IIT JAM Mathematics Practice Test- 13 - Mathematics MCQ

Since the order of a differential equation is defined as the order of the highest derivative occurring in the differential equation, i.e for nth derivative dⁿy / dxⁿ if n = 1.
It has order 1→ differential equation contains only dy / dx derivative with variables and constants.

We have x²y”+xy’-y = 0, ...(1)

Compare this with

and given y = x is a solution of (1)

Then it’s 2nd L.I. solution = 

2nd L.l. solution = 1/x

(we can remove constant in (*))

We have (D² + 9)y = 0

⇒ y(x) = c₁cos 3x + c₂sin3x

y(0) = 0 ⇒c₁ = 0

y(2π) = 1 ⇒ c₁ = 1

⇒ DE has no solution.

Here

M  = xy²sinxy + y cosxy

So thus the integrating factor will be given by

⇒ f(x) is an increasing function .

⇒ max f(x) =f(3) and min f(x)= f(2)

⇒ Difference = f(3) -f(2)

hence ,minimum value of f(x) is 0 at x = 0. 

hence, no of points =1.

There is only one function in option(A), where critical point  but in other part critical point . then we can say that function in option (B),(C) and (D) are continuous on [0,1] and differentiable in (0,1).

now for 

here

⇒ f is not differentiable at 

f is not differentiable at  

⇒ LMVT is not applicable to f(x) in [0,11.

The equation of family of circles touching y-axis at the origin is

Here we know

global maxima of f(x) is 1.

then sol.

 the sign of f has no changes

 be a point of infection.

 the sign of f has changes from +ve to -ve

 be the point of local maxima and clearly it be the only point at which f has local maximum value.

here f(x)= - 20 +(x-a)ⁿ can not be for n ≥4 and n ≥ 5

Since if n = 4 then   

f(x) = -2 0 + ( x - a )⁴

similarly if n = 5 then

when x > 1 then f(x) = 0 , so by (1) 

be the continuous solution

Given problem is,

C le a rly the v alue o f (x₀,y₀) can not be (1,1), (0,1),(0,1) 

T h u s the p ro b le m has unique solution if (x₀, y₀) =(2 -1)

Given equation is,

Given y (1) = 1

The limit can only exist if f(x) is having least degree as that of 4.

Given that f(x) has local extremum at x = 0,1,2

Now f"(x) = -6x Now for minimum of f, f" should be greater than 0. 

and P.l. = 2e^x

= general soln : y(x) = c₁xe^x + c₂x² + 2e^x

Now y(0) = 0 =>0 = 2 Not possible

=> No particular solution exist w.r. to given conditions

Given equation is,

solving above equation we have, 

So the minimum value of y(x) is given by,

The value of α for which infimum is greater than equal to 1 is given by

We have y² = 4a(x + a) ...(1) diff.(1)

w.r. to x

Now for differentiable at (0,0)

  and 

Clearly when (h,k) - > (0,0) then 

f is differentiable at (0,0)

we have

now take

=> f adnd f' both are continous and differentiable on [0,1]

also given f(0) = f(1)

=> function f satisfy all the conditions of Rolle's theorem in [0,1] so 

Case-II: Now f' is continous in [0,1] and f' is diff. in (0,1) so we can apply firstly rake x> C, then apply LMVT on f' in [c,x] then we have

Now if take x<c then by LMVT on [x,c]

so by given y(0) = 0

⇒ C = o

 ⇒ y is continous at x = 0

⇒ y is continous everywhere on R

Now differentiable at x = 0

f is not differentiable at x = 0

now the relative minima of  y at x = 0

we have 

for maxima and minima  a = 0

y is minimum for all b and a = 0