Linear MCQ Quiz in मराठी - Objective Question with Answer for Linear - मोफत PDF डाउनलोड करा

Last updated on Apr 24, 2025

पाईये Linear उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Linear एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Linear MCQ Objective Questions

Top Linear MCQ Objective Questions

Linear Question 1:

One of the points which lies on the solution curve of the following differential equation

2xydx+(x2+y2)dy=0

with the initial condition y(1) = 1 is

  1. (-1, 1)
  2. (0, 0)
  3. (0, 1)
  4. (2, 1)

Answer (Detailed Solution Below)

Option 1 : (-1, 1)

Linear Question 1 Detailed Solution

2xydx+(x2+y2)dy=0

dydx=2xyx2+y2

Put y = vx

dydx=v+xdvdx

v+xdvdx=2x(vx)x2+(vx)2

v+xdvdx=2v1+v2

xdvdx=2v1+v2v

xdvdx=(2v+v+v31+v2)

xdvdx==(3v+v31+v2)

(1+v23v+v3)dv=dxx

13(3+3v23v+v3)dv=dxx

By integrating on both sides,

lnx=13ln(3v+v3)+c

lnx=13ln(3yx+(yx)3)+c

Given that, y(1) = 1

ln1=13ln(3(1)+(1)3)+c

c=13ln4

Now, the solution becomes:

lnx=13ln(3yx+(yx)3)+13ln4

x3(3yx+(yx)3)=4

From the options, (-1, 1) satisfies the above condition.

Linear Question 2:

A differential equation dvdt0.4v=0 is applicable over -10 < t < 10. If v(5) = 10, the v(-4) is

Answer (Detailed Solution Below) 0.2 - 0.3

Linear Question 2 Detailed Solution

Given differential equation,

dvdt0.4v=0

(D – 0.4) v(t) = 0

⇒ D = 0.4

⇒ v(t) = K e+0.4t

Given that, v(5) = 10

⇒ K e+(0.4 × 5) = 10

⇒ k = 1.35

⇒ v(t) = 1.35 e+0.4t

⇒ v(-4) = 1.35 e+0.4(-4) = 0.27

Linear Question 3:

Consider the differential equation  dxdt=sin(x), with the initial condition x(0) = 0. The solution to this ordinary differential equation is

  1. x(t) = 0
  2. x(t) = sin (t)
  3. x(t) = cos (t)
  4. x(t) = sin (t) – cos (t)

Answer (Detailed Solution Below)

Option 1 : x(t) = 0

Linear Question 3 Detailed Solution

Given the differential equation is,

dxdt=sin(x)

dxsin(x)=dt

By integrating on both the sides,

dxsin(x)=dt

log |cosec x – cot x| = t + c

Taking antilog on both sides, we get

⇒ cosec x – cot x = et + c

⇒ tan(x/2) = ket

⇒ x(t) = 2 tan-1(ket)

Now, putting the initial condition x(0) = 0 i.e. at t = 0, x = 0.

⇒ 0 = 2 tan-1(ket)

⇒ k = 0

Now, the solution becomes,

x(t) = 2 tan-1(ket) = 0

Linear Question 4:

The curve y = f(x) is such that the tangent to the curve at every point (x, y) has a Y-axis intercept c, given by c = -y. Then, f(x) is proportional to

  1. x-1
  2. x2
  3. x3
  4. x4

Answer (Detailed Solution Below)

Option 2 : x2

Linear Question 4 Detailed Solution

Let the equation of tangent is,

y = mx + c

slope, m=dydx

y-intercept = c = - y

y=dydxxy

y=dydxxy

2y=dydxx

dyy=2dxx

⇒ lny = 2 lnx + lnk

⇒ y = kx2

So, y is proportional to x2.

⇒ f(x) ∝ x2

Linear Question 5:

The solution of the equation y' sin x = y ln y satisfying the initial condition y(π2)=e, is

  1. etan(x/2)
  2. ecot(x/2)
  3. lntan(x2)
  4. lncot(x2)

Answer (Detailed Solution Below)

Option 1 : etan(x/2)

Linear Question 5 Detailed Solution

ysinx=ylny

dydxsinx=ylny

dyylny=cosecxdx

By integrating on both sides,

=dyylny=cosecxdx

ln(lny)=ln(cosecx+cotx)+c

Given that, y(π/2) = e

⇒ ln (ln e) = -ln (cosec π/2 + cot π/2) + c

⇒ c = 0

The solution is,

ln (ln y) = -ln (cosec x + cot x)

lny=1cosecx+cotx

lny=11sinx+cosxsinx=sinx1+cosx=2sinx2cosx22cos2x2

lny=tanx2

y=etanx2

Linear Question 6:

The solution for the differential equation dydx=11+x2(etan1xy) is

  1. y=etan1x+Cetan1x
  2. y=12e2tan1x+Cetan1x
  3. y=12etan1x+Cetan1x
  4. y=12etan1x+Cetan1x

Answer (Detailed Solution Below)

Option 3 : y=12etan1x+Cetan1x

Linear Question 6 Detailed Solution

dydx=11+x2(etan1xy)

dydx+y1+x2=etan1x1+x2

Now, it is in the form of first order linear differential equation dydx+Py=Q

Integrating factor IF=ePdx=e11+x2dx=etan1x

Solution is: y(IF)=IF.Qdx

y(etan1x)=etan1xetan1x1+x2dx

Put etan1x=t

etan1x1+x2dx=dt

Now, the solution becomes

yt=tdt

yt=t22+C

y(etan1x)=12e2tan1x+C

y=12etan1x+Cetan1x

Linear Question 7:

If y=xx,x>0, then dydx is

  1. xlogx
  2. 1+logx
  3. xxlogx
  4. xx(1+logx)

Answer (Detailed Solution Below)

Option 4 : xx(1+logx)

Linear Question 7 Detailed Solution

Given y=xx,x>0

logy=xlogx

Differentiate with respect to x on both sides then we get

1y.dydx=x.1x+logx.1dydx=y(1+logx)dydx=xx(1+logx)

Linear Question 8:

Consider the equation dydx=y2x with the boundary y(1) = 1. Which one of the following is the correct range of x for which y is real and finite?

  1. (x3)(3x)
  2. -3 ≤ x ≤ 0
  3. 0 ≤ x ≤ 3
  4. 3x

Answer (Detailed Solution Below)

Option 4 : 3x

Linear Question 8 Detailed Solution

dydx=y2x

dyy2=dxx

=dyy2=dxx

1y=ln|x|+c

put y(1) = 1

1(1)=ln(1)+c

⇒ c = - 1

1y=ln|x|1

y=11ln|x|

For y to be finite

1ln|x|0

⇒ ln |x| ≠ 1

⇒ |x| ≠ e1

⇒ x ≠ ± e

⇒ x ≠ ±  2.718

So the correct range of x is: x ≠ ±  2.718 and x ≤ ∞

From the given options, 3 ≤ x ≤ ∞ satisfies the above conditions.

Linear Question 9:

Consider the differential equation dxdt=100.2x with initial condition x(0)=1. The response x(t) for t>0 is

  1. 2e0.2t
  2. 2e0.2t
  3. 5049e0.2t
  4. 5049e0.2t

Answer (Detailed Solution Below)

Option 3 : 5049e0.2t

Linear Question 9 Detailed Solution

dxdt=100.2x110x5.dx=dt5ln(10x5)=t+c(10x5)=e15(c+t)x5=10ec+t5x=505ec+t5

Given x(0)=11=505ec5

5ec5=49ec5=9.8x(t)=505ec5.et5=5049e0.2t

Linear Question 10:

The solution of the differential equation dydx+2xy=ex2 with y (0) = 1 is :

  1. (1+x)e+x2
  2. (1+x)ex2
  3. (1x)e+x2
  4. (1x)ex2

Answer (Detailed Solution Below)

Option 2 : (1+x)ex2

Linear Question 10 Detailed Solution

Explanation:

dydx+2xy=ex2

It is a linear differential equation of   dydx+Py=Q  form

Hence

P=2x,Q=ex2
Now, we have IF=e2xdx=ex2

Then, the solution of the equation is given by,

y×IF=(Q×IF)dx+Cy(ex2)=ex2ex2dx+Cy(ex2)=dx+Cy(ex2)=x+C

y(ex2)=x+C.................... (1)

From the given condition y(0) = 1, we have

1×e02=0+CC=1

Then from equation (1), we have

y(ex2)=x+1y=(1+x)ex2

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