NEB Class 12 Math Model Question 2080 With Solution

$$r=−1.7132

Now, Regression Coefficient of x on
y is

$$bxy=n∑xy−∑x∑y(∑y)2−n∑y2

$$∴bxy=0.12

16) a) Define Hospital’s rule.

b) Write the slope of the tangent and normal to the curve
y =f(x) at (x₁,y₁).

c) Write the integral of $\int \frac{1}{x^2+a^2}dx$.

d)What is the integral of $\int \sinh x.dx$

Solution:

(a) If f(x) and g(s) are two function then their
derivatives f’(x) and g’(x) are continuous at x=0 and if f(a) =g(a), then,

$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=\frac{f^{\prime }(a)}{g^{\prime }(a)}$ When g’(a)≠0.

(b) The slope of tangant to the curve y=f(x) at (x₁,y₁)
is

$$dydx=f′(x)

$$⇒dydx(x1,y1)=f′(x1)

Then, Slope of normal to y=f(x) at ((x₁,y₁)) is

F’(x). m = -1

Thus, m=$-\frac{dx}{dy}$

(c) $\int \frac{1}{x^2+a^2}dx$=$\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C$

(d) $\int \sinh x.dx$ = coshx +C

17) a) Solve: $\frac{dy}{dx}=\frac{y}{x}$

b) Verify Rolle’s Theorem for f(x) = x²+3x-4
in [-4,1].

 Solution:

$$dydx=yx

$$dyy=dxx

Integrating both sides:

ln(y) = ln(x) +C

or, ln(y) = ln(x) + ln(C)

Thus, y=xc

(b) f(x) = x²+3x-4 in [-4,1]

(i) Here f(x) is a polynomial function so it is continuous at
[-4,1].

(ii) f’(x) = 2x+3

f(x) is differentiable in (-4, 1).

(iii) f(-4) = 0, f(1) = 0

Thus, f(a) = f(b).

All the condition of Rolle’s Theorem are verified. So there
exists at least a point C ε (-4, 1).

Such that: f’(C) =0.

18) Using simplex method, maximize P
(x, y) = 15x+10y subject to 2x+y≤10, x+3y≤10, x, y≥0.

 Solution:

Maximize P (x, y) = 15x+10y

Subject to:

2x+y≤10

x+3y≤10

x, y≥0

Let S₁ and S₂ be two non-negative
slack variables. Then, the standard form of the above LPP is:

2x+y+r =10

x+3y+s=10

-15x-10y+P=0

The initial simplex tableau for above LLP is

In the initial Simplex tableau the most negative entry is
-15 so the x-column is pivot column. Also, $\frac{10}{2}$=5 is the smallest positive
value. Thus, r-row with element 2 on x column is the pivot element.

Dividing First Row by 2

R₂à
R₂-R₁ and R₃à R₃+15R₁

Here last row still has -ve entry. So y is pivot column and
s-row is the pivot row with $\frac{5}{2}$ as pvot element.

R₂à
$\frac{2}{5}$R₂

Again, R₃ à
R₃+$\frac{5}{2}$R₂ and R₁àR₁– ½ R₂

Since there is no negative entry in last row. Thus Maximum
Solution is obtained.

P_max=80 at x=4 and y=2.

19) A particle is projected with a velocity ‘v’ and
greatest height is ‘H’, prove the horizontal range R is: R = 4$\sqrt{H\left(\frac{v^2}{2g}-H\right)}$.

Solution:

Let β be angle of projection. Then,

$$H=v2sin2β2g

$$R=v2sin⁡2βg

Now,

$$v22g−H

=$\frac{v^2}{2g}-\frac{v^2{\sin }^2\beta }{2g}$

=$\frac{v^2}{2g}(1-{\sin }^2\beta )$

=$\frac{v^2}{2g}{\cos }^2\beta$

And,

R.H.S of Question:

4$\sqrt{H\left(\frac{v^2}{2g}-H\right)}$ = $4.\frac{v\sin \beta .\cos \beta }{2g}$

$$4v2sin2β2g.v22gcos2β

$$4.v2sin⁡β.cos⁡β2g

$\frac{}{}$v2.2sin⁡β.cos⁡βg

$\frac{}{}$v2sin⁡2βg

=R

=L.H.S

OR

The cost function C(x) in thousands of rupees for
producing x units of math’s textbooks is given by C(x)=30+20x-0.5x²
, 0≤x≤15.

a) Find the marginal cost of Function.

b) Find the marginal cost for producing 12000 math’s
textbooks.

 Solution:

C(x)=30+20x-0.5x², x, y≥0

(a) Marginal Cost of Function = C‘(x)=0+20-x

(b) Marginal cost = 20-x=20-12=Rs. 8 (in thousands)

Group “C”

20) a) Using matrix methods, solve the following system of
linear equations: x+y+z = 4, 2x+y-3z = -9,2x-y+z = -1

b) Apply De-Moiver’s theorem to find the value of [2(𝑐𝑜𝑠15°
+ 𝑖𝑠𝑖𝑛15°)] ⁶

c) Prove that: $\left(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\right)\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...\right)=1$

 Solution:

(a)

(b) De Moiver’s Theorem states that, {r(cosθ+isinθ)}ⁿ=rⁿ(cosnθ+isinnθ).

∴[2(𝑐𝑜𝑠15° + 𝑖𝑠𝑖𝑛15°)]
⁶ = 2⁶ [cos90 + i sin90] = 64i

(c) We Know  e^x= 
$1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

Put x = 1

e¹=$1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...$(i)

Put x = -1

e^-1=$1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...$(ii)

Taking LHS of Question:

$$(1+11!+12!+13!+...)(1−11!+12!−13!+...)

=e × e^–

=1

=R.H.S

21) a) Find the direction cosines of the line joining the
points (4,4,-10) and (-2,2,4).

b) Find the angle between the two diagonals of a cube.

c) Find the vertices of the conic section: 16(y-1)²
-9(x-5)² = 144.

 Solution:

(a) Given Points: P(4,4,-10) and Q(-2,2,4)

Direction ratios of PQ = -2-4, 2-4, 4-(-10)=-6, -2, 14

PQ=$\sqrt{{(-6)}^2+{(-2)}^2+{(14)}^2}=2\sqrt{59}$

Thus, Direction cosines are: $\frac{-6}{2\sqrt{59}},\frac{-2}{2\sqrt{59}},\frac{14}{2\sqrt{59}}$

(b) Let OABCDEFG be a cube with vertices as
below 

O(0,0,0),A(a,0,0),B(a,a,0),C(0,a,0),D(0,a,a),E(0,0,a),F(a,0,a),G(a,a,a)

There are four diagonals OG,CF,AD  and BE for
the cube.

Let us consider any two say OG and AD

We know that if A(x1​,y1​,z1​) and B(x2​,y2​,z2​) are
two points in space then

AB=(x₂​−x₁​)i+(y₂​−y₁​)j+(z₂​−z₁​)k

⇒OG=(a−0)i+(a−0)j+(a−0)k=ai+aj+ak

and AD=(0−a)i+(a−0)j+(a−0)k=−ai+aj+ak.

Therefore ∣$\overrightarrow{OG}$∣=$\sqrt{a^2+a^2+a^2}=a\sqrt{3}$

And,

$\overrightarrow{AD}$ = $\sqrt{{(-a)}^2+a^2+a^2}=a\sqrt{3}$

$$OG→.AD→=−a2+a2+a2=a2

We know that angle between two vectors $\overrightarrow{OG},\overrightarrow{AD}$ is given by θ = ${\cos }^{-1}\frac{\overrightarrow{OG}.\overrightarrow{AD}}{|\overrightarrow{OG}|.|\overrightarrow{AD}|}$

=${\cos }^{-1}\frac{a^2}{a\sqrt{3}.a\sqrt{3}}$

=${\cos }^{-1}\frac{1}{3}$

Thus, angle between two vectors $\overrightarrow{OG},\overrightarrow{AD}$ is ${\cos }^{-1}\frac{1}{3}$

(c) Given:

16(y-1)² -9(x-5)² = 144….(i)

Dividing Both Sides by 144.

$$(y−1)29−(x−5)216=1

Comparing with $\frac{{\left(y-h\right)}^2}{a^2}-\frac{{\left(x-k\right)}^2}{b^2}=1$

We Get:

h=1, k=5, a=3, b=4

Thus, Vertex = (h±a, k) = (1±3, 4) =(4,4) 0r (-2, 4)

22) a) If the limiting value of $\frac{f(x)-5}{x-3}$ at x= 3 is 2 by using L’ Hospital’ rule, find the appropriate value of
f(x).

b) Write any one homogeneous differential equation in
(x,y) and solve it.

c) The concept of anti-derivative is necessary for
solving a differential equation. Justify this statement with example.

 Solution:

(a)$\begin{array}{l}\frac{\underset{x\to 3}{lim}f(x)-5}{x-3}=2\\ or,\underset{x\to 3}{lim}f(x)-5=2x-6\\ or,\underset{x\to 3}{lim}f(x)=2x-1\end{array}$

(b)  Homogenous
differential Equation in (x,y) is $\frac{dy}{dx}=\frac{y}{x}+1$

Solving This Equation:

$$dydx=yx+1

Put y =vx, then $\frac{dy}{dx}=v+x\frac{dv}{dx}$

$\begin{array}{}\end{array}$yx+1=v+xdvdxor,v+1=v+xdvdxor,1=xdvdxor,dxx=dv

Integrate both sides

v = log x + log c

or, v = log cx

Replacing v

$$yx=log⁡cx

∴ y = x log cx 

(c) The concept of anti-derivatives is necessary for
solving differential equation. Let’s understand this statement through the following
example:

Consider a differential Equation $\frac{dy}{dx}=\frac{x}{y}$

i.e., y.dy = x.dx

Let’s integrate this on both sides:

$$∫y.dy=∫x.dx

$$y22=x22+C

Or, y²=x²+2C

i.e., y²=x²+C