NEB – GRADE XI

Mathematics
New Model Questions – 2078 Solution
Set 1
Group ‘A’
Rewrite the correct option in your own answer sheet. (1 X 11) = 11
1. The range of sine function is
a) (−∞, ∞) b) [−∞, ∞] c) [-1, 1] d) (-1, 1)
 [-1, 1]
𝑎 1 1
2. If a, b and c are in A.P, then , and will be in:
𝑏𝑐 𝑐 𝑏
a) A.P. b) G.P. c) H.P. d) none of these
 A.P.
Hint: a, b and c are in A.P.
𝑎 𝑏 𝑐
⇒ , , are also in A.P.
𝑏𝑐 𝑏𝑐 𝑏𝑐
𝑎 1 1
⇒ , , are also in A.P.
𝑏𝑐 𝑐 𝑏
3 4
3. If a triangle ABC, if sinA = , sinB = , then a:b:c =
5 5
a) 1:2:3 b) 2:3:4 c) 3:4:5 d) 6:8:10
 3:4:5
Hint: Given triangle is right angle triangle with sides a = 3, b = 4 and c = 5 so a:b:c = 3:4:5
Also, you can use sine law to find the ratio.
4. The length of the perpendicular drawn from the point (2, 1) to the line x – 2y – 5 = 0 is:
a) 5 units b) √5 units c) 6 units d) √6 units
 √5 units
2−2(1)−5 5
Hint: | |= = √5 units
√12 +(−2)2 √5

5. A conic section in which the value of eccentricity is unity is known as

a) circle b) parabola c) hyperbola d) ellipse
 parabola
6. If 𝑎⃗ and 𝑏⃗⃗ are non-zero and non collinear vectors and x𝑎⃗ + 𝑦𝑏⃗⃗ = 0, then
a) x =0, y≠0 b) x≠ 0, y =0 c) x≠ 0, y≠0 d) x = 0, y =0
 x = 0, y =0
7. In a leap year, the probability of having 53 Sundays is
1 2 3 4
a) b) c) 7 d)
7 7 7
2

7
Hint: 1 leap year = 52 weeks + 2 days
In remaining 2 days the event of happening Sunday arise 2 times and total sample space is 7.
2
∴ P (53 Sundays) =
7

Prepared by: Prabin Poudel

 8.The curve of a function y =f(x) is concave upward if
a) 𝑦 ′′ < 0 b) 𝑦 ′′ > 0 c) 𝑦 ′′ = 0 d) none of these
 𝑦 >0′′

9. ∫ 𝑙𝑜𝑔𝑥𝑑𝑥 = ?
a) log x +c b) x log x +c c) x log x +x +c d) x log x – x + c
 x log x – x + c
10. The rate of convergence in bisection method is
a) constant b) linear c) quadratic d) cubic
 linear
11. The magnitude of resultant of two forces of equal magnitude is equal to the magnitude of either force.
The angle between two forces is:
a) 30o b) 60o c) 120o d) 150o
 120o
1
Hint: F = √𝐹 2 + 𝐹 2 + 2𝐹. 𝐹𝑐𝑜𝑠𝜃 ⇒ 𝐹 2 = 2𝐹 2 + 2𝐹 2 𝑐𝑜𝑠𝜃 ⇒ cos𝜃 = − 2 ⇒ 𝜃 = 120o
OR
When % change in quantity demanded is equal to the % change in price, then elasticity of demand is
a) >1 b) <1 c) 1 d) 0
 1

Group ‘B’ (5 × 8 =40)

12. (a) Define tautology. Show that 𝑝 ∧ 𝑞 → 𝑝 ∨ 𝑞 is a tautology. ( 1 + 2)
 Solution:
First Part: A compound statement is said to be tautology if it is true for any set of given compound
statement.

p q 𝑝∧𝑞 𝑝∨𝑞 𝑝∧𝑞 →𝑝∨𝑞

T T T T T
T F F T T
F T F T T
F F F F T

Compound statement 𝑝 ∧ 𝑞 → 𝑝 ∨ 𝑞 has truth value in all cases. Hence it is a tautology.

12. (b) Express the complex number 1 - √3𝑖 into polar form. (2)
 Solution:
Let 1 - √3𝑖 = x + iy = r (cos𝜃 + isin𝜃) so we have x = 1 and y = -√3 .
Since x is positive and y is negative so complex number lies in fourth quadrant.
2
Then, r = √𝑥 2 + 𝑦 2 = √(1)2 + (− √3 ) = 2
𝑦 √3
tan𝜃 = =− = -√3 = tan (360 – 60) = tan300o ∴ 𝜃 = 300o
𝑥 1
∴1 - √3𝑖 = r (cos𝜃 + isin𝜃) = 2 (cos300o + isin300o)
Prepared by: Prabin Poudel
 1 1 1
13. If (b – c)2, (c – a)2, (a – b) 2 are in A.P. Then prove that , , are in A.P. (5)
𝑏−𝑐 𝑐−𝑎 𝑎−𝑏
 Solution:
Here, (b – c)2, (c – a)2, (a – b) 2 are in A.P
⇒ 𝑏 2 − 2𝑏𝑐 + 𝑐 2 , 𝑐 2 − 2𝑎𝑐 + 𝑎2 , 𝑎2 − 2𝑎𝑏 + 𝑏 2 are in A.P.
Subtracting (𝑎2 + 𝑏 2 + 𝑐 2 ) from each term.
⇒ 𝑏 2 − 2𝑏𝑐 + 𝑐 2 − (𝑎2 + 𝑏 2 + 𝑐 2 ), 𝑐 2 − 2𝑎𝑐 + 𝑎2 − (𝑎2 + 𝑏 2 + 𝑐 2 ), 𝑎2 − 2𝑎𝑏 + 𝑏 2 − (𝑎2 + 𝑏 2 + 𝑐 2 )are in A.P.
⇒ −𝑎2 − 2𝑏𝑐, −𝑏2 − 2𝑎𝑐, −𝑐2 − 2𝑎𝑏 are in A.P
Multiply each term by – sign, the terms will again be in A.P.
⇒ 𝑎2 + 2𝑏𝑐, 𝑏 2 + 2𝑎𝑐, 𝑐 2 + 2𝑎𝑏 are in A.P
Subtracting (𝑎𝑏 + 𝑏𝑐 + 𝑎𝑐) from all terms. Again, the terms will be in A.P.
⇒ 𝑎2 + 2𝑏𝑐 − (𝑎𝑏 + 𝑏𝑐 + 𝑎𝑐), 𝑏 2 + 2𝑎𝑐 − (𝑎𝑏 + 𝑏𝑐 + 𝑎𝑐), 𝑐 2 + 2𝑎𝑏 − (𝑎𝑏 + 𝑏𝑐 + 𝑎𝑐) are in A.P
⇒ 𝑎2 + 𝑏𝑐 − 𝑎𝑏 − 𝑎𝑐, 𝑏 2 + 𝑎𝑐 − 𝑎𝑏 − 𝑏𝑐, 𝑐 2 + 𝑎𝑏 − 𝑏𝑐 − 𝑎𝑐 are in A.P
⇒ (𝑎 − 𝑏)(𝑎 − 𝑐), (𝑏 − 𝑐)(𝑏 − 𝑎), (𝑏 − 𝑐)(𝑎 − 𝑐) are in A.P
Dividing by (a-b) (b-c) (a-c) from all terms
1 𝑏−𝑎 1
⇒ , , are in A.P.
𝑏−𝑐 (𝑎−𝑏)(𝑎−𝑐) 𝑎−𝑏
1 1 1
⇒ , , are in A.P.
𝑏−𝑐 𝑐−𝑎 𝑎−𝑏
Another method
Here, (b – c)2, (c – a)2, (a – b) 2 are in A.P. So, we can write
⇒ (𝑐 − 𝑎)2 − (𝑏 − 𝑐)2 = (𝑎 − 𝑏)2 − (𝑐 − 𝑎)2
⇒ (c – a +b – c) (c – a – b + c) = (a – b + c – a) (a – b – c + a)
⇒ (b – a) (2c – a – b) = (c – b) (2a – b – c) ................................(*)
Now,
1 1 1
To prove: 𝑏−𝑐 , , are in A.P.
𝑐−𝑎 𝑎−𝑏
1 1 1 1
∴ - = -
𝑐−𝑎 𝑏−𝑐 𝑎−𝑏 𝑐−𝑎
𝑏−𝑐−𝑐+𝑎 𝑐−𝑎−𝑎+𝑏
⇒ =
(𝑐−𝑎)(𝑏−𝑐) (𝑐−𝑎)(𝑎−𝑏)
−2𝑐+𝑏+𝑎 −2𝑎+𝑏+𝑐
⇒ =
(𝑏−𝑐) (𝑎−𝑏)
⇒ -(a – b) (2c - b - a) = - (b – c) (2a – b – c)
⇒ (b – a) (2c – b – a) = (c – b) (2a – b – c) ...............................(**)
Equation (*) and (**) are same.
1 1 1
∴ , , are in A.P.
𝑏−𝑐 𝑐−𝑎 𝑎−𝑏

1 1 3
14. (a) In any ∆ ABC, if + = , show that C = 60o. What will be the relation of a, b and c,
𝑎+𝑐 𝑏+𝑐 𝑎+𝑏+𝑐
if C = 90o? (3+1)
 Solution:
First Part:
𝑏+𝑐+𝑎+𝑐 3
⇒ =
(𝑎+𝑐)(𝑏+𝑐) 𝑎+𝑏+𝑐

⇒ (b + a + c + c) (a +b +c) = 3 (a + c) (b + c)
⇒ (b + a + c) (a + b + c) + c (a + b + c) = 3(ab + ac + bc + c2)
Prepared by: Prabin Poudel
 ⇒ a2 + b2 + c2 + 2ab + 2bc + 2ac + ac + bc + c2 = 3ab + 3ac +3c2 + 3bc
⇒ a2 + b2 + 2c2 + 2ab + 3bc + 3ac = 3ab + 3ac +3c2 + 3bc
⇒ a2 + b2 + 2c2 + 2ab = 3ab +3c2
⇒ a2 + b2 - c2 = ab
𝑎2 +𝑏2 −𝑐 2 1
⇒ =
2𝑎𝑏 2
o
⇒ cos C= cos60
∴ C = 60 o
Second Part:
If C = 90 o the relation between a, b and c will be c = √𝑎2 + 𝑏 2.

14. (b) Write the condition that the general equation of second degree may represent a pair of straight line (1)
𝑎 ℎ 𝑔
 Solution: The required condition is 𝑎𝑏𝑐 + 2𝑓𝑔ℎ − 𝑎𝑓 − 𝑏𝑔 − 𝑐ℎ = 0 or |ℎ 𝑏 𝑓 |=0.
2 2 2

𝑔 𝑓 𝑐

15. Following are the marks obtained by student A and B in 6 tests of 100 marks each. (5)

Test 1 2 3 4 5 6
A 56 72 48 69 64 81
B 63 74 45 57 82 63

i) If the consistency of the performance is the criteria for awarding prize, who should get the prize?
 Solution:
For test A:
n=6
∑X = 56 + 72 + 48 + 69 + 64 + 81 = 390
∑𝑋 390
𝑋= = = 65
𝑛 6
∑X 2 = 562 + 722 + 482 + 692 + 642 + 812 = 26042
∑𝑋 2 ∑𝑋 2 26042 390 2
𝜎𝐴 = √ −( ) =√ −( ) = 10.739
𝑛 𝑛 6 6
𝜎𝐴 10.739
𝐶. 𝑉.𝐴 = × 100% = × 100% = 16.522 %
𝑋 65
For test B:
n=6
∑Y = 63 + 74 + 45 + 57 + 82 + 63 = 384
∑𝑌 384
𝑌= = = 64
𝑛 6
∑Y 2 = 632 + 742 + 452 + 572 + 822 + 632 = 25412
∑𝑌 ∑𝑌 2 25412 384 2 2
𝜎𝐵 = √ −( ) =√ − ( ) = 11.804
𝑛 𝑛 6 6
Prepared by: Prabin Poudel
 𝜎𝐵 11.804
𝐶. 𝑉.𝐵 = × 100% = × 100% = 18.444%
𝑌 64
Since, 𝐶. 𝑉.𝐴 < 𝐶. 𝑉.𝐵 , A is more consistent than B. Therefore, A should get the prize.
Note: A distribution having less C.V. is said to be less variable or more homogeneous or more consistent or
more uniform or more stable or more stationary or more equitable than others.
If 𝑋𝐴 > 𝑋𝐵 , A is better than B.
If 𝑀𝑑𝐴 > 𝑀𝑑𝐵 , A is more intelligent.
𝑥−√2−𝑥 2
16. Define indeterminate form. Give examples. Evaluate: lim (1+4)
𝑥→1 2𝑥− √2+2𝑥 2
 Solution:
First Part:
An expression which has no quantitative or finite value are called indeterminate form.
0 ∞
For example: , , ∞ , -∞ , 0o, ∞𝑜 , 1∞ , ∞ − ∞, ∞ + ∞ , 0 x ∞ etc.
0 ∞
Second Part:
0
The given function takes the form , which is indeterminate.
0
𝑥−√2−𝑥 2 𝑥−√2−𝑥 2 𝑥+√2−𝑥 2 2𝑥+ √2+2𝑥 2
∴ lim = lim × ×
𝑥→1 2𝑥− √2+2𝑥 2 𝑥→1 2𝑥− √2+2𝑥 2 𝑥+√2−𝑥 2 2𝑥+ √2+2𝑥 2
𝑥 2 −(2−𝑥 2 ) 2𝑥+√2+2𝑥 2
= lim (2𝑥)2 ×
𝑥→1 −(2+2𝑥 2 ) 𝑥+√2−𝑥 2
2𝑥 2 −2 2𝑥+√2+2𝑥 2
= lim ×
𝑥→1 2𝑥 2 −2 𝑥+√2−𝑥 2
2𝑥+√2+2𝑥 2
= lim
𝑥→1 𝑥+√2−𝑥 2
2(1)+√2+2(1)2
=
1+√2−(1)2
2+2
=
1+1
4
=
2
=2
𝑥
17. Evaluate: ∫ dx. (5)
√9+𝑥 2
 Solution:
𝑥
Let I = ∫ dx.
√32 +𝑥 2
𝑥
Let x = 3tan𝜃 ⇒ 𝑡𝑎𝑛𝜃 =
3
𝑑𝑥
p=x

= 3 sec 𝜃 ⇒ 𝑑𝑥 = 3sec2 𝜃 𝑑𝜃
2
𝑑𝜃
3𝑡𝑎𝑛𝜃
∴I=∫ × 3 sec 2 𝜃 𝑑𝜃
√9+9 tan2 𝜃
9𝑡𝑎𝑛𝜃 sec2 𝜃 𝜃
=∫ d𝜃
3√1+tan2 𝜃
9𝑡𝑎𝑛𝜃 sec2 𝜃 b= 3
=∫ d𝜃
3𝑠𝑒𝑐𝜃
= ∫ 3𝑡𝑎𝑛𝜃𝑠𝑒𝑐𝜃 𝑑𝜃
Prepared by: Prabin Poudel
 = 3𝑠𝑒𝑐𝜃 + c
ℎ
=3 +c
𝑏
3√𝑥 2 +9
= +c
3
= √𝑥 2 + 9 + c
18. Show that the equation 𝑥 3 − 𝑥 − 4 = 0 has one positive root and two negative roots and find the positive
root correct to 3 place of decimals using bisection method. (1 + 4)
 Solution:
First part:
Using descartes’ rule of signs for positive root
f(x) = 𝑥 3 − 𝑥 − 4 =+ 𝑥 3 − 𝑥 − 4

Here, the sign changes one times, so there is one positive root.
Using descartes’ rule of signs for negative root
f(-x) = −𝑥 3 + 𝑥 − 4

Here, sign changes two times, so there are two negative roots.
Second part:
Since no interval is given, we have to find it by ourselves.
f (1) = (1)3 − (1) − 4 = −4
f (2) = (2)3 − (2) − 4 = 2
Since, f (1) and f (2) have opposite sign. So, one root lies between 1 and 2.
Mid-point of 1 and 2 is 1.5
f (1.5) = (1.5)3 − (1.5) − 4 = −2.125 < 0
Since, f (1.5) and f (2) have opposite sign. So, root lies between 1.5 and 2.
Mid-point of 1.5 and 2 is 1.75
f (1.75) = (1.75)3 − (1.75) − 4 = −0.391 < 0
Since, f (1.75) and f (2) have opposite sign. So, root lies between 1.75 and 2.
Mid-point of 1.75 and 2 is 1.875
f (1.875) = (1.875)3 − (1.875) − 4 = 0.717 > 0
Since, f (1.75) and f (1.875) have opposite sign. So, root lies between 1.75 and 1.875
Mid-point of 1.75 and 1.875 is 1.8125
f (1.8125) = (1.8125)3 − (1.8125) − 4 = 0.142 > 0
Since, f (1.75) and f (1.8125) have opposite sign. So, root lies between 1.75 and 1.8125.
Mid-point of 1.75 and 1.8125 is 1.78125.
f (1.78125) =(1.78125)3 − (1.78125) − 4 = −0.1296 < 0
Since, f (1.78125) and f (1.8125) have opposite sign. So, root lies between 1.78125 and 1.8125.
Mid-point of 1.78125 and 1.8125 is 1.796875.
f (1.796875) = (1.796875)3 − (1.796875) − 4 = 0.0048 >0
Since, f (1.78125) and f (1.796875) have opposite sign. So, root lies between 1.78125 and 1.796875.
Mid-point of 1.78125 and 1.796875 is 1.789063
f (1.789063) = (1.789063)3 − (1.789063) − 4 = -0.062726
Here, f (1.796875) is close to 0 than f (1.789063).
∴ The positive root correct to 3 place of decimal is 1.796.

Prepared by: Prabin Poudel

 19. i) State and prove triangle law of forces. (3)
ii) O is the circumcenter of the triangle ABC and forces P, Q and R are acting along OA, OB and OC
𝑃 𝑄 𝑅
are in equilibrium, show that = = (2)
𝑠𝑖𝑛2𝐴 𝑠𝑖𝑛2𝐵 𝑠𝑖𝑛2𝐶
 Solution:
i) Statement:
Triangle law of forces states that “if three forces acting on a point can be represented in magnitude
and direction by the sides of a triangle, taken in order, they are in equilibrium.”
Proof:

D
A
⃗⃗
𝑄
𝑃⃗⃗ ⃗⃗
𝑄 𝑅⃗⃗
⃗⃗
𝑄
𝑅⃗⃗ O
B C
𝑃⃗⃗
Let sides BC, CA and AB of a triangle ABC represents in magnitude and direction the three-given force
P, Q and R respectively acting at O. We have to prove that forces are in equilibrium i.e.,
𝑃⃗⃗ + 𝑄 ⃗⃗ + 𝑅⃗⃗ = 0
Complete the parallelogram BCAD as in the figure. Then
⃗⃗⃗⃗⃗⃗⃗ = 𝐶𝐴
𝐵𝐷 ⃗⃗⃗⃗⃗⃗ = 𝑄
⃗⃗
From parallelogram law of forces
𝑃⃗⃗ + 𝑄 ⃗⃗ + 𝑅⃗⃗ = 𝐵𝐶⃗⃗⃗⃗⃗⃗ + 𝐶𝐴⃗⃗⃗⃗⃗⃗ + 𝐴𝐵⃗⃗⃗⃗⃗⃗
= ⃗⃗⃗⃗⃗⃗
𝐵𝐴 + ⃗⃗⃗⃗⃗⃗ 𝐴𝐵
= −𝐴𝐵 ⃗⃗⃗⃗⃗⃗ + 𝐴𝐵⃗⃗⃗⃗⃗⃗
=0
A
Hence, the forces are in equilibrium.
ii) O is the circumcenter of the triangle.
Here Force P is along OA, Q is along OB, and R along OC.
From figure,
< BOC = 2A, < BOA = 2C and < AOC = 2B
( Angle at the center is double of the angle at circumference
standing on the same arc)
Since all forces are in equilibrium, by Lami’s theorem.
𝑃 𝑄 𝑅
= = B
𝑠𝑖𝑛𝐵𝑂𝐶 𝑠𝑖𝑛𝐴𝑂𝐶 𝑠𝑖𝑛𝐵𝑂𝐴 C
𝑃 𝑄 𝑅
⇒ = =
𝑠𝑖𝑛2𝐴 𝑠𝑖𝑛2𝐵 𝑠𝑖𝑛2𝐶

OR,
19. A firm has a demand function P = 200 – 3Q and the cost function C = 2Q2 – 3Q. Find the marginal
cost, marginal revenue and the marginal profit when the output is 12. (5)

Prepared by: Prabin Poudel

 Solution:
Given, C = 2Q2 – 3Q, P = 200 – 3Q
Let R be the revenue function. Then
R = P x Q = (200 – 3Q) x Q =200Q – 3Q2
Profit function (𝜋) = R – C = 200Q – 3Q2 – (2Q2 – 3Q) = 203Q – 5Q2
Now,
𝑑𝐶
Marginal cost = = 4Q – 3
𝑑𝑄
𝑑𝐶
When Q = 12, = 4(12) – 3 = 45
𝑑𝑄
𝑑𝑅
Marginal Revenue = = 200 – 6Q
𝑑𝑄
𝑑𝑅
When Q = 12, = 200 – 6(12) = 200 – 72 = 128
𝑑𝑄
𝑑𝜋
Marginal profit = = 203 – 10Q
𝑑𝑄
𝑑𝜋
When Q = 12, = 203 – 10(12) = 203 – 120 = 83
𝑑𝑄

Group ‘C’ (8 x 3 = 24)

20. a) Let R is a relation on A = {1, 3, 5, 7} is defined by x = 3y. List all the elements of R. (2)
 Solution:
Given, A = {1, 3, 5, 7}
∴ A × A = {1, 3, 5, 7} × {1, 3, 5, 7}
= {(1, 1), (1, 3), (1, 5), (1, 7), (3, 1), (3, 3), (3, 5), (3, 7), (5, 1), (5, 3), (5, 5), (5, 7)
(7, 1), (7, 3), (7, 5), (7, 7)}
∴R = {(x, y): x = 3y} = {(3, 1)}.

20. b) Find domain and range of f (x) = x2 + 1. (2)

 Solution:
For domain:
Let y = x2 + 1
For every x 𝜖 R, y is defined. So, D(f) = R = (−∞, ∞).
For Range:
x2 = y – 1
Since, x2 ≥ 0. Therefore y – 1 ≥ 0 ⇒ y ≥ 1
So, R(f) = [1, ∞)
𝑥
20. c) Test the periodicity of the function f (x) = tan . (2)
4
 Solution:
1
Comparing with f(x) = a tan bx we get a = 1 and b = .
4
𝑥 𝜋 𝜋
∴ Periodicity of tan = = 1 = 4𝜋.
4 |𝑏|
4

Prepared by: Prabin Poudel

 2 4 3
20. d) Define symmetric matrix. If A = (2 3 4), prove that A+AT is symmetric. (2)
5 2 6
 Solution:
First Part:
A square matrix A is called symmetric if it equal to its transpose i.e., A =AT.
Second Part:
2 4 3
Given, A = (2 3 4)
5 2 6
2 2 5
T
∴ A = (4 3 2 )
3 4 6
Now,
2 4 3 2 2 5 4 6 8
A+AT = (2 3 4) + (4 3 2) = (6 6 6)
5 2 6 3 4 6 8 6 12
Again,
4 6 8 𝑇 4 6 8
(6 6 6 = 6 6 6)
) (
8 6 12 8 6 12
∴ A+AT is a symmetric matrix.
21. a) Find the value of k if the length of the perpendicular from point (2, -3) on the line 2x+ky +3=0
is 3. (2)
 Solution:
We have,
𝑎𝑥1 +𝑏𝑦1 +𝑐
p=| |
√𝑎2 +𝑏2
Given, a = 2, b = k, c = 3, 𝑥1 = 2 and 𝑦1 = -3
2(2)+𝑘(−3)+3
∴3= ±
√22 +𝑘 2
Squaring both sides, we get
⇒9 (4 + k2) = (7 – 3k)2
⇒ 36 + 9k2 = 49 – 42k + 9k2
⇒ 42k = 13
13
⇒k=
42
13
∴ The required value of k is .
42
21. b) Show that the points (1, 2) and (3, 2) are on the same side of the line x + 3y + 1 =0 (2)
 Solution:
Let f (x, y) = x + 3y + 1
∴ f (1, 2) = 1 + 3(2) + 1 = 8 > 0
∴ f (3, 2) = 3 + 3(2) + 1 = 10 > 0
This shows that (1, 2) and (3, 2) are on same side of the line x + 3y +1 = 0
Prepared by: Prabin Poudel
 21. c) Define parabola. Derive the equation of parabola in standard form. (1 + 3)
 Solution:
First Part:
The conic section whose eccentricity is 1(e = 1) is
called parabola.
Second part: M P(x,y)
Let us consider a parabola with vertex at origin and )
focus at F (a, 0), where a > 0. The axis of parabola is
X-axis and the directrix of parabola is parallel to D(-a,0) F(a,0)
Y-axis. Let P (x, y) be any point on the parabola.
Draw PM perpendicular to directrix and join PF.
The coordinate of M will be (-a, y).
By definition of parabola, we have
PF = PM
⇒ PF2 = PM 2
⇒ (x – a)2 + (y – 0)2 = (x + a)2 + (y – y)2
⇒ y2 = (x + a)2 - (x – a)2
⇒ y2 = (x + a + x – a) (x + a – x + a)
⇒ y2 = 2x (2a)
⇒ y2 = 4ax
22. a) Find the absolute maximum and minimum values of the function on the given interval. (4)
f (x) = x3 – 6x2 + 9x on [0,5]
 Solution:
Given f(x) = x3 – 6x2 + 9x
𝑓′(x) = 3x2 – 12x + 9
For stationary points
𝑓′(x) = 0
⇒ 3x2 – 12x + 9 = 0
⇒ x2 – 4x + 3 = 0
⇒ x2 – 3x -x + 3 = 0
⇒ x (x – 1) – 1(x – 3) =0
⇒ (x – 3) (x – 1) = 0
Either x = 3 or x = 1
When x = 1, f (1) = (1)3 – 6(1)2 + 9(1) = 4
When x = 3, f (3) = (3)3 – 6(3)2 + 9(3) = 0
When x = 0, f (0) = (0)3 – 6(0)2 + 9(0) = 0
When x = 5, f (5) = (5)3 – 6(5)2 + 9(5) = 20
Hence, the absolute maximum value of f(x) is 20 at x = 5 and minimum value is 0 at x = 0 and 3.
𝑥2 𝑦2
22. b) Find the area of the ellipse + = 1 using method of integration. (4)
36 25
 Solution:
𝑥2 𝑦2 25(36−𝑥 2 ) 5√36−𝑥 2
Given equation of ellipse is + = 1 i.e., 𝑦 2 = ⇒ 𝑦=± .... (i)
36 25 36 6
This ellipse is symmetric about both the axes.
To find the area of whole ellipse, first we have to find the area of portion lying on first quadrant.
Prepared by: Prabin Poudel
 When x = 0, y = ± 5 and when y = 0, x = ± 6
The area of ellipse = 4 × Area of ellipse lying on first quadrant.
To find the area of ellipse lying on first quadrant we have to integrate equation (i) from x= 0 to x =6.
6 5√36−𝑥2
∴ The area of ellipse = 4 × |∫0 ± 6
𝑑𝑥|
6 5√36−𝑥2
= 4 × ∫0 dx
6
20 6
= × ∫0 √36 − 𝑥 2 dx
6
Let x = 6 sin𝜃 .......(ii). Then, dx = 6 cos𝜃 d𝜃.
𝜋
When x = 0, then 𝜃 = 0 and when x = 6, then 𝜃 = .
2
𝜋
10
= × ∫2 √36
0
− 36 sin2 𝜃 × 6 cos𝜃 d𝜃.
3
𝜋
60
= ∫02 √36(1 − sin2 𝜃) × cos𝜃 d𝜃
3
𝜋
= 20 ∫02 6 𝑐𝑜𝑠𝜃 × cos𝜃 d𝜃
𝜋
= 60 ∫02 2 cos2 𝜃 d𝜃
𝜋
= 60 ∫02(1 + 𝑐𝑜𝑠2𝜃) d𝜃
𝜋
𝑠𝑖𝑛2𝜃 2
= 60 [𝜃 + ]0
2
𝜋 𝑠𝑖𝑛𝜋 𝑠𝑖𝑛0
= 60 [ 2 + 2 – (0 + )]
2
𝜋
= 60 2
= 30𝜋 sq. units

Prepared by: Prabin Poudel