SEE Optional Mathematics New Specification Grid 2080
MODEL QUESTION 2080
Subject: Optional Mathematics
Class: 10
Full Marks: 75
Time: 3 hrs
All the questions are compulsory.
Note: Due to limitation of blogger, I was unable to use the vector symbol on question no. 9, 17 and 32. So, add vector symbols accordingly while you practice this set.
Group A [10 × 1 = 10]
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Define trigonometric function.
✎ Trigonometric function is the function of an angle expressed as the ratio of two sides of a right-angled triangle. The most common trigonometric functions are sine, cosine, and tangent.
For example:
sin θ = P/h, cos θ = b/h, and tan θ = P/b -
What is arithmetic mean between two numbers a and b?
✎ The arithmetic mean (AM) between two numbers a and b is:
AM = (a + b) / 2 -
Write the name of the set of numbers which is continuous.
✎ The set of real numbers is continuous.
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If matrix A = [[a, b], [c, d]], what is the value of |A|?
✎ The value of |A| is:
|A| = ad - bc -
If the angle between two straight lines is θ and their slopes are m1 and m2 respectively, write the formula for tan θ.
✎ Here, the formula for tan θ is:
tan θ = ±(m1 - m2) / (1 + m1m2) -
Which geometric figure will be formed if a plane intersects a cone parallel to its base?
✎ A circle will be formed if a plane intersects a cone parallel to its base.
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Express sin 2A in terms of tan A.
✎ sin 2A can be expressed in terms of tan A as:
sin 2A = (2 tan A) / (1 + tan² A)
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Define angle of elevation.
✎ The angle of elevation is the angle between a horizontal line from the observer and the line of sight to an object that is above the horizontal line.
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What is the scalar product of two vectors a and b if the angle between them is θ?
✎ Here, the scalar product of given vectors is:
a · b = |a||b|cosθ -
If P' is image of P and r is radius of inversion circle with center O in an inversion transformation, write the relations of OP, OP', and r.
✎ The relations of OP, OP', and r is:
OP × OP' = r²
Group ‘B’ [8 × 2 = 16]
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If
2x³ − 7x² + x + 10 = (x − 1)Q(x) + R, find the remainder R and quotient Q(x).✎ Here,
2x³ − 7x² + x + 10 = (x − 1)Q(x) + RComparing
(x − 1)with(x − a), we geta = 1.Now, using Synthetic Division:
x³ x² x x° 2 −7 1 10 ↓ 2 −5 −4 2 −5 −4 6 = R x² x x°Now,
Quotient =
Q(x) = 2x² − 5x − 4Remainder =
R = 6 -
Write down the inequality represented by the shaded region in the adjoining figure.
✎ From the given figure, the boundary line passes through the points (4, 0) and (0, −3). So, the equation of boundary line is:
y − y₁ = (y₂ − y₁) / (x₂ − x₁) (x − x₁)or,
y − 0 = (−3 − 0) / (0 − 4) (x − 4)or,
y = (3/4)(x − 4)or,
4y = 3x − 12∴
3x − 4y = 12Since, the half plane with the boundary line contains origin, so the inequality represented by the shaded region is:
3x − 4y ≤ 12
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Find the determinants D and D₂ of coefficients of x and y by using Cramer's rule from the equations 4x − 5y = 2 and 3x + 4y = 48.
✎ Here,
4x − 5y = 2 ...(i)
3x + 4y = 48 ...(ii)
Now, arranging the coefficients and constant term, we get:
Coeff. of x Coeff. of y Constant 4 -5 2 3 4 48 Now,
D:
| 2 -5 |
| 48 4 |
= 2×4 − 48×(−5) = 8 + 240 = 248
D₂:
| 4 2 |
| 3 48 |
= 4×48 − 3×2 = 192 − 6 = 186
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Find the slopes of two straight lines 3x + 4y = 5 and 6x + 8y + 7 = 0 and write the relationship between them.
✎ Slope of 3x + 4y = 5 is m₁ =
-coeff. of x / coeff. of y=-3 / 4Slope of 6x + 8y + 7 = 0 is m₂ =
-coeff. of x / coeff. of y=-6 / 8=-3 / 4Since m₁ = m₂, the given lines are parallel.
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Convert sin6A.cos4A into sum or difference of sine or cosine.
✎ We know that, 2sinAcosB = sin(A+B) + sin(A−B)
Now,
sin6Acos4A = (1/2) × 2sin6Acos4A
= (1/2)[sin(6A + 4A) + sin(6A − 4A)]
= (1/2)[sin10A + sin2A]
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If 2sinθ = √3, find the value of θ. (0° ≤ θ ≤ 180°)
✎ Here,
2sinθ = √3
or, sinθ = √3 / 2
or, sinθ = sin60°
∴ θ = 60°
Again,
sinθ = sin60° = sin(180° − 60°) = sin120°
∴ θ = 120°
Thus, the values of θ are 60° and 120°.
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O is the origin in the given figure. If a and b are the position vectors of the points A and B, show that the position vector of the point P is p = 1/2 (a + b).
✎ Here,
OA = a
,OB = b
, andOP = p
Now,
AP = PB
or,
AO + OP = PO + OB
or,
-OA + OP = -OP + OB
or,
-a + p = p + b
or,
p + p = a + b
or,
2p = a + b
or,
p = (a + b)/2
∴p = (a + b)/2 -
In a series, the first quartile (Q1) = 35 and third quartile (Q3) = 75, find the quartile deviation and its coefficient.
✎ Here, Q1 = 35 and Q3 = 75
Now,
Quartile Deviation (QD) = (Q3 - Q1) / 2
= (75 - 35) / 2
= 40 / 2
= 20
Coefficient of Quartile Deviation = (Q3 - Q1) / (Q3 + Q1)
= (75 - 35) / (75 + 35)
= 40 / 110
= 0.36Group ‘C’ [11 × 3 = 33]
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If two functions are f(x) = (2x + 5) / 8 and g(x) = 3x - 4, find (f ∘ g)-1(3).
✎ Here, f(x) = (2x + 5) / 8 and g(x) = 3x - 4
Now,
f ∘ g(x) = f(g(x))
= f(3x - 4)
= [2(3x - 4) + 5] / 8
= (6x - 8 + 5) / 8
= (6x - 3) / 8
Let f ∘ g(x) = (6x - 3) / 8 = y
To find (f ∘ g)-1(x), interchanging the roles of x and y and solving for y,
6y - 3 / 8 = x
or, 6y - 3 = 8x
or, y = 8x + 3 / 6
or, (f ∘ g)-1(x) = 8x + 3 / 6
So, (f ∘ g)-1(3) = 8 x 3 + 3 / 6 = 9 / 2
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Solve by graphical method: 2x² + x - 6 = 0
✎ Here, 2x² + x - 6 = 0
or, 2x² = 6 - x
Let y = 2x² = 6 - x
Then, y = 2x² ... (i)
y = 6 - x ... (ii)
Equation (i) represents the parabola with vertex at origin and equation (ii) represents a straight line. The point of intersection of the parabola and straight line gives the solution.
From equation (i), y = 2x²
x -2 -1 0 1 2 y 8 2 0 2 8
Plotting these points on the graph and joining them, we get a parabola with vertex at origin.
From equation (ii), y = 6 - x
x 0 2 y 6 4
Plotting these points on the graph and joining them, we get a straight line.
From the graph, the points of intersection of the parabola and straight line are (-2, 8) and (1.5, 4.5).
∴ x = -2, 1.5 -
For a real valued function f(x) = 2x + 3, find the values of f(2.99), f(3.01), and f(3). Is this function continuous at x = 3?
✎ Here, f(x) = 2x + 3
When x = 2.99 → 3⁻, f(x) = 8.98 → 9
Now,
f(2.99) = 2 × 2.99 + 3 = 8.98
f(3.01) = 2 × 3.01 + 3 = 9.02
f(3) = 2 × 3 + 3 = 9
∴ limx→3⁻ f(x) = 9
When x = 3.01 → 3⁺, f(x) = 9.02 → 9
∴ limx→3⁺ f(x) = 9
When x = 3, f(3) = 9
∴ limx→3⁻ f(x) = limx→3⁺ f(x) = f(3) = 9
Thus, function f(x) is continuous at x = 3. -
By using the matrix method solve the following systems of equations:
3x + 5y = 11, 2x - 3y = 1
✎ Here,
3x + 5y = 11
2x - 3y = 1
Writing in matrix form, x
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Find the equations of pair of lines represented by equation 6x² - xy - y² = 0 and also find the angle between them.
✎ Here,
6x² - xy - y² = 0
or, 6x² - (3 - 2)xy - y² = 0
or, 6x² - 3xy + 2xy - y² = 0
or, 3x(2x - y) + y(2x - y) = 0
or, (2x - y)(3x + y) = 0
Thus, the two separate equations are: 2x - y = 0 and 3x + y = 0Comparing 6x² - xy - y² = 0 with ax² + 2hxy + by² = 0, we get:
a = 6, 2h = -1 → h = -1/2 and b = -1
Now, The angle between the pair of lines is given by:
tan θ = ± 2√(h² - ab) / (a + b)
Substituting values:
tan θ = ± 2√((-1/2)² - 6(-1)) / (6 - 1)
= ± 2√(1/4 + 6) / 5
= ± 2√(25/4) / 5
= ± (2 × 5 / 2) / 5
= ± 1Taking Positive Sign
tan θ = 1
θ = tan⁻¹(1)
θ = 45°Taking Negative Sign
tan θ = -1
θ = tan⁻¹(-1)
θ = tan⁻¹(tan (180° - 45°))
θ = 135°Thus, the angle between them is 45° and 135°.
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Prove that: tan A + 2tan 2A + 4cot 4A = cot A
✎ SolutionLHS:
tan A + 2tan 2A + 4cot 4A
= tan A + 2tan 2A + 4 / tan 4A
= tan A + 2tan 2A + 4 / tan 2(2A)
= tan A + 2tan 2A + 4 / (2tan 2A / (1 - tan² 2A))
= tan A + 2tan 2A + (2(1 - tan² 2A)) / (tan 2A)
= tan A + (2tan² 2A + 2 - 2tan² 2A) / (tan2A)
= tan A + (1 - tan² A) / tan A
= (tan² A + 1 - tan² A) / tan A)
= 1 / tan A
= cot A
= RHS
Proved. -
If A + B + C = π, prove that: sin²A − sin²B + sin²C = 2sinA cosB sinC
✎ Solution:
Here:
A + B + C = π
⇒ A + B = π − C
⇒ sin(A + B) = sin(π − C)
⇒ sin(A + B) = sin CNow:
LHS = sin²A − sin²B + sin²C
= (1/2)[2sin²A − 2sin²B] + sin²C
= (1/2)[(1 − cos2A) − (1 − cos2B)] + sin²C
= (1/2)[−cos2A + cos2B] + sin²C
= (1/2)[cos2B − cos2A] + sin²C
= (1/2). 2sin[(2A + 2B)/2] sin[(2A − 2B)/2] + sin²C
= sin(A + B) sin(A − B) + sin²C
= sin C sin(A − B) + sin²C
= sinC [sin(A − B) + sinC]
= sinC [sin(A − B) + sin(A + B)]
= sinC . 2sinA cosB
= 2sinA cosB sinC
= RHS
Proved. -
From a place at the ground level in front of a tower, the angles of elevation of the top and bottom of a flagstaff 6m high situated at the top of the tower are observed to be 60° and 45°, respectively. Find the height of the tower and the distance between the base of the tower and the point of observation.
✎ Solution:
Let BD be the tower, AD is the flagstaff, and BC is the distance between the base of the tower and the point of observation C.
Given:
AD = 6m
∠BCA = 60° (angle of elevation of the top of the flagstaff)
∠BCD = 45° (angle of elevation of the bottom of the flagstaff)
BD = x (say) Now, in ∆ DCB:
tan 45° = x / BC
⇒ 1 = x / BC
⇒ BC = xAgain, in ∆ACB:
tan 60° = (x + 6) / x
⇒ √3 = (x + 6) / x
⇒ 1.732x = x + 6
⇒ 1.732x − x = 6
⇒ 0.732x = 6
⇒ x = 6 / 0.732
⇒ x = 8.20Thus, the height of the tower is 8.20 m, and the distance between the base of the tower and the point of observation is 8.20 m.
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Find a 2×2 matrix which transforms the unit square
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Find the mean deviation from mean and its coefficient from given data:
Marks Obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 No. of students 2 3 6 5 4 ✎ Here,
Marks No. of students (f) Mid-value (x) fx |D| = |x – x̄| f|D| 0 – 10 2 5 10 23 46 10 – 20 3 15 45 13 39 20 – 30 6 25 150 3 18 30 – 40 5 35 175 7 35 40 – 50 4 45 180 17 68 N = Σf = 20 Σfx = 560
Now,
∴ Mean = x̄ = Σfx / N = 560 / 20 = 28
∴ Mean Deviation = M.D. = Σf|D| / N = 206 / 20 = 10.3
∴ Coefficient of Mean Deviation = M.D. / x̄ = 10.3 / 28 = 0.37
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Find the standard deviation and coefficient of variation from given data:
✎ Here, taking a = 25 and h = 10
Age No. of persons (f) Mid-value (x) d = (x – a) / h fd fd² 0 – 10 4 5 -2 -8 16 10 – 20 6 15 -1 -6 6 20 – 30 10 25 0 0 0 30 – 40 20 35 1 20 20 40 – 50 6 45 2 12 24 50 – 60 4 55 3 12 36 N = Σf = 50 Σfd = 30 Σfd² = 102
Now,
Standard Deviation (σ) = h √(Σfd² / N - (Σfd / N)²)
= 10 √(102 / 50 - (30 / 50)²)
= 10 √(2.04 - 0.36)
= 10 √1.68
= 10 × 1.296
= 12.96
Group ‘D’ [4 × 4 = 16]
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The sum of three terms in an arithmetic series is 24. If 1, 6, and 18 are added to them respectively, the results are in geometrical series, find the terms.
✎ Let the three consecutive terms in an arithmetic be a – d, a, a + d.
It is given that the sum of these numbers is 24.
a – d + a + a + d = 24
or, 3a = 24
or, a = 24 / 3
∴ a = 8
If 1, 6, and 18 are added to these terms respectively, then a – d + 1, a + 6, a + d + 18 are in geometric series.
Substituting a = 8, we get:
a – d + 1 = 8 – d + 1 = 9 – d
a + 6 = 8 + 6 = 14
a + d + 18 = 8 + d + 18 = 26 + d
i.e., 9 – d, 14, 26 + d are in geometric series.
So, 14² = (9 – d)(26 + d)
or, 196 = 234 + 9d – 26d – d²
or, 196 = 234 – 17d – d²
or, d² + 17d + 196 – 234 = 0
or, d² + 17d – 38 = 0
or, (d + 19)(d – 2) = 0
Either, (d + 19) = 0 ⇒ d = –19
Or, (d – 2) = 0 ⇒ d = 2
When d = –19:
a – d = 8 – (–19) = 27
a = 8
a + d = 8 – 19 = –11
When d = 2:
a – d = 8 – 2 = 6
a = 8
a + d = 8 + 2 = 10
Thus, the required three terms are 27, 8, –11 or 6, 8, 10.
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In the given figure, X and Y are the center of circles A and B respectively. Circle A passes through center Y of circle B. If the equation of circle B is x² + y² - 4x + 6y - 12 = 0 and the coordinate of X is (-4, 5), find the equation of circle A.
✎ The equation of circle B is:
x² + y² - 4x + 6y - 12 = 0 ... (i)
Comparing equation (i) with x² + y² + 2gx + 2fy + c = 0, we get
2g = -4, 2f = 6, c = -12
∴ g = -2, f = 3
Center of circle B = Y(-g, -f) = Y(2, -3)
Center of circle A = X(-4, 5) = (h, k)
Now,
Radius of circle A (r) = XY
r = √((x₂ - x₁)² + (y₂ - y₁)²)
r = √((2 + 4)² + (-3 - 5)²)
r = √(36 + 64)
r = √100
r = 10
The equation of circle A is:
(x - h)² + (y - k)² = r²
or, (x + 4)² + (y - 5)² = 10²
or, x² + 2.x.4 + 4² + y² - 2.y.5 + 5² = 100
or, x² + y² + 8x - 10y + 16 + 25 - 100 = 0
∴ x² + y² + 8x - 10y - 59 = 0
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By using vector method, prove that the quadrilateral formed by joining the midpoints of adjacent sides of a quadrilateral is a parallelogram.
✎ Given:
ABCD is a quadrilateral. The midpoints of sides AB, BC, CD, DA are P, Q, R, S respectively.
To prove: PQRS is a parallelogram.
Construction: Join A and C.
Proof:
Statements Reasons 1. PQ = 1/2 AC and PQ // AC 1. In △ABC, the line segment joining the midpoint of two sides of triangle is half of third side and parallel to third side. 2. SR = 1/2 AC and SR // AC 2. In △ADC, the line segment joining the midpoint of two sides of triangle is half of third side and parallel to third side. 3. PQ = SR and PQ // SR 3. From statements (1) and (2). 4. PQRS is a parallelogram 4. From statement (3), being opposite side of quadrilateral equal and parallel. Proved
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The image of the triangle A is A' and image of A' is A'' in the given graph.
- a) By what transformation the image of the triangle A is A'? Write with reason.
- b) By what transformation the image of the triangle A' is A''? Write with reason.
- c) Write the name of transformation which denotes the combined transformation of the above two transformations? Write with reason.
✎ The vertices of triangle A are: (1, 0), (5, 1) and (5, 4).
The vertices of triangle A' are: (-1, 0), (-5, -1) and (-5, -4).
The vertices of triangle A'' are: (0, 1), (1, 5) and (4, 5).
a) When triangle A is transformed into A':
- (1, 0) → (-1, 0)
- (5, 1) → (-5, -1)
- (5, 4) → (-5, -4)
The transformation of triangle A into triangle A' is the rotation through 180° about origin because it follows the rule of the rotation through 180° about origin i.e., (x, y) → (-x, -y).
b) When triangle A' is transformed into A'':
- (-1, 0) → (0, 1)
- (-5, -1) → (1, 5)
- (-5, -4) → (4, 5)
The transformation of triangle A' into triangle A'' is the reflection on y = -x because it follows the rule of the reflection on y = -x i.e., (x, y) → (-y, -x).
c) When triangle A is transformed into A'' (combined transformation):
- (1, 0) → (0, 1)
- (5, 1) → (1, 5)
- (5, 4) → (4, 5)
The combined transformation is the reflection on y = x because it follows the rule of the reflection on y = x i.e., (x, y) → (y, x).
THE END
