Summative Assessment I

A rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q.

Since for option D numerator, p = -3 and denominator q = 5 both are integers.

-3/5 is a rational number.

If 3 is the solution for the equation. It must satisfy the expression.

So, putting x = 3 it must be zero.

33 - 4 × 32 + 2 × 3 + k = 0

27 – 4 × 9 + 6 + k = 0

k - 3 = 0

k = 3

We need to do hit and trial to find root of a cubic equation.

If it is a root of equation, it must satisfy the equation.

So, let’s start with option A.

(-2)³ + 2(-2)² - 5(-2)-6 = -8 + 8 + 10 - 6 = 4

Let’s try option B

(2)³ + 2(2)² - 5(2)-6 = 8 + 8 – 10 - 6 = 0

Let’s try option C

(-3)³ + 2(-3)² - 5(-3)-6 = -27 + 18 + 15 - 6 = 0

For option D

(3)³ + 2(3)² - 5(3)-6 = 27 + 18 – 15 - 6 = 24

Hence Option B and C are correct

Verifying –

Factors of the given equation is (x-2)(x + 3)(x + 1) = x³ + 2x² - 5x - 6.

-x² + 7x - 12 can be factorized as-

-x² + 4x + 3x - 12

-x(x - 4) + 3(x - 4)

(x - 4)(3 - x)

Also recheck by-

Sum of roots = 7 {-coefficient of x/ coefficient of x²}

Product of roots = 12 {constant/ coefficient of x²}

Sum of angles in a straight line is 180°

So, ∠AOD + ∠DOC + ∠BOC = 180°

3x + 5x + 4x = 180

12x = 180

x = 15

∠BOC = 4x = 4×15 = 60°.

Since we know all the angles in an equilateral triangle is of 60°.

So, ∠ABC = ∠ACB = ∠CAB = 60° …(i)

Also for an isosceles triangle, the angles opposite to equal sides are equal.

So, ∠DBC = ∠DCB = x (let’s say)

Also sum of all angles in a triangle = 180°.

So, in ΔBDC,

∠DBC + ∠DCB + ∠BDC = 180°

x + x + 90 = 180 {since ∠BDC = 90°}

2x = 90

x = 45°

so ∠DCB = 45 …(ii)

And ∠ACD = ∠ACB + ∠DCB = 60° + 45° = 105° {from (i) and (ii)}

Applying heron’s formula-

We know,

here a, b and c are sides of a triangle

So, Area =

Hence Area =

= √3600

= 60 square units

For a right-angled triangle,

Applying Pythagoras theorem,

(hypotenuse)² = (base)² + (perpendicular)²

Since triangle is isosceles.

So, base = perpendicular = x (let’s say)

Hence (hypotenuse)² = (x)² + (x)²

(4√2)² = 2x²

32 = 2x²

x² = 16

so, x = 4 cm.

Let to be y.

So y =

Squaring both sides,

Also, x = 7 + 4√3

So

= 16

So, y = √16 = 4

Hence

(7a³ + 56b³)

= 7(a³ + 8b³)

= 7(a³ + (2b)³)

= 7(a + (2b))(a² + (2b)² - a(2b))

[since a³ + b³ = (a + b)(a² + b² - ab)]

= 7(a + 2b)(a² + 4b² - 2ab)

If (x - 1) is a factor of the polynomial (a²x³ - 4ax + 4a - 1).

then it must satisfy it.

So, putting x = 1 the polynomial must be zero.

Putting x = 1 and equating to zero.

= (a²(1)³ - 4a(1) + 4a - 1)

= a² - 4a + 4a - 1 = 0

= a² = 1

So, a = �1.

Given- AC = BD

Subtracting BC on both sides-

(AC - BC) = (BD - BC)

AB = CD

In a triangle sum of all angles = 180°

So, ∠A + ∠B + ∠C = 180°

It is given that-

∠A = 3/2 ∠B

∠C = � ∠B

So, ∠A + ∠B + ∠C = (3/2) ∠B + ∠B + (1/2) ∠B = 180°

3∠B = 180°

∠B = 60°

In ΔABC sum of all angles = 180°.

So, ∠BAC + ∠ABC + ∠ACB = 180°

30 + 50 + ∠ACB = 180

∠ACB = 100°

Since BCD represents a straight line ∠ACB + ∠ECD = 180°

So, ∠ECD = 80°

In ΔECD sum of all angles = 180°

So, ∠ECD + ∠EDC + ∠CED = 180°

60 + 40 + ∠CED = 180

∠CED = 80°

Since AEC represents a straight line, ∠CED + ∠AED = 180°

So, ∠AED = 120°

(on rationalizing we get)

= 4 + (√5)(√3)

= 4 + √15

Similarly (rationalising)

= (5 + 3-2(√5)(√3))/2

= 4- (√5) (√3)

= 4- √15

So,

= 32 + 30

= 62

(II)

Taking LCM as (3 + √5)( 3-√5)

(since (a + b)(a - b) = a² - b²)

We know for a cubic polynomial, sum of roots

Let the third root be x.

If we divide f(x) = 4x² - 12x² + 14x - 3 by (2x - 1) remainder can be find at value of –

(2x-1) = 0

Or x = 1/2

So, we will put x = 1/2 in f(x) = 4x² - 12x² + 14x – 3

We know that –

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca).

here if a + b + c = 0

a³ + b³ + c³ = 3abc.

So, (p-q)³ + (q-r)³ + (r-p)³ = 3(p-q)(q-r)(r-p) {since (p-q) + (q-r) + (r-p) = 0}

Since DE || BC and AB acts as transversal.

So, ∠ADE = ∠ABC {corresponding angles}

since ∠ABC = 40°

So, ∠ADE = 40°

Since EF || AB and DN acts as transversal.

So, ∠ADE = ∠MEN {corresponding angles}

∠MEN = 40°

Hence, ∠ADE + ∠MEN = 80°

(ii) 140°

Since AB represents a straight line. Sum of angles in line AB = 180°

So, ∠BDE + ∠ADE = 180°

since, ∠ADE = 40°

So, ∠BDE = 140°

(iii) 140°

Since DE || BC and FM acts as transversal.

So, ∠EFC = ∠ MEN = 40°

And BC represents a straight line. Sum of angles in line BC = 180°

= ∠EFC + ∠BFE = 180°

= ∠BFE = 140°

Taking ΔABC and ΔABD in consideration-

AD = BC

Since, it is given that

∠1 = ∠2 and ∠3 = ∠4

Adding them –

∠1 + ∠3 = ∠2 + ∠4.

= ∠DAB = ∠ABC

And AB is the common side on both triangle.

So, by side angle side(SAS) criteria-

Triangle ΔABC and ΔABD are congruent.

So, BD = AC (by congruency criteria).

Since C is the mid-point of AB.

So, AC = BC.

Taking ΔACE and ΔBCD in consideration-

∠DBC = ∠EAC

AC = BC

Also ∠DCA = ∠ECB

Adding ∠DCE on both sides-

∠DCB = ∠ECA

So, by Angle side Angle(ASA) criteria ΔACE and ΔBCD are congruent.

And hence DC = EC (by congruency criteria).

Sum of all angles in a triangle = 180°

∠A + ∠B + ∠C = 180°

∠A = 2 ∠CAM = 2 ∠MAB {since AM is bisector of ∠A}

= 2∠CAM + ∠B + ∠C = 180°

= 2∠CAM = 180 - (∠B + ∠C)

∠AML = ∠CAM + ∠C {Exterior Angle theorem}

In Triangle ΔALM, Sum of all angles must be 180°

So, ∠LAM + ∠AML + 90 = 180

∠LAM + ∠AML = 90

∠LAM = 90 -∠AML

Since AH || EC

So, ∠GAE = ∠AEC = 30° {alternate angle}

Also ∠BAG = 100°-∠GAE

∠BAG = 70°

Here also, AB || DC and GH acts as transversal.

So, ∠BAG = ∠DHA = 70° {corresponding angles}

Similarly,

AH || EC and DC acts as transversal.

So, ∠DCE = ∠DHA = 70° {corresponding angles}

a³ – b³ + 1 + 3ab

= a³ + (–b)³ + 13 – 3{1 × a × (–b)}

= {a + (–b) + 1} {a² + (–b)² + 12– a(–b) – (–b)1 – 1a}

using identity {a³ + b³ + c³ – 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)} + (a-b + 1)( a² + b² + 1 + ab + b – a)

Now, x² = (2 + √3)² = 4 + 3 + 4√3 = 7 + 4√3

Also, x³ = x × x² = (2 + √3)(7 + 4√3)

= 2(7) + 7(√3) + 2(4√3) + (√3)(4√3)

= 14 + 15√3 + 12

= 26 + 15√3

Put all the values in the expression: x³ - 2x² - 7x + 5

= (26 + 15√3)-2(7 + 4√3)-7(2 + √3) + 5

= 3

rationalize-

= √2-1 + √3-√2 + …√8-√7 + √9-√8

= √9-1

= 3-1

= 2

= 4b²x² + a² - 4abx = a² - 4b²

= 4b²x² - 4abx + 4b² = 0 {rearranging terms and cancelling a²}

Dividing the expression by 4b - bx² - ax + b = 0

If (x - 2) is a factor of the polynomial (x³ + mx² – x + 6) then it must satisfy it.

So, putting x = 2 the polynomial must be zero.

Putting x = 2 and equating to zero.

= (23 + m2² –2 + 6)

= 4m + 12 = 0

= m = -3

If we divide f(x) = (x³ + mx² –x + 6) by (x - 3) remainder can be find at value of –

(x - 3) = 0

Or x = 3

So we will put x = 3 in f(x) = (x³ + mx² – x + 6)

f(3) = (3³ + m3² – 3 + 6)

= 30 + 9m

So remainder = 30 + 9m

= 30 + 9(-3) = 30 - 27 = 3

So, r = 3.

If we divide f(x) = (x³ + 2x² - 5ax - 7) by (x + 1) remainder can be find at value of –

(x + 1) = 0

Or x = -1

So, we will put x = -1 in f(x) = (x³ + 2x² - 5ax - 7)

f(-1) = ((-1)³ + 2(-1)² - 5a(-1)-7)

= -6 + 5a

So, remainder = r = -6 + 5a

Also if we divide f(x) = (x³ + ax² - 12x + 6) by (x - 2) remainder can be find at value of –

(x - 2) = 0

Or x = 2

So we will put x = 2 in f(x) = (x³ + ax² - 12x + 6)

f(2) = (2³ + a2² - 12(2) + 6)

= 4a - 10

So, remainder = s = 4a - 10

Also it is given that 2r + s = 6

So putting r and s from above expressions-

2(-6 + 5a) + (4a - 10) = 6

= 14a = 28

= a = 2

We know that –

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² – ab – bc - ca).

So applying the theorem here,

(a + b)³ + (b + c)³ + (c + a)³ - 3(a + b)(b + c)(c + a) = ((a + b) + (b + c) + (c + a))((a + b)² + (b + c)² + (c + a)² - (a + b)(b + c) - (b + c)(c + a)-(c + a)(a + b))

= (2(a + b + c))((a + b)² + (b + c)² + (c + a)² - (a + b)(b + c)-(b + c)(c + a)-(c + a)(a + b))

{since ((a + b)² + (b + c)² + (c + a)² - (a + b)(b + c) - (b + c)(c + a) - (c + a)(a + b))

= (a² + b² + c² – ab – bc - ca)}

= 2 (a³ + b³ + c³ - 3(a)(b)(c))

{using this theorem again: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)}

It is clear from the graph A and E are mirror image wrt. x-axis and there is no mirror image points wrt. y-axis.

We know that,

Sum of all angles in a triangle = 180°

So, in ΔBEC

= 40 + x + 90 = 180

So, x = 50°

Now, in ΔADC-

= 50 + 30 + ∠ADC = 180

= ∠ADC = 100°

Since BC represents a straight line, sum of angles = 180°.

So, ∠ADC + y = 180

hence y = 80° since ∠ADC = 100°

By exterior angle sum theorem of the smaller triangle formed-

z = ∠DAE + ∠BEA = 90° + 30° = 120°

In ΔBDC ∠DBC = ∠DCB so

BD = DC …(i)

{sides opposite to equal angles in a triangle are equal}

Now let’s consider that ΔABD and ΔADC –

AB = AC {given}

AD is a common side.

And BD = DC {from equation (i)}

Hence ΔABD and ΔADC are congruent.

So ∠BAD = ∠DAC (congruency criteria)

Hence AD bisects ∠BAC.

Since diagonal of square bisects the angles.

So, ∠CBD = ∠CDB = 45° [ Also all angles of square are right angles i.e. half of all is 45°] (1)

Also similarly ∠ABD = ∠ADB = 45°

Since lines EF || BD

By corresponding angles-

∠CEF = ∠CDB = 45°

Also ∠CFE = ∠CBD = 45°

So, CE = CF {since sides opposite to equal angles are equal} …(i)

And CD = BC {sides of a square are equal} …(ii)

Subtracting I from II

CD-CE = BC-CF

So, BF = DE

Also let’s consider ΔADX and ΔABX {where X is intersection point of AM and BD}

∠ABD = ∠ADB = 45°

AX is a common side.

AD = AB {sides of a square are equal}

The triangles are congruent by SAS (side angle side) criteria.

So, ∠DAM = ∠MAB (congruency criteria)

Hence AM bisects ∠BAD.

Draw one line EF || CD and AB.

Since EF || CD and CE is transversal.

∠FEC + ∠ECD = 180°

∠FEC = 60° {since ∠ECD = 120°}

Also, EF || AB and AE is transversal.

∠FEA + ∠BAE = 180°

∠FEA = 80° {since ∠BAE = 100°}

And x = ∠FEC + ∠FEA

= 60° + 80°

= 140°

Irrational numbers are numbers which cannot be expressed as simple fraction or simple ratios of two integers. That leaves us with just two options A and C. So, only √5 comes in between 2 and 2.5.

A polynomial in one variable is an algebraic expression that consists of terms in the form of axⁿ, where n is either zero or positive only. Given the options all expressions except D has the value of n as negative.

Given,

Rationalising the above term,

Using the formula (a + b) (a – b) = a² – b² for the denominator,

Given, p(x) = (x⁴ – x² + x)

Substituting the value of 1/2 in place of will give,

Given, p(x) = x³ + x² + ax + 115

(x³ + x² + ax + 115) is exactly divisible by (x + 5)

Hence, substituting x = -5 will give us the value of a

⇒ (-5)³ + (-5)² + a (-5) + 115 = 0

⇒ -125 + 25 – 5a + 115 = 0

⇒ 5a = 15

⸫ a = 3

We know that, the value of x is always zero on the y-axis.

According to the figure,

⇒ 4x + 5x = 180° [Angle on a straight line]

⇒ 9x = 180°

⸫ x = 20°

Given,

∠BAC = 40°

∠ACB = 65°

According to figure,

⸫ ∠ACE = 40° [Alternate angles]

⸫ ∠ACB + ∠ACE = x° [Alternate angles]

⇒ x° = 65° + 40°

⸫ x = 105°

Given, √2x² + 3x + √2

By splitting the middle term,

⇒ √2 x² + 2x + x + √2

⇒ √2 x(x + √2) + 1(x + √2)

⸫ (x + √2)(√2 x + 1)

Let’s assume that √5 is a rational number.

Hence, √5 can be written in the form a/b [where a and b (b ≠ 0) are co-prime (i.e. no common factor other than 1)]

⸫ √5 = a/b

⇒ √5 b = a

Squaring both sides,

⇒ (√5 b)² = a²

⇒ 5b² = a²

⇒ a²/5 = b²

Hence, 5 divides a²

By theorem, if p is a prime number and p divides a², then p divides a, where a is a positive number

So, 5 divides a too

Hence, we can say a/5 = c where, c is some integer

So, a = 5c

Now we know that,

5b² = a²

Putting a = 5c,

⇒ 5b² = (5c)²

⇒ 5b² = 25c²

⇒ b² = 5c²

⸫ b²/5 = c²

Hence, 5 divides b²

By theorem, if p is a prime number and p divides a², then p divides a, where a is a positive number

So, 5 divides b too

By earlier deductions, 5 divides both a and b

Hence, 5 is a factor of a and b

⸫ a and b are not co-prime.

Hence, the assumption is wrong.

⸫ By contradiction,

⸫ √5 is irrational

Given, x = (3 + √8)

Let us calculate 1/x,

Rationalising the above term,

Using the formula (a + b) (a – b) = (a² – b²),

Now,

On squaring both sides, we get

Given, three sides of a triangle 52 cm, 56 cm, 60cm

Area of a triangle is given by,

where,

and a, b, c are the sides of the triangle

⸫ Area of triangle =

Lets draw another line XY || AB and CD.

According to the figure,

⇒ ∠a = 40° [Alternate angles]

⇒ ∠b = 35° [Alternate angles]

⸫ ∠x + ∠a + ∠b = 360° [Angle at a point = 360°]

⸫ ∠x = 360° - 40° - 35° = 285°

Given, x⁴ + ax³ – 7x² - 8x + b = 0

⸫ x = -2, -3 are a root of the above equation (⸪ they are exactly divisible)

Substituting the value -2 and -3 in place of x will give,

⇒ (-2)⁴ + a (-2)³– 7(-2)² - 8(-2) + b = 0

⇒ 16 – 8a – 28 + 16 + b = 0

⸫ 8a – b = 4 …. (i)

⇒ (-3)⁴ + a (-3)³– 7(-3)² - 8(-3) + b = 0

⇒ 81 – 27a – 63 + 24 + b = 0

⸫ 27a – b = 42 …. (ii)

Simultaneously solving eq(i) and eq(ii) we get,

⸫ a = 2

⸫ b = 12

Given, p(x) = x³ – 3x² + 4x + 50

Divisor, (x + 3)

⸫ x = -3

Substituting -3 in place of x gives us,

⇒ (-3)³– 3(-3)² + 4(-3) + 50

= -27 – 27 – 12 + 50 = -16

Given, (2x³ + 54)

Taking common terms out,

⇒ 2 (x³ + 27)

Using the formula, (a³ + b³) = (a + b) (a² – ab + b²)

⇒ 2 (x + 3) (x²– 3x + 3²)

⸫ 2 (x + 3) (x² – 3x + 9)

Given, (a – b – c) (a2 + b² + c² + ab + ac – bc)

= a³ + ab² + ac² + a²b + a²c – abc – a²b – b³ – bc² – ab² – abc + b²c – a²c – b²c – c³ – abc – ac² – bc²

Cancelling the terms with opposite signs,

= a³ – b³ – c³ – 3 abc

Let the three angles of a triangle be ∠A, ∠B, ∠C

Given, ∠A - ∠B = 33°

⇒ ∠A = ∠B + 33°

∠B - ∠C = 18°

⇒ ∠C = ∠B – 18°

Now,

∠A + ∠B + ∠C = 180° [Sum of all angles of a triangle = 180°]

⇒ ∠B + 33° + ∠B + ∠B – 18° = 180°

⇒ 3∠B = 180° - 15°

⸫ ∠B = 55°

⸫ ∠A = ∠B + 33° = 88°

⸫ ∠C = ∠B – 18° = 37°

Given, ∠A = 70°

Let the two angles ∠B = 2x and ∠C = 2y.

Then, angle bisector of B, ∠OBC = x and angle bisector of C, ∠OCB = y

⸫ ∠A + ∠B + ∠C = 180° [Sum of all angles of a triangle = 180°]

⇒ 70° + 2x + 2y = 180°

⇒ 2x + 2y = 110°

⸫ x + y = 55° …. (i)

Now,

∠BOC + x + y = 180° [Sum of all angles of a triangle = 180°]

⇒ ∠BOC = 180° - (x + y)

⇒ ∠BOC = 180° - 55° [from eq. (i)]

⸫ ∠BOC = 125°

Given, ∠BAO = 110°, ∠AOC = 20°

∠CEO = 110° [Corresponding angles]

⸫ x° = 110° + 20° [Exterior angle = Sum of two opposite interior angles]

⸫ x° = 130°

Given, ΔABC is a right- angled triangle at B i.e. ∠B = 90°

To prove AC is the longest side of ΔABC

Proof:

In ΔABC,

∠A + ∠B + ∠C = 180° [Sum of all angles of a triangle = 180°]

∠A + 90° + ∠C = 180° [Given ∠B = 90°]

∠A + ∠C = 180° - 90°

⸫ ∠A + ∠C = 90°

Hence, ∠A < 90°

∠A < ∠B

BC < AC [Side opposite to a larger angle is longer]

Similarly,

∠C < 90°

∠C < ∠B

AB < AC [Side opposite to a larger angle is longer]

Hence,

⸫ AC is the longest side of ΔABC i.e. the hypotenuse.

In ΔABC,

∠A + ∠B + ∠C = 180° [Sum of all angles of a triangle = 180°]

According to the figure,

⇒ ∠B + (α + ∠DAC) + (γ + ∠DCA) = 180°

⇒ ∠DAC + ∠DCA + α + β + γ = 180°

⇒ ∠DAC + ∠DCA = 180° - (α + β + γ) …. (i)

In ΔADC,

⇒ x + ∠DAC + ∠DCA = 180° [Sum of all angles of a triangle = 180°]

⇒ x = 180° - ∠DAC - ∠DCA

⇒ x = 180° - 180° + (α + β + γ)

⸫ x = (α + β + γ)

Hence proved.

Since, we want six numbers, we write 1 and 2 as rational numbers with denominator 6 + 1 = 7

So, multiply in numerator and denominator by 7, we get

and

We know that, 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28

Hence, six rational numbers between are

Given,

Rationalising the above term,

Using the formula (a + b) (a - b) = (a² – b²)

⸫ 4 + √15

Comparing with a + √15 b,

⸫ a = 4, b = 1

OR

Solution: Given, (5a – 7b)³ + (9c – 5a)³ + (7b – 9c)³

Using the formula, (a + b + c)³ = a³ + b³ + c³ + 3(a + b) (b + c) (c + a)

⇒ a³ + b³ + c³ = (a + b + c)³ - 3(a + b) (b + c) (c + a)

⇒ (5a – 7b)³ + (9c – 5a)³ + (7b – 9c)³ = (5a - 7b + 9c – 5a + 7b – 9c)³ – 3(5a – 7b + 9c – 5a) (9c – 5a + 7b – 9c) (7b - 9c + 5a – 7b)

⇒ (5a – 7b)³ + (9c – 5a)³ + (7b – 9c)³ = 0³– 3(-7b + 9c) (-5a + 7b) (-9c + 5a)

⸫ (5a – 7b)³ + (9c – 5a)³ + (7b – 9c)³ = 3(5a – 7b) (7b – 9c) (9c – 5a)

Given, 12(x² + 7x)² – 8(x² + 7x) (2x – 1) – 15(2x – 1)²

By splitting the middle term i.e. 8(x² + 7x) (2x – 1), we get

= 12(x² + 7x)² – 18(x² + 7x) (2x – 1) + 10(x² + 7x) (2x – 1) – 15(2x – 1)²

= 6(x² + 7x) [2(x² + 7x) – 3(2x – 1)] + 5(2x - 1) [2(x² + 7x) – 3(2x – 1)]

= [2(x² + 7x) – 3(2x – 1)] [6(x² + 7x) + 5(2x – 1)]

= (2x² + 14x – 6x + 3) (6x² + 42x + 10x – 5)

= (2x² + 8x + 3) (6x² + 52x - 5)

Given, (x³ + ax² + bx + 6) exactly divisible by (x – 2)

⸫ x = 2 is a root of the above equation.

⇒ 2³ + a (2)² + b (2) + 6 = 0

⇒ 8 + 4a + 2b + 6 = 0

⸫ 4a + 2b = -14

Given, (x³ + ax² + bx + 6) divided by (x – 3) leaves a remainder 3

⸫ 3³ + a (3)² + b (3) + 6 = 3

⇒ 27 + 9a + 3b + 6 = 3

⸫ 9a + 3b = -30 …. (ii)

Put value of b from (i) in this equation to get,



18a - 42 - 12 a = - 60

6a - 42 = -60

6a = -60 + 42

6a = -18

a = -3

Put the value of a in (i) to get:

Solving simultaneously eq (i) and eq (ii), we get

a = -3, b = -1

Let’s find the roots of the equation (x² + 2x - 3)

⇒ x² + 3x – x – 3 = 0

⇒ x(x + 3) – 1(x + 3) = 0

⸫ (x + 3) (x – 1)

Hence, if (x + 3) and (x – 1) satisfies the equation x³ – 3x² – 13x + 15 = 0, then (x³ – 3x² – 13x + 15) will be exactly divisible by (x² + 2x - 3).

For x = -3,

⇒ (-3)³– 3(-3)²– 13(-3) + 15

⇒ -27 – 27 + 39 + 15 = 0

For x = 1,

⇒ 13 – 3(1)²– 13(1) + 15

⇒ 1 – 3 – 13 + 15 = 0

Hence proved.

Given, a³ – b³ + 1 + 3ab

⇒ a³ + (-b)³ + 1³– 3(1 * a * (-b))

⇒ [a + (-b) + 1] [a² + (-b)² + 1² – a(-b) – (-b)1 – 1a]

⸫ (a – b + 1) (a² + b² + 1 + ab + b – a)

Given, ∠ECD = 100°, ∠EAB = 50°

∠COB = 100°

⸫ x = 100° - 50° [Exterior angle = Sum of two opposite interior angles of a triangle]

⸫ x = 50°

Given, ∠ACD and ∠BCD are linear pairs

CE and CF bisect ∠ACD and ∠BCD respectively

To prove:

∠ECF = 90°

⸫ ∠ACD + ∠BCD = 180° [Angle on a straight line]

⇒ ∠ACD/2 + ∠BCD/2 = 180°/2 = 90°

⇒ ∠ECD + ∠DCF = 90° [⸪ CE and CF bisect ∠ACD and ∠BCD respectively]

⸫ ∠ECD + ∠DCF = ∠ECF = 90°

Hence Proved.

Let the ratio be y

⸫ ∠DAB = y

⸫ ∠DAC = 3y

⸫ y + 3y + 108° = 180° [Angle on a straight line]

⇒ 4y = 72°

⸫ y = 18°

⸫ ∠DAC = 3y = 54°

∠ABD = 18° [⸪ AD = DB, ΔABD is an isosceles triangle]

In ΔABC,

⇒ x + ∠A + ∠B = 180° [Sum of all angles of a triangle = 180°]

⇒ x = 180° - 72° - 18°

⸫ x = 90°

In ΔABC,

∠A = 180° - 70° - 20° [Sum of all angles of a triangle = 180°]

⸫ ∠A = 90°

⸫ ∠BAN = 45° [⸪ AN is the bisector of ∠A]

In ΔABN,

∠N = 180° - 70° - 45° [Sum of all angles of a triangle = 180°]

⸫ ∠N = 65°

In ΔAMN,

∠MAN = 180° - 90° - 65° [Sum of all angles of a triangle = 180°]

⸫ ∠MAN = 25°

Given,

In ΔPQR,

PS bisects ∠QPR and QS = SR

To prove:

PQ = PR

In ΔPQS and ΔPRS

QS = SR [Given]

∠QPS = ∠RPS [Given]

PS = PS [Common]

ΔPQS is congruent to ΔPRS [S.A.S]

⸫ PQ = PR [C.P.C.T.C]

Hence Proved.