एकात्मक फलनको परिभाषा लेख्नुहोस् ।
Write the definition of identity function.
SEE 2082 (2026) · RE-2021'A'
ऐच्छिक प्रथम गणित
Optional Mathematics — First Paper
RE-2021'A'
समूह 'क' (Group 'A')
Short Answer Questions — 1 mark each
1
Model Answer
A function f: A → A is called an identity function if every element maps to itself.
Definition: f(x) = x for all x ∈ A
The domain and codomain are the same set, and every element is its own image.
एकात्मक फलनमा प्रत्येक तत्वको प्रतिबिम्ब आफैँ हुन्छ। f(x) = x
2
यदि a र b दुई धनात्मक सङ्ख्याहरू भए तिनीहरूको समान्तरीय मध्यमा (AM) र गुणोत्तर मध्यमा (GM) बिचको सम्बन्ध लेख्नुहोस् ।
If a and b are two positive numbers, write the relation between their arithmetic mean (AM) and geometric mean (GM).
Model Answer
AM = (a + b)/2 GM = √(ab)
AM ≥ GM i.e. (a+b)/2 ≥ √(ab)
Equality holds when a = b.
समान्तरीय मध्यमा सधैँ गुणोत्तर मध्यमा भन्दा ठूलो वा बराबर हुन्छ।
3
फलन f(x) बिन्दु x = a मा निरन्तर हुने अवस्था लेख्नुहोस् ।
Write the condition of the continuity of the function f(x) at the point x = a.
Model Answer
f(x) is continuous at x = a if and only if all three conditions hold:
1. f(a) is defined
2. limx→a f(x) exists (LHL = RHL)
3. limx→a f(x) = f(a)
2. limx→a f(x) exists (LHL = RHL)
3. limx→a f(x) = f(a)
बायाँ सीमा = दायाँ सीमा = फलनको मान — यी तीनै समान भए मात्र निरन्तर।
4
मेट्रिक्स [1 0 / 0 1] को विपरीत मेट्रिक्स के हुन्छ, लेख्नुहोस् ।
What is the inverse matrix of matrix I = [[1,0],[0,1]]? Write it.
Model Answer
The given matrix is the Identity Matrix I.
det(I) = (1)(1) − (0)(0) = 1 ≠ 0 → Inverse exists
For Identity Matrix: I⁻¹ = I
I⁻¹ = [[1, 0], [0, 1]]
एकात्मक मेट्रिक्सको विपरीत मेट्रिक्स आफैँ हुन्छ।
5
समीकरण ax² + 2hxy + by² = 0 ले जनाउने जोडा रेखाहरू कुन अवस्थामा सम्पाती हुन्छन् ? लेख्नुहोस् ।
Under what condition the pair of straight lines represented by equation ax² + 2hxy + by² = 0 are coincident? Write it.
Model Answer
The pair of lines ax² + 2hxy + by² = 0 are coincident when the discriminant of the quadratic is zero:
h² − ab = 0 i.e. h² = ab
सम्पाती रेखाका लागि: h² = ab (विभेदक = 0)
6
कुन अवस्थामा एउटा सोलीलाई समतलीय सतहले प्रतिच्छेदन गर्दा वृत्त बन्छ, लेख्नुहोस् ।
In which condition a circle is formed when a plane surface intersects a cone? Write it.
Model Answer
A circle is formed when the cutting plane is:
Perpendicular to the axis of the cone (plane ⊥ axis)
i.e., the plane makes an angle of 90° with the axis of the cone.
जब समतल र सोलीको अक्ष लम्ब हुन्छ, वृत्त बन्छ।
7
sinA लाई tan(A/2) को रूपमा व्यक्त गर्नुहोस् ।
Express sinA in terms of tan(A/2).
Model Answer
Let t = tan(A/2)
Using half-angle identities:
sin A = 2sin(A/2)cos(A/2) = 2·[tan(A/2)·cos²(A/2)]
= 2t/(1 + t²)
sin A = 2tan(A/2) / [1 + tan²(A/2)]
8
Cos(A−B) − Cos(A+B) लाई गुणनफलको रूपमा व्यक्त गर्नुहोस् ।
Express Cos(A−B) − Cos(A+B) as a product.
Model Answer
Using sum-to-product formula:
cos(A−B) = cosAcosB + sinAsinB
cos(A+B) = cosAcosB − sinAsinB
cos(A−B) − cos(A+B) = (cosAcosB + sinAsinB) − (cosAcosB − sinAsinB)
cos(A−B) − cos(A+B) = 2 sinA sinB
9
भेक्टरहरू a⃗ र b⃗ बिचको कोण θ पत्ता लगाउने सूत्र लेख्नुहोस् ।
Write the formula to find the angle θ between the vectors a⃗ and b⃗.
Model Answer
cos θ = (a⃗ · b⃗) / (|a⃗| |b⃗|)
Or equivalently: θ = cos⁻¹ [(a⃗ · b⃗) / (|a⃗| |b⃗|)]
where a⃗ · b⃗ = a₁b₁ + a₂b₂ (dot product)
10
विपरीत स्थानान्तरणमा यदि बिन्दु P उत्क्रम वृत्तको केन्द्रबिन्दुमा पर्छ भने P को प्रतिबिम्ब P' कहाँ पर्दछ, लेख्नुहोस् ।
In an inversion transformation, if a point P lies at the centre of inversion circle then where does its image P' lie? Write it.
Model Answer
In inversion with centre O and radius r: OP · OP' = r²
If P = O (centre), then OP = 0, so OP' → ∞
P' lies at infinity (अनन्तमा)
उत्क्रम केन्द्रको प्रतिबिम्ब अनन्तमा हुन्छ।
समूह 'ख' (Group 'B')
Short Answer Questions — 2 marks each
11
यदि बहुपदीय p(y) = y³ − 2ay² + ay − 2 को एउटा गुणन खण्ड y − 2 भए a को मान पत्ता लगाउनुहोस् ।
If y − 2 is a factor of polynomial p(y) = y³ − 2ay² + ay − 2, find the value of a.
Model Answer
By Factor Theorem: if (y − 2) is a factor, then p(2) = 0
p(2) = (2)³ − 2a(2)² + a(2) − 2 = 0
8 − 8a + 2a − 2 = 0
6 − 6a = 0
6a = 6
8 − 8a + 2a − 2 = 0
6 − 6a = 0
6a = 6
∴ a = 1
12
असमानता 2x + y ≤ 2 लाई लेखाचित्रमा देखाउनुहोस् ।
Show the inequality 2x + y ≤ 2 in the graph paper.
Model Answer
Step 1: Find intercepts of boundary line 2x + y = 2
x-intercept: set y=0 → x = 1 → Point (1, 0)
y-intercept: set x=0 → y = 2 → Point (0, 2)
y-intercept: set x=0 → y = 2 → Point (0, 2)
Step 2: Draw line through (1,0) and (0,2). Use solid line (≤ includes boundary).
Step 3: Test origin (0,0): 2(0)+0 = 0 ≤ 2 ✓ → Shade the region containing origin.
Shaded region: below and including the line 2x + y = 2
13
यदि समीकरणहरू ax − 5y = 3 र 4x + 3y = 4 मा D = 26 भए क्रामरको नियम प्रयोग गरी 'a' को मान पत्ता लगाउनुहोस् ।
If the equations ax − 5y = 3 and 4x + 3y = 4 have D = 26, find the value of 'a' using Cramer's rule.
Model Answer
System: ax − 5y = 3 and 4x + 3y = 4
D = |a −5| = a(3) − (−5)(4) = 3a + 20
|4 3|
Given D = 26:
3a + 20 = 26
3a = 6
|4 3|
Given D = 26:
3a + 20 = 26
3a = 6
∴ a = 2
14
समीकरण 6x² + 5xy − 6y² − 3x + 2y = 0 ले प्रतिनिधित्व गर्ने जोडा रेखाहरूको छुट्टा छुट्टै समीकरण पत्ता लगाउनुहोस् ।
Find the separate equation of the pair of lines represented by the equation 6x² + 5xy − 6y² − 3x + 2y = 0.
Model Answer
Factorise: 6x² + 5xy − 6y² − 3x + 2y = 0
Treat as quadratic in x:
6x² + x(5y − 3) + (−6y² + 2y) = 0
6x² + x(5y − 3) − 2y(3y − 1) = 0
Factor: (2x + 3y − 1)(3x − 2y) = 0
Check: (2x)(3x) = 6x², (2x)(−2y) + (3y)(3x) = −4xy + 9xy = 5xy,
(3y)(−2y) = −6y², (−1)(3x) = −3x, (−1)(−2y) = 2y ✓
6x² + x(5y − 3) + (−6y² + 2y) = 0
6x² + x(5y − 3) − 2y(3y − 1) = 0
Factor: (2x + 3y − 1)(3x − 2y) = 0
Check: (2x)(3x) = 6x², (2x)(−2y) + (3y)(3x) = −4xy + 9xy = 5xy,
(3y)(−2y) = −6y², (−1)(3x) = −3x, (−1)(−2y) = 2y ✓
2x + 3y − 1 = 0 and 3x − 2y = 0
15
क्याल्कुलेटर र त्रिकोणमितीय तालिका प्रयोग नगरी प्रमाणित गर्नुहोस् : sin75°·cos15° = (2+√3)/4
Without using calculator and trigonometric table prove that: sin75°·cos15° = (2+√3)/4
Model Answer
Note: cos15° = sin75° (complementary angles)
So: sin75°·cos15° = sin75°·sin75° = sin²75°
sin75° = sin(45°+30°) = sin45°cos30° + cos45°sin30°
= (1/√2)(√3/2) + (1/√2)(1/2)
= (√3 + 1)/(2√2)
sin²75° = (√3+1)²/8 = (3 + 2√3 + 1)/8 = (4 + 2√3)/8 = (2+√3)/4
So: sin75°·cos15° = sin75°·sin75° = sin²75°
sin75° = sin(45°+30°) = sin45°cos30° + cos45°sin30°
= (1/√2)(√3/2) + (1/√2)(1/2)
= (√3 + 1)/(2√2)
sin²75° = (√3+1)²/8 = (3 + 2√3 + 1)/8 = (4 + 2√3)/8 = (2+√3)/4
∴ sin75°·cos15° = (2+√3)/4 ✓ Proved
16
हल गर्नुहोस् (Solve): sin2x − sinx = 0 (0° ≤ x ≤ 90°)
Model Answer
sin2x − sinx = 0
2sinx·cosx − sinx = 0
sinx(2cosx − 1) = 0
Case 1: sinx = 0 → x = 0°
Case 2: 2cosx − 1 = 0 → cosx = 1/2 → x = 60°
2sinx·cosx − sinx = 0
sinx(2cosx − 1) = 0
Case 1: sinx = 0 → x = 0°
Case 2: 2cosx − 1 = 0 → cosx = 1/2 → x = 60°
x = 0° or x = 60°
17
भेक्टरहरू 3i⃗ + 4j⃗ र 8i⃗ − 6j⃗ बिचको कोण पत्ता लगाउनुहोस् ।
Find the angle between the vectors 3i⃗ + 4j⃗ and 8i⃗ − 6j⃗.
Model Answer
Let a⃗ = 3i+4j, b⃗ = 8i−6j
a⃗ · b⃗ = (3)(8) + (4)(−6) = 24 − 24 = 0
|a⃗| = √(9+16) = √25 = 5
|b⃗| = √(64+36) = √100 = 10
cos θ = 0/(5×10) = 0
θ = cos⁻¹(0)
|a⃗| = √(9+16) = √25 = 5
|b⃗| = √(64+36) = √100 = 10
cos θ = 0/(5×10) = 0
θ = cos⁻¹(0)
θ = 90° (Vectors are perpendicular)
18
कुनै एउटा निरन्तर श्रेणीमा Σf|m−x̄| = 1134, N = 100 र मध्यक (x̄) = 40, भए मध्यक भिन्नता र यसको गुणाङ्क पत्ता लगाउनुहोस् ।
In a continuous series if Σf|m−x̄| = 1134, N = 100 and mean (x̄) = 40, find the mean deviation and its coefficient.
Model Answer
Mean Deviation (MD) = Σf|m−x̄| / N = 1134 / 100 = 11.34
Coefficient of MD = MD / x̄ = 11.34 / 40 = 0.2835
Coefficient of MD = MD / x̄ = 11.34 / 40 = 0.2835
MD = 11.34 | Coefficient of MD = 0.2835
समूह 'ग' (Group 'C')
Long Answer Questions — 3 marks each
19
यदि f(x) = (2x+5)/3 र g(x) = (x+5)/2 मा g∘g(x) = f⁻¹(x) भए x को मान पत्ता लगाउनुहोस् ।
If f(x) = (2x+5)/3 and g(x) = (x+5)/2 such that g∘g(x) = f⁻¹(x), find the value of x.
Model Answer
Find f⁻¹(x):
y = (2x+5)/3 → 3y = 2x+5 → x = (3y−5)/2
∴ f⁻¹(x) = (3x−5)/2
∴ f⁻¹(x) = (3x−5)/2
Find g∘g(x):
g(x) = (x+5)/2
g∘g(x) = g[g(x)] = g[(x+5)/2] = [(x+5)/2 + 5]/2 = (x+5+10)/4 = (x+15)/4
g∘g(x) = g[g(x)] = g[(x+5)/2] = [(x+5)/2 + 5]/2 = (x+5+10)/4 = (x+15)/4
Set g∘g(x) = f⁻¹(x):
(x+15)/4 = (3x−5)/2
x+15 = 2(3x−5)
x+15 = 6x−10
25 = 5x
x+15 = 2(3x−5)
x+15 = 6x−10
25 = 5x
∴ x = 5
20
लेखाचित्र विधिद्वारा हल गर्नुहोस् (Solve graphically): y = x² and y = 2x + 3
Model Answer
Find intersection algebraically:
x² = 2x + 3
x² − 2x − 3 = 0
(x−3)(x+1) = 0
x = 3 or x = −1
When x=3: y = 9 → Point (3, 9)
When x=−1: y = 1 → Point (−1, 1)
x² − 2x − 3 = 0
(x−3)(x+1) = 0
x = 3 or x = −1
When x=3: y = 9 → Point (3, 9)
When x=−1: y = 1 → Point (−1, 1)
Table of values:
| x | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| y = x² | 4 | 1 | 0 | 1 | 4 | 9 |
| y = 2x+3 | −1 | 1 | 3 | 5 | 7 | 9 |
Intersection points: (−1, 1) and (3, 9)
21
फलन f(x) निम्नानुसार परिभाषित छ:
f(x) = x+2 for x<2; 4 for x=2; 3x−2 for x>2
के फलन f(x) बिन्दु x=2 मा निरन्तर छ ? कारण दिनुहोस् ।
f(x) = x+2 for x<2; 4 for x=2; 3x−2 for x>2
के फलन f(x) बिन्दु x=2 मा निरन्तर छ ? कारण दिनुहोस् ।
Is the function f(x) continuous at x=2? Give reason.
Model Answer
LHL = lim(x→2⁻) f(x) = lim(x→2) (x+2) = 2+2 = 4
RHL = lim(x→2⁺) f(x) = lim(x→2) (3x−2) = 6−2 = 4
f(2) = 4 (given directly)
RHL = lim(x→2⁺) f(x) = lim(x→2) (3x−2) = 6−2 = 4
f(2) = 4 (given directly)
Since LHL = RHL = f(2) = 4:
✓ f(x) IS continuous at x = 2
बायाँ सीमा = दायाँ सीमा = फलनको मान = 4 ∴ निरन्तर छ।
22
मेट्रिक्स विधिको प्रयोग गरी हल गर्नुहोस् (Solve by using matrix method): 2x − 3y = 7 and 3x − 4y = 10
Model Answer
Matrix form: AX = B
A = [[2,−3],[3,−4]], X = [[x],[y]], B = [[7],[10]]
det(A) = (2)(−4) − (−3)(3) = −8 + 9 = 1
A⁻¹ = (1/1)·[[−4, 3],[−3, 2]] = [[−4, 3],[−3, 2]]
X = A⁻¹·B:
x = (−4)(7) + (3)(10) = −28 + 30 = 2
y = (−3)(7) + (2)(10) = −21 + 20 = −1
det(A) = (2)(−4) − (−3)(3) = −8 + 9 = 1
A⁻¹ = (1/1)·[[−4, 3],[−3, 2]] = [[−4, 3],[−3, 2]]
X = A⁻¹·B:
x = (−4)(7) + (3)(10) = −28 + 30 = 2
y = (−3)(7) + (2)(10) = −21 + 20 = −1
x = 2 , y = −1
23
यदि वृत्त x²+y²−2x−10y+1=0 को एउटा व्यासको एक छेउको निर्देशाङ्कहरू (1,0) भए अर्को छेउको निर्देशाङ्कहरू पत्ता लगाउनुहोस् ।
If the co-ordinates of one end of a diameter of the circle x²+y²−2x−10y+1=0 is (1,0), find the co-ordinates of the other end.
Model Answer
Standard form x²+y²+2gx+2fy+c=0:
x²+y²−2x−10y+1=0
2g = −2 → g = −1
2f = −10 → f = −5
Centre = (−g, −f) = (1, 5)
2g = −2 → g = −1
2f = −10 → f = −5
Centre = (−g, −f) = (1, 5)
Centre is the midpoint of the diameter. One end = (1, 0), centre = (1, 5).
Midpoint formula: centre = [(x₁+x₂)/2, (y₁+y₂)/2]
(1+x₂)/2 = 1 → x₂ = 1
(0+y₂)/2 = 5 → y₂ = 10
(1+x₂)/2 = 1 → x₂ = 1
(0+y₂)/2 = 5 → y₂ = 10
Other end of diameter = (1, 10)
24
प्रमाणित गर्नुहोस् (Prove that): 4sin³θ·cos3θ + 4cos³θ·sin3θ = 3sin4θ
Model Answer
Use identities: sin3θ = 3sinθ − 4sin³θ and cos3θ = 4cos³θ − 3cosθ
LHS = 4sin³θ·cos3θ + 4cos³θ·sin3θ
= 4sin³θ(4cos³θ − 3cosθ) + 4cos³θ(3sinθ − 4sin³θ)
= 16sin³θcos³θ − 12sin³θcosθ + 12sinθcos³θ − 16sin³θcos³θ
= 12sinθcosθ(cos²θ − sin²θ)
= 6sin2θ·cos2θ [since sin2θ=2sinθcosθ, cos2θ=cos²θ−sin²θ]
= 3·(2sin2θcos2θ)
= 3sin4θ = RHS
= 4sin³θ(4cos³θ − 3cosθ) + 4cos³θ(3sinθ − 4sin³θ)
= 16sin³θcos³θ − 12sin³θcosθ + 12sinθcos³θ − 16sin³θcos³θ
= 12sinθcosθ(cos²θ − sin²θ)
= 6sin2θ·cos2θ [since sin2θ=2sinθcosθ, cos2θ=cos²θ−sin²θ]
= 3·(2sin2θcos2θ)
= 3sin4θ = RHS
∴ 4sin³θ·cos3θ + 4cos³θ·sin3θ = 3sin4θ ✓ Proved
25
यदि P+Q+R = 180° भए प्रमाणित गर्नुहोस् :
sin(P+Q−R) + sin(Q+R−P) + sin(R+P−Q) = 4sinP·sinQ·sinR
sin(P+Q−R) + sin(Q+R−P) + sin(R+P−Q) = 4sinP·sinQ·sinR
If P+Q+R = 180°, prove that: sin(P+Q−R) + sin(Q+R−P) + sin(R+P−Q) = 4sinP·sinQ·sinR
Model Answer
Since P+Q+R=180°, we have P+Q=180°−R, Q+R=180°−P, R+P=180°−Q
sin(P+Q−R) = sin(180°−R−R) = sin(180°−2R) = sin2R
sin(Q+R−P) = sin(180°−P−P) = sin(180°−2P) = sin2P
sin(R+P−Q) = sin(180°−Q−Q) = sin(180°−2Q) = sin2Q
LHS = sin2P + sin2Q + sin2R
Using known identity: sin2P + sin2Q + sin2R = 4sinP·sinQ·sinR (when P+Q+R=180°)
Proof of this identity:
sin2P + sin2Q + sin2R
= 2sin(P+Q)cos(P−Q) + 2sinRcosR
= 2sinRcos(P−Q) + 2sinRcosR [since sin(P+Q)=sin(180°−R)=sinR]
= 2sinR[cos(P−Q) + cosR]
= 2sinR[cos(P−Q) + cos(P+Q)] [since R=180°−P−Q → cosR=−cos(P+Q)]
Wait: cosR = cos(180°−P−Q) = −cos(P+Q)
= 2sinR[cos(P−Q) − cos(P+Q)]
= 2sinR·2sinP·sinQ
= 4sinP·sinQ·sinR = RHS
sin(Q+R−P) = sin(180°−P−P) = sin(180°−2P) = sin2P
sin(R+P−Q) = sin(180°−Q−Q) = sin(180°−2Q) = sin2Q
LHS = sin2P + sin2Q + sin2R
Using known identity: sin2P + sin2Q + sin2R = 4sinP·sinQ·sinR (when P+Q+R=180°)
Proof of this identity:
sin2P + sin2Q + sin2R
= 2sin(P+Q)cos(P−Q) + 2sinRcosR
= 2sinRcos(P−Q) + 2sinRcosR [since sin(P+Q)=sin(180°−R)=sinR]
= 2sinR[cos(P−Q) + cosR]
= 2sinR[cos(P−Q) + cos(P+Q)] [since R=180°−P−Q → cosR=−cos(P+Q)]
Wait: cosR = cos(180°−P−Q) = −cos(P+Q)
= 2sinR[cos(P−Q) − cos(P+Q)]
= 2sinR·2sinP·sinQ
= 4sinP·sinQ·sinR = RHS
∴ sin(P+Q−R) + sin(Q+R−P) + sin(R+P−Q) = 4sinP·sinQ·sinR ✓
26
एउटा धरहराको टुप्पोबाट त्यसको ठिक अगाडि रहेको 20 मिटर अग्लो एउटा खम्बाको टुप्पो र फेदमा हेर्दा अवनती कोणहरू क्रमशः 30° र 60° पाइयो भने धरहराको उचाइ पत्ता लगाउनुहोस् ।
The angles of depression of the top and bottom of a pole 20 m high observed from the top of a tower are found to be 30° and 60° respectively. Find the height of the tower.
Model Answer
Let tower height = H, distance between tower and pole = d, pole height = 20 m.
Angle of depression to bottom of pole = 60°:
tan60° = H/d → √3 = H/d → d = H/√3 ...(i)
Angle of depression to top of pole = 30°:
tan30° = (H−20)/d → 1/√3 = (H−20)/d
d = √3(H−20) ...(ii)
From (i) and (ii):
H/√3 = √3(H−20)
H = 3(H−20)
H = 3H − 60
60 = 2H
tan60° = H/d → √3 = H/d → d = H/√3 ...(i)
Angle of depression to top of pole = 30°:
tan30° = (H−20)/d → 1/√3 = (H−20)/d
d = √3(H−20) ...(ii)
From (i) and (ii):
H/√3 = √3(H−20)
H = 3(H−20)
H = 3H − 60
60 = 2H
∴ Height of tower H = 30 m
27
शीर्षविन्दुहरू A(3,6), B(4,2) र C(2,2) भएको △ABC लाई x-अक्षमा परावर्तन गर्दा प्राप्त हुने प्रतिबिम्ब त्रिभुजको शीर्षविन्दुहरूका निर्देशाङ्कहरू मेट्रिक्स विधिद्वारा पत्ता लगाउनुहोस् ।
Find the co-ordinates of the vertices of image triangle of △ABC with vertices A(3,6), B(4,2) and C(2,2) under the reflection on the x-axis by matrix method.
Model Answer
Reflection matrix for x-axis: M = [[1, 0],[0, −1]]
Object matrix P = [[3, 4, 2],[6, 2, 2]]
Image P' = M × P
= [[1,0],[0,−1]] × [[3,4,2],[6,2,2]]
= [[(1)(3)+(0)(6), (1)(4)+(0)(2), (1)(2)+(0)(2)],
[(0)(3)+(−1)(6), (0)(4)+(−1)(2), (0)(2)+(−1)(2)]]
= [[3, 4, 2],[−6, −2, −2]]
Image P' = M × P
= [[1,0],[0,−1]] × [[3,4,2],[6,2,2]]
= [[(1)(3)+(0)(6), (1)(4)+(0)(2), (1)(2)+(0)(2)],
[(0)(3)+(−1)(6), (0)(4)+(−1)(2), (0)(2)+(−1)(2)]]
= [[3, 4, 2],[−6, −2, −2]]
A'(3, −6) , B'(4, −2) , C'(2, −2)
x-अक्षमा परावर्तनमा x-निर्देशाङ्क अपरिवर्तित, y-निर्देशाङ्क विपरीत हुन्छ।
28
तल दिइएको तथ्याङ्कको पहिलो चतुर्थांश 16.25 भए चतुर्थांशीय भिन्नता पत्ता लगाउनुहोस् :
Calculate the quartile deviation of the data given below where the first quartile is 16.25.
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| No. of Students | 6 | 8 | 12 | 11 | 7 |
Model Answer
N = 6+8+12+11+7 = 44
| Class | f | cf |
|---|---|---|
| 0-10 | 6 | 6 |
| 10-20 | 8 | 14 |
| 20-30 | 12 | 26 |
| 30-40 | 11 | 37 |
| 40-50 | 7 | 44 |
Q₁ is given = 16.25
Find Q₃:
3N/4 = 3×44/4 = 33 → Q₃ class = 30-40 (cf=26 < 33 ≤ 37)
Q₃ = L + [(3N/4 − cf)/f] × h
= 30 + [(33−26)/11] × 10
= 30 + (7/11) × 10
= 30 + 6.36 = 36.36
Quartile Deviation (QD):
QD = (Q₃ − Q₁)/2 = (36.36 − 16.25)/2 = 20.11/2
Find Q₃:
3N/4 = 3×44/4 = 33 → Q₃ class = 30-40 (cf=26 < 33 ≤ 37)
Q₃ = L + [(3N/4 − cf)/f] × h
= 30 + [(33−26)/11] × 10
= 30 + (7/11) × 10
= 30 + 6.36 = 36.36
Quartile Deviation (QD):
QD = (Q₃ − Q₁)/2 = (36.36 − 16.25)/2 = 20.11/2
QD = 10.055 ≈ 10.06
समूह 'घ' (Group 'D')
Long Answer Questions — 4 marks each
29
दिइएको तथ्याङ्कबाट स्तरीय भिन्नता निकाल्नुहोस् (cumulative data):
Calculate the standard deviation from the given data (cumulative frequency).
| Marks | 0-10 | 0-20 | 0-30 | 0-40 | 0-50 |
| No. of Students (cf) | 5 | 15 | 35 | 40 | 50 |
Model Answer
Convert cumulative to simple frequencies: f = 5, 10, 20, 5, 10. N=50
| Class | f | m | fm | m² | fm² |
|---|---|---|---|---|---|
| 0-10 | 5 | 5 | 25 | 25 | 125 |
| 10-20 | 10 | 15 | 150 | 225 | 2250 |
| 20-30 | 20 | 25 | 500 | 625 | 12500 |
| 30-40 | 5 | 35 | 175 | 1225 | 6125 |
| 40-50 | 10 | 45 | 450 | 2025 | 20250 |
| Total | 50 | 1300 | 41250 |
Mean (x̄) = Σfm/N = 1300/50 = 26
σ² = Σfm²/N − (x̄)² = 41250/50 − 26² = 825 − 676 = 149
σ = √149 ≈ 12.21
σ² = Σfm²/N − (x̄)² = 41250/50 − 26² = 825 − 676 = 149
σ = √149 ≈ 12.21
Standard Deviation σ ≈ 12.21
30
गुणोत्तर श्रेणीमा रहेका तीनओटा क्रमिक सङ्ख्याहरूको योगफल 56 छ । यदि ती सङ्ख्याहरूबाट क्रमशः 1, 7 र 21 घटाउँदा सङ्ख्याहरू समान्तरीय श्रेणीमा हुन्छन् भने ती सङ्ख्याहरू पत्ता लगाउनुहोस् ।
The sum of three consecutive numbers of a geometric series is 56. If 1, 7 and 21 be subtracted from the numbers respectively, they form an arithmetic series. Find the numbers.
Model Answer
Let the three numbers in GP be: a/r, a, ar
Sum: a/r + a + ar = 56 ...(i)
After subtracting 1,7,21 they form AP:
(a/r − 1), (a − 7), (ar − 21) are in AP
∴ 2(a−7) = (a/r−1) + (ar−21)
2a−14 = a/r + ar − 22
2a + 8 = a/r + ar ...(ii)
From (i): a/r + ar = 56 − a
From (ii): a/r + ar = 2a + 8
So: 56 − a = 2a + 8
48 = 3a → a = 16
Substitute back: a/r + ar = 56 − 16 = 40
16/r + 16r = 40
16 + 16r² = 40r
16r² − 40r + 16 = 0
2r² − 5r + 2 = 0
(2r−1)(r−2) = 0
r = 1/2 or r = 2
After subtracting 1,7,21 they form AP:
(a/r − 1), (a − 7), (ar − 21) are in AP
∴ 2(a−7) = (a/r−1) + (ar−21)
2a−14 = a/r + ar − 22
2a + 8 = a/r + ar ...(ii)
From (i): a/r + ar = 56 − a
From (ii): a/r + ar = 2a + 8
So: 56 − a = 2a + 8
48 = 3a → a = 16
Substitute back: a/r + ar = 56 − 16 = 40
16/r + 16r = 40
16 + 16r² = 40r
16r² − 40r + 16 = 0
2r² − 5r + 2 = 0
(2r−1)(r−2) = 0
r = 1/2 or r = 2
When r=2: terms = 8, 16, 32 | When r=1/2: terms = 32, 16, 8
The three numbers are 8, 16, 32
Check: 8+16+32=56 ✓ ; (8−1),(16−7),(32−21) = 7,9,11 — AP ✓
31
एउटा वर्ग PQRS को शीर्षविन्दुहरू P र R का निर्देशाङ्कहरू क्रमशः (3, 7) र (−1, 3) छन् भने विकर्ण QS को समीकरण पत्ता लगाउनुहोस् ।
The co-ordinates of vertices P and R of a square PQRS are (3, 7) and (−1, 3) respectively. Find the equation of diagonal QS.
Model Answer
In a square, diagonals bisect each other at 90°.
Step 1: Midpoint of PR (= midpoint of QS):
M = ((3+(−1))/2, (7+3)/2) = (1, 5)
Step 2: Slope of PR:
m₁ = (3−7)/(−1−3) = −4/−4 = 1
Step 3: Since QS ⊥ PR, slope of QS:
m₂ = −1/m₁ = −1/1 = −1
Step 4: Equation of QS through M(1,5) with slope −1:
y − 5 = −1(x − 1)
y − 5 = −x + 1
x + y = 6
M = ((3+(−1))/2, (7+3)/2) = (1, 5)
Step 2: Slope of PR:
m₁ = (3−7)/(−1−3) = −4/−4 = 1
Step 3: Since QS ⊥ PR, slope of QS:
m₂ = −1/m₁ = −1/1 = −1
Step 4: Equation of QS through M(1,5) with slope −1:
y − 5 = −1(x − 1)
y − 5 = −x + 1
x + y = 6
Equation of diagonal QS: x + y = 6 or x + y − 6 = 0
32
समबाहु चतुर्भुज MNOP को विकर्णहरू आपसमा लम्ब हुन्छन् भनी भेक्टरविधिद्वारा प्रमाणित गर्नुहोस् ।
Prove by vector method that the diagonals of rhombus MNOP are perpendicular to each other.
Model Answer
Let MNOP be a rhombus with all sides equal: |MN⃗| = |MN⃗| = |NO⃗| = |OP⃗| = |PM⃗| = a
Let MN⃗ = b⃗ and MP⃗ = d⃗
Diagonal MO⃗ = MN⃗ + NO⃗ = b⃗ + d⃗ (since NO⃗ = MP⃗ = d⃗ in rhombus)
Diagonal NP⃗ = NM⃗ + MP⃗ = −b⃗ + d⃗ = d⃗ − b⃗
Dot product of diagonals:
MO⃗ · NP⃗ = (b⃗ + d⃗)·(d⃗ − b⃗)
= d⃗·d⃗ − b⃗·b⃗
= |d⃗|² − |b⃗|²
Since MNOP is a rhombus: |MN| = |MP| → |b⃗| = |d⃗|
∴ MO⃗ · NP⃗ = |d⃗|² − |d⃗|² = 0
Diagonal NP⃗ = NM⃗ + MP⃗ = −b⃗ + d⃗ = d⃗ − b⃗
Dot product of diagonals:
MO⃗ · NP⃗ = (b⃗ + d⃗)·(d⃗ − b⃗)
= d⃗·d⃗ − b⃗·b⃗
= |d⃗|² − |b⃗|²
Since MNOP is a rhombus: |MN| = |MP| → |b⃗| = |d⃗|
∴ MO⃗ · NP⃗ = |d⃗|² − |d⃗|² = 0
Since MO⃗ · NP⃗ = 0, the diagonals are perpendicular. ✓ Proved
33
R ले रेखा y = −x मा हुने परावर्तन र S ले उद्गमबिन्दु वरिपरि 180° मा हुने परिक्रमणलाई जनाउँछ भने संयुक्त स्थानान्तरण RoS ले जनाउने एकल स्थानान्तरण पत्ता लगाउनुहोस् । सोही एकल स्थानान्तरण प्रयोग गरी शीर्षविन्दुहरू A(1,2), B(−4,3) र C(3,5) भएको △ABC लाई स्थानान्तरण गर्दा बन्ने प्रतिबिम्बको शीर्षविन्दुहरूका निर्देशाङ्कहरू पत्ता लगाउनुहोस् ।
R denotes the reflection on the line y = −x and S denotes the rotation about origin through 180°. Find the single transformation represented by combined transformation RoS. Using the same single transformation, find the co-ordinates of the vertices of the image of △ABC having vertices A(1,2), B(−4,3) and C(3,5).
Model Answer
Transformation matrices:
Reflection on y = −x: R = [[0, −1],[−1, 0]]
Rotation 180° about origin: S = [[−1, 0],[0, −1]]
Combined RoS = R × S (apply S first, then R):
RoS = [[0,−1],[−1,0]] × [[−1,0],[0,−1]]
= [[(0)(−1)+(−1)(0), (0)(0)+(−1)(−1)],
[(−1)(−1)+(0)(0), (−1)(0)+(0)(−1)]]
= [[0, 1],[1, 0]]
Rotation 180° about origin: S = [[−1, 0],[0, −1]]
Combined RoS = R × S (apply S first, then R):
RoS = [[0,−1],[−1,0]] × [[−1,0],[0,−1]]
= [[(0)(−1)+(−1)(0), (0)(0)+(−1)(−1)],
[(−1)(−1)+(0)(0), (−1)(0)+(0)(−1)]]
= [[0, 1],[1, 0]]
This is reflection on the line y = x.
Object matrix = [[1, −4, 3],[2, 3, 5]]
Image = [[0,1],[1,0]] × [[1,−4,3],[2,3,5]]
= [[(0)(1)+(1)(2), (0)(−4)+(1)(3), (0)(3)+(1)(5)],
[(1)(1)+(0)(2), (1)(−4)+(0)(3), (1)(3)+(0)(5)]]
= [[2, 3, 5],[1, −4, 3]]
Image = [[0,1],[1,0]] × [[1,−4,3],[2,3,5]]
= [[(0)(1)+(1)(2), (0)(−4)+(1)(3), (0)(3)+(1)(5)],
[(1)(1)+(0)(2), (1)(−4)+(0)(3), (1)(3)+(0)(5)]]
= [[2, 3, 5],[1, −4, 3]]
Single transformation: Reflection on y = x
A'(2, 1) , B'(3, −4) , C'(5, 3)
y=x मा परावर्तनले (x,y) → (y,x) बनाउँछ।