IIT JAM Mathematics Mock Test- 1 - Mathematics MCQ

Given that f : (0, 2) → R then

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

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

Given differential equestion is 

multiply bothe of side of equation by e^y we get

Take e^y = t

Integrating factor = 

solutions is

where c is an arbitrary constant

where u = e^x

so 

now 

when x = log2

= not defind

The given differential equation is:

x log x (dy/dx) + y = log x²

First, rewrite the equation in standard linear form by dividing through by x log x:

(dy/dx) + (1 / (x log x)) y = 2 / x

Compute the integrating factor μ(x) which is

μ(x) = e^{∫ (1 / (x log x)) dx} = log x

Multiply through by μ(x):

log x * (dy/dx) + (1 / x) y = (2 log x) / x

Recognize the left-hand side as the derivative of y log x:

d/dx [y log x] = (2 log x) / x

Integrate both sides:

y log x = (log x)² + C

Solve for y:

y = log x + C / log x

This matches option b. Verifying by substitution confirms the solution satisfies the original equation.

Answer: b

Putting x = 1 - x

Multiplying equation(2) by 2 and subtracting from (1)

Since T :  

Such that

But let 

on solving -1 = 2a + b
-2  = a + 0
a + b = 1
⇒ b = 3
So applying transformation T on (1) bothside

We have,

y² = 2x and x² + y² = 4x

y² = 2x is parabola opening to the right of positive direction x-axis 

x² + y² = 4x

⇒ (x - 2)² + y² = 4 which is circle having center at (2,0) and radius 2 units

Solving the curves we get, 

When x = 0, y = 0 and when x = 2, y = ±2
Thus points of intersection are (0,0), (2,2) and (2,-2)

The graph is as shown in the following figure:

Let,

Applying the general Leibnity's rule we obtain

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 , 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 approxi-
mately equal to 0.632 120559.

2014 starts with a Wednesday

1234567

WTFSSM T

So if Wednesday is day n = 1 then Sundays will come on days n = 5,12,19,…96, total 15 Sundays.

Similarly if we start with-

Thursday - 14 sundays 

Friday - 14 sundays 

Saturday - 15 sundays 

Sunday - 15 sundays 

Monday - 14 sundays 

Tuesday - 14 sundays

So, probability 

⇒ P = 2/7

Since here a_n = p(n)

If let p(x) = 

Radius of convergence of power series is given by

Let i of 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 map-ping since matrix is 3 x 2 the column of A span R³ if Ahas 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

Since  be two solutions of,

Then will satisfy (1),

So if 

On solving them we get,

Differential equation becomes,

And solution is of the form,

Take options,

On solving equation set (1) we get no solution.

Thus for the conditions  differential equation has no solution.

Condition 1: 

(It is Homogeneous)

Condition 2: 

Equation (1) can be written as 

It is not a linear form.

It is in linear form.

Condition 3: 

So, it is an exact equation.

We find that

for 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 differen-tial equation on every real interval.

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.

Adding (i) and (ii), we get

The LCM of 4,5,6 and 7 = 420

so, the required number = 420k + 3

which is exactly divisible by 9 for some value of k

Now, 420K + 3 = 46 x 9k + 6k + 3

When k = 1, 6k + 3 = 9, which is divisible by 9

So, the required number = 420 x 1 + 3 = 423

We know that

Now by differentiating equation (1) with respect to x, we get:

Hence 

In this case the region D will be the region beWeen these tvvo circles and that will only change the limits in the double integral.
Here is the work for this integral.

We have, Cartesian sub tangent + abscissa = constant

Integrating, we get 

As the curve passes through the point (2a, a), we have c = a²

The required curve is .

We know that for maximum or minimum value of 

Therefore f(x) has maximum value at .

Given,

The given equation can be solved using the variable separable method as,

Integrating both sides, we get,

Now, from equation (1), we get

∴ y = constant = 1  is also a solution to the given differential equation.

Let n be the number of children in each family 1/12 = 

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

the gradient eq. in eq (1) gives

from which we find

and
so y = 0 or λ = ±2. we now consider these Wvo cases
case I if
y = O 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
given