 

the gradient eq. in eq (1) gives

from which we find 

so y = 0 or λ = ± 2. we now consider these two cases 

case I if y = 0 then x = y = 0 But (0, 0) is not on the ellipse therefore y + 0. 

case II If y ≠ 0 then λ = ± 2 and x = substituting this in the eq. g(x,y) = 0

The function f(x,y) = xy therefore takes on its extrem values on the ellipse at the four points (± 2, 1), (± 2, – 1) 

The extreme values are xy = 2 and xy = – 2. 

Here S_n = Sum of first n terms of the given series 

Let f : (0, 2) → R be defined by

⇒ f(x) is differentiable only when x = 1  

i.e., f(x) is differentiable, exactly at one point

Since T : R³ → R³

such that T(2, 1, 1) = (3, 2, 1) 

T(1, 0, 1) = (0, 1 , –1) 

But let (–1, –2 ,1) = a (2, 1, 1) + b(1, 0, 1) 

on solving – 1 = 2a + b 

– 2 = a + 0 

a+ b = 1

⇒ b = 3 

so on applying transformation T on (1) bothsides 

T(1, -2, 1) = T(–2, (2, 1,1) + 3(1, 0, 1) = –2 T(2, 1, 1) + 3T(1, 0, 1) 

= –2 (3, 2, 1)+ 3(0, 1, –1) = (–6, –1, –5) 

(a) Let  denote the zero polynomial, so  for every value of x. 

 ∈ W, since  Suppose f. g ∈ W. Then f(1) = 0 and g(1) = 0. Also, for scalars a and b, we have 

(af + bg) (1) = af(1) + bg(1) = a0 + b0 = 0 

Thus af + bg ∈ W, and hence W is a subspace. 

(b) . Suppose f, g ∈ W. Them f(3) = f(1) and g(3) g(1). Thus, for any scalars a and b, we have

(af + bg)(3) = af(3) + bg(3) = af(1) + bg(1) = (af + bg) (1) 

Thus af + bg ∈ W, and hence W is a subspace.

 (c) 0 ∈ W, . Suppose f . g ∈ W. Then f(–x) = –f(x) and g(–x) = –g(x). 

Also, for scalar a and b,

(af + bg)(–x) = af (–x) + bg(–x) = –af(x) – bg(x) = –(af + bg)(x) 

Thus ab + gf ∈ W, and hence W is a subspace of V. 

f₁g ∈ W

f(4) = 3 + f(2) 

and g(4) = 3 + g(2) 

If a, b ∈ ℝ, then 

af + bg)(4) = af (4) + bg(4) = a[3 + f(2)] + b[3 + g(2)], from (i) 

= af(2) + bg(2) + 3a + 3b = (af + bg) (2) + 3(a + b) 

≠ 3+( af bg )(2) 

∴ (af+ bg)( 4) ∉ W

Hence, W is not a subspace of V. 

The definition of lipschitz function is given by 

Let A R ⊆ and let f : A → ℝ if there exists a constant k > 0 such that 

so we have lipschitz function in our question asked (as it satisfies lipschitz condition) now we will show uniform continuity of lipschitz function 

Since 

The given ε > 0 we can take δ = ε/k if x. y ∈ ℝ satisfy ��-�� < δ

then, 

Therefore f is uniformly continuous on R 

But for differentiability

Since 

⇒ f is differentiable at y ∀ ∈ x,y ℝ

⇒ f is differentiable function to 0

By a well known result we know that if A be an n × n matrix with coefficients in F, having rows {a₁, a₂, ....., a_n }, then the following statements are true. 

 (a) if A’ be a matrix obtained from A by an elementary row operation (interchanging two rows). Then 

 D(A’) = – D(A) 

 (b) if A’ be a matrix obtained from a by an elementary row operation (replacing the row a_i by λa_j , with λ ∈ F, i ≠ j). Then 

D(A’) = D(A) 

(c) if A’ be a matrix obtained from A by an elementary row operation (replacing ai by µai , for µ ≠ 0 in F). Then 

 D(A’) = µD(A)

i.e. all the three options are correct. 

Since y₁ = 1 + x and y₂ = e^x be two solutions of 

y’’(x) + P(x) y’(x) + Q(x) y(x) = 0 ............(1) 

then will satisfy (1) 

so if y₁ = 1 + x if y₂ = e^x

and P(x) + Q(1 + x) = 0 

(1 + P + Q)e^x = 0 

⇒ P + Q(1 + x) = 0 

P + Q + 1 = 0 

on solving them we get 

on solving equation set (1) 

we get no solution 

Thus for the conditions y(0) = 2, y’(0) = 1 

differential equation has no solution

Since, we know that every group of order p² is cyclic, where p is a prime integer.

∴  Group of order p² is abelian also

∴  Group of order 7² i.e, 49 is abelian and cyclic.

It is a very well known theorem that if f and g be continuous real – valued function on the metric space M and A be the set of all x ∈ M s.t. f(x) < g(x) then A is open set. 

Let L(w) = w 

and x,y ∈ w,   α, β ∈ FE 

then x,y∈ L(w) 

x,y are linear combination of members of w. 

⇒ αx + βy is a linear combination of members of w 

⇒ αx + βy ∈ L(w) 

⇒ αx + βy ∈w 

⇒ w is a subspace of v. 

A.E. is m² + 4 = 0 m = 2i, -2i 

C.F. is y = C₁cos 2x + c₂ sin 2x 

In this case the region D will be the region between these two circles and that will only change the limits in the double integral. 

Here is the work for this integral. 

Given differential equation is:

Here 

which shows that the transformation  reduces the given differential equation into the form .