IIT JAM Mathematics Practice Test- 2 - Mathematics MCQ

Hence , f is increasing 

So, x= 1/2 is the point of inflection as concavity changes at x = 1/2.

Let therefore 

y" indicates that it is of second order. Degree of y", y' and y is one. Hence the given differential equation is second order linear equation.

Putting y = vx

Now, putting all those values in equation (1), we get

   ..(1)

Let  then (1) gives 

which is a linear equation, of the form    whose   then

∴ Solution is 

On integrating, we get

be second linearly independent solution.

Given y = x³ is a solution.
So let y = zx³,

b the complete solution of the given equation, where z is a function of x to be determined.

From (i), we get y₁ = z₁x³ + 3x² z, y₂ = z₂ x³ + 6x² z₁ + 6xz.
Substituting these values of y₂, y₁ and y in the given equation we get x²( x³z₂ + 6x² z₁ + 6xz) + x(x³ z₁ + 3x² z) - 9x³ z = 0 or x⁵ Z₂ + 7x⁴ z₁ = 0 or XZ₂ + 7z₁ = 0

Integrating, log z₁ + 7 log x = log c₁ or z₁x⁷ = c₁

or

Integrating, z = c₁ (x^-6/ (-6) ) + c₂

∴ From (i), the required solution is 

Let f(x) = ax³ + bx² + cx + d

f(x) vanishes at x = -2 ,

thus

- 8a + 4b -2c + d= 0     (1)

Now f'(x) = 3ax² + 2bx + c

f(x) has maxima / minima at x = 1 /3 , x =(-1)

f'(-1)= 0 = f'(1/3)

or a + 2b + 3c =0

3a - 2b + c =0

after solving these eqⁿ we get
a=1,b = 1,c = -1,d = 2

eqⁿ is x³ + x² - x + 2

Given equation can be written in the standard form as

Then y = x is an integral belonging to the C.F. of the given equation.

we have

Now

Let cosπx= t

Let z = log x => x = e²

Put these values in given DE. we have,

Case -1 when 25 - 24λ > Qthen solⁿ is given by,

Where c₁ and c₂ are arbitrary finite  constants. Now we want when x → ∞  then y(x) -> 0 .
Now take

So in this case limit not exist.

So now no need to consider further case.

Now we can say no % exist for which all the solⁿ tends to 0 as   x → ∞ 

By (1) and (2) 

replace  in (3) we have

x² + 3y² = c² w here c² = -2c₁ ,where c₁ < 0 => Both I and ii are true and ii is the correct explanation of I.

we have y² = 4ax + 4a²    ...(1)

⇒ put it in (1), we have

⇒ y = 2 xy’ + y(y')²    ...(2) ;  DE o f given family

now replace  in (2) , to get the DE. of orthogonal trajectories,

which is the same as the differential equation (2) of the given system (1). Hence the system of parabolas (1) is self orthogonal i.e. each member of the given family of parabola's intersects its own member orthogonally.

we lave (x³ - 2y²) dx + 2xy dy = 0 ....(1) compare it with M dx + N dy = 0, we have

Mult, given eqⁿ (1) by its I.F. we have 

given y(1) = 1 ⇒c = 2

⇒ x³ + y² = 2x²

⇒y³ = 2x² = x³ = x³ (2 - x)

⇒ y(x) exist for all x < - 2.

⇒y(x) exist in the neighbourhood of 0. 

On solving (1) and (2) we get,

 E has minimum value . 

y has minimum values.

Then solution be,

Putting x = 1 , we get 

Putting x = 2 , we get

Now | f(x) - g(x) | < 2 

also f(2) = g(2)

=> 2x + 2 = 0

=> x = -1

f(x) - g(x) = 2x has no solⁿ.
Now | f(x) - g(x) |< 4 

In order that y = C₁y₁ + C₂y₂ may be the G.S. o f the given equation , y₁ and y₂ must b e L.l. Hence the wronskian o f y, and y2 m ust b e non zero

So w(y₁ ,y₂) ≠ 0 showing that  m≠n.

Let logx = z => x = e^z, then given equation can be written as,

So the solution be

Clearly both parts of C.F. be the solutions.

Given differential equation is y(xy + 2x²y²)dx + x(xy - x²y²)dy = 0....(1)

which is in the form Mdx + Ndy = 0 since 

[it is an exact differential equation]

On integrating both sides, we get

Substituting all these values in the first mean value theorem

Given that function is clearly continuous at any point (x, y) ≠ (0, 0) so for g to be continuous c m ust be in limit

We compute

(Put x = r cos θ, y = r sin θ) = 2.
so c = 2 is the value that makes g continuous.

u = log ( tan x + tan y)

and

add (1) and (2) 

Since f(x,y) is a homogeneous function of degree 2 then by the Euler’s theorem of homogeneous function we know that

(Here n is the degree o f the function)

[∵ degree o f function is two]

Given y(x) = v(x) secx be solⁿ, so it must satisfied the given DE. 

⇒ y' = V' sec x + v sec x tanx 

y” = v” secx + 2v' secx tanx + v [secx tan²x + sec³x]

Put these values in given DE, we have, 

v” secx + 2v' secx tanx + v [secx tan²x + sec³x] 

- 2 tanx [v' secx + v secx tanx] + 5 v secx = 0 

v” secx + v [sec³x - secx tan²x + 5secx] = 0 

⇒ v” + v [sec²x - tan²x + 5] = 0 

⇒ v" + 6v = 0

The operator form of differential equation is (x²D² + KxD + 1)y = 0 ...(1) 
Let x = e^t ⇒ t = log x 
x²D² ≡ D′(D′ – 1) D′ ≡ d / dt xD ≡ D′
then subsituting x2D2 and xD in equation (1), we get 
[D′(D – 1) + KD′ + 1]y = 0 
[D′² + (K – 1) D′ + 1]y = 0 
A.E. is m² + (K – 1)m + 1 = 0 
Solution of (1) is 
y = (c¹ + c² log x) / x = (c¹ + c² t) e ^{– t} 
⇒ K = – 1.

Re-write the given cliff. eqn as

put these values in (1) we have

Given

y(1) = 0 => 2 = 1 + 2 c

also (2,3) lies on y² = ax³ + b

⇒ 9 = 8a + b ⇒ b = - 7

⇒a + b = -5