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NEB Class 12 Science Mathematics (NEB Released Items) Question Paper 2081 (Set B) Nepal

NEB Class 12SCIENCEmodel
marks
min
4questions

This is the official NEB Class 12 (Science stream) Mathematics (NEB Released Items) question paper for 2081 Set B, as set in the model model examination. It carries — full marks, across 4 questions. On Kekkei you can attempt this Mathematics (NEB Released Items) past paper online with a timer, get instant AI feedback and step-by-step solutions, and track the topics where you lose marks — completely free. Whether you are revising for your NEB Class 12 Mathematics (NEB Released Items) exam or solving previous years' question papers, this 2081 paper is a great way to practise under real exam conditions.

A

Group A

Multiple Choice Questions carrying 1 mark each. Copy the correct option in the answer sheet.

2 questions·1 marks each
1mcq1 marks

If 1,ω,ω21, \omega, \omega^2 are the cube roots of unity, then which of the following is true?

  • A

    ω+ω2=1\omega + \omega^2 = 1

  • B

    ω3=1\omega^3 = 1

  • C

    ω2=1\omega^2 = 1

  • D

    1−ω=ω21 - \omega = \omega^2

complex-numberscube-roots-of-unityalgebra
2mcq1 marks

If the line y=mx+cy = mx + c touches the circle x2+y2=a2x^2 + y^2 = a^2, then what is the value of cc?

  • A

    c=±1+m2c = \pm\sqrt{1 + m^2}

  • B

    c=±a2+m2c = \pm\sqrt{a^2 + m^2}

  • C

    c=±1+a2m2c = \pm\sqrt{1 + a^2 m^2}

  • D

    c=±a1+m2c = \pm a\sqrt{1 + m^2}

conic-sectionscircletangent-line
B

Group B

Short Answer Questions carrying 5 marks each. There may be sub-questions within a question.

1 questions·5 marks each
3short5 marks

a) In an examination, a candidate has to pass in each of the four subjects. In how many ways can the candidate fail? [2]

b) If y=x+x22+x33+x44+⋯y = x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \dfrac{x^4}{4} + \cdots, show that x=y−y22!+y33!−y44!+⋯x = y - \dfrac{y^2}{2!} + \dfrac{y^3}{3!} - \dfrac{y^4}{4!} + \cdots. [3]

a) A candidate fails if they fail in at least one of the four subjects. The number of ways of failing is the number of ways of choosing 1, 2, 3 or 4 subjects to fail in:

C(4,1)+C(4,2)+C(4,3)+C(4,4)=4+6+4+1=15.C(4,1) + C(4,2) + C(4,3) + C(4,4) = 4 + 6 + 4 + 1 = 15.

So the candidate can fail in 15 ways. (Equivalently, total subsets 24=162^4 = 16 minus the one way of passing all = 15.)

b) Given

y=x+x22+x33+x44+⋯y = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots

Recall the logarithmic series −ln⁡(1−x)=x+x22+x33+⋯-\ln(1-x) = x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots. Hence

y=−ln⁡(1−x)⇒−y=ln⁡(1−x).y = -\ln(1-x) \quad\Rightarrow\quad -y = \ln(1-x).

Taking exponentials,

1−x=e−y=1−y+y22!−y33!+y44!−⋯1 - x = e^{-y} = 1 - y + \frac{y^2}{2!} - \frac{y^3}{3!} + \frac{y^4}{4!} - \cdots

Therefore

x=y−y22!+y33!−y44!+⋯x = y - \frac{y^2}{2!} + \frac{y^3}{3!} - \frac{y^4}{4!} + \cdots

as required.

combinationlogarithmic-seriesalgebra
C

Group C

Long Answer Questions carrying 8 marks each. There may be sub-questions within a question.

1 questions·8 marks each
4long8 marks

a) A particle moves along the curve 6y=x3+26y = x^3 + 2. Find the point on the curve at which the yy-coordinate is changing 8 times as fast as the xx-coordinate. [3]

b) Integrate ∫dx1+e2x\displaystyle\int \frac{dx}{1 + e^{2x}} by reducing it to a standard integral form. [2]

c) Write an example of a homogeneous differential equation. Solve it. [3]

a) Differentiating 6y=x3+26y = x^3 + 2 with respect to time tt:

6dydt=3x2dxdt.6\frac{dy}{dt} = 3x^2 \frac{dx}{dt}.

We require dydt=8dxdt\dfrac{dy}{dt} = 8\dfrac{dx}{dt}. Substituting,

6(8)dxdt=3x2dxdt  ⇒  48=3x2  ⇒  x2=16  ⇒  x=±4.6(8)\frac{dx}{dt} = 3x^2 \frac{dx}{dt} \;\Rightarrow\; 48 = 3x^2 \;\Rightarrow\; x^2 = 16 \;\Rightarrow\; x = \pm 4.

For x=4x = 4: 6y=64+2=66⇒y=116y = 64 + 2 = 66 \Rightarrow y = 11, point (4,11)(4, 11). For x=−4x = -4: 6y=−64+2=−62⇒y=−3136y = -64 + 2 = -62 \Rightarrow y = -\frac{31}{3}, point (−4,−313)\left(-4, -\frac{31}{3}\right).

So the required points are (4,11)(4, 11) and (−4,−313)\left(-4, -\dfrac{31}{3}\right).

b) ∫dx1+e2x\displaystyle\int \frac{dx}{1 + e^{2x}}. Put ex=ye^{x} = y, so ex dx=dye^{x}\,dx = dy, i.e. dx=dyydx = \dfrac{dy}{y}. Then

∫dx1+e2x=∫11+y2⋅dyy=∫dyy(1+y2).\int \frac{dx}{1 + e^{2x}} = \int \frac{1}{1 + y^2}\cdot\frac{dy}{y} = \int \frac{dy}{y(1 + y^2)}.

Using partial fractions 1y(1+y2)=1y−y1+y2\dfrac{1}{y(1+y^2)} = \dfrac{1}{y} - \dfrac{y}{1+y^2}:

∫dyy(1+y2)=ln⁡∣y∣−12ln⁡(1+y2)+c=ln⁡ex−12ln⁡(1+e2x)+c=x−12ln⁡(1+e2x)+c.\int \frac{dy}{y(1+y^2)} = \ln|y| - \frac{1}{2}\ln(1 + y^2) + c = \ln e^{x} - \frac{1}{2}\ln(1 + e^{2x}) + c = x - \frac{1}{2}\ln(1 + e^{2x}) + c.

c) Example of a homogeneous differential equation:

dydx=x+yx.\frac{dy}{dx} = \frac{x + y}{x}.

It is homogeneous (degree 0). Put y=vxy = vx, so dydx=v+xdvdx\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}. Then

v+xdvdx=x+vxx=1+v  ⇒  xdvdx=1  ⇒  dv=dxx.v + x\frac{dv}{dx} = \frac{x + vx}{x} = 1 + v \;\Rightarrow\; x\frac{dv}{dx} = 1 \;\Rightarrow\; dv = \frac{dx}{x}.

Integrating, v=ln⁡x+cv = \ln x + c, and since v=yxv = \dfrac{y}{x},

y=x(ln⁡x+c).y = x(\ln x + c).
calculusderivativesintegrationdifferential-equations

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