The document Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.

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**SECTION -A**

**Questions 1 to 20 carry 1 mark each.**

**Q.1. The value of ****(a) 9(c) 3(b) 6(d) none of theseAns:** Choice (c) is correct.

(a) 0

(b) 1

(c) 4√2

Or

If (x-2) is factor of p(x) = x

(a) 0

(b) 7/6

(c) -7/6

(d) None of these

Ans:

Explanation :

On putting x = 2-√2 in given equation.

Or

Correct option: (b)

Explanation : ∴ (x - 2) is a factor of p(x)

⇒ By Factor Theorem, p(2) = 0

⇒ (2)

⇒ 8- 8a + 2a - 1 = 0

⇒ - 6a = -7

⇒ a=7/6

(a) 1

(b) 2

(c) 3

(d) 4

Ans:

(a) 5 cm

(b) 6 cm

(d) 7 cm

Ans:

As seen, ODB is a right angled triangle and by property, we know that, perpendicular from the centre to the chord bisects the chord in equal part.

(a) 0.8

(b) 0.04

(c) 0.08

(d) 0.6

Ans:

[Using Demorgan’s law of sets]

= 1 - P(A ∩ B) = 1 - 0.4 = 0.6

(a) 90°

(b) 95°

(c) 105°

(d) 120°

Ans:

Explanation: We know that the sum of angles in a quadrilateral = 360°

Therefore,

Fourth angle of the quadrilateral = 360° - (75° + 90° + 75°)

= 360° - 240°

= 120°

If ar (BDEF) = x ar (ΔAFE), then the value of x is :

(b) 1

(c) 2

(d) 4

(a) 22 cm

(b) 33 cm

(c) 44 cm

(d) 55 cm

Ans:

= 44 cm

(a) Suresh does not live in Bhopal or he does not live in Mumbai.

(b) Suresh does not live in Bhopal and he does not live in Mumbai.

(c) Suresh lives in Bhopal and he lives in Mumbai.

(d) Suresh neither lives in Bhopal nor lives in Mumbai.

Ans:

The negation of the given statement is Suresh does not live in Bhopal and he does not live in Mumbai.

(a) 17 cm

(b) 15 cm

(c) 4 cm

(d) 8 cm

Ans:

Explanation :

Draw OP ⊥ AB

We know that the perpendicular from the centre of the circle bisects the chord.

In right ΔOPA, we have

**Questions 11 to 15 carry one mark each.**

**Q.11. 325 x 325 - 25 x 25 = .............****Ans:** 105000

[∵ (325)^{2} - (25)^{2} = (325 - 25) (325 + 25) = (300) (350) = 105000]**Q.12. Two circles of the same radii are.........****Ans:** Two circles of the same radii are congruent,**Q.13. 1 ^{2} + 2^{2} + 3^{2} + ........... + n^{2} = ________.**

Ans:

is the sum of n terms of the special series 1

Ans:

**Questions 16 to 20 carry one mark each**

**Q.16. If (x-3) is one factor of x ^{2} + ax -4 = 0 and then find the value of a.OrIf x = k^{2} and y = k is a solution of the equation x - 7y + 12 =0, find the value of k.Ans: **(x-3) is a factor of x

x

So, x=3, put in polynomials.

∴ (3)2 + a x 3-4=0

⇒ 9+3a-4=0

⇒ 3a+5=0

⇒

Or

Since, x = k

∴ k

⇒ k2-4k-3k+12 = 0

⇒ k(k-4)-3(k-4) = 0

⇒ (k-4) (k-3) = 0

⇒ k-4 = 0 or k-3 = 0

∴ k=3, 4

Ans:

Ans:

∴ 2x° + 3x° + 6x° + 7x° = 360°

[Angle sum property of quadrilateral]

or, 18x° = 360°

or, x°= 20°

∴ Largest angle - 7x = 7 x 20° = 140°

Since (x + 2) divides p(x) exactly

∴ P(-2) = 0

⇒ 2(-2)

⇒ -16 - 4k - 6 + 10 - 0

⇒ - 4k - 12 = 0

⇒

⇒

Volume of hemisphere,

**SECTION - B**

**Questions 21 to 26 carry two marks each**

**Q.21. Express in the form a + ib :Ans:**

Ans:

and BX = BY ... (ii)

Subtract (ii) from (i)

⇒ AB - BX = BC - BY

Now, by Euclid's axiom 3, we have If equals are subtracted from equals, the remainders are equal.

Hence, AX = CY

Or

In the given figure, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4, write the relation between ∠1 and ∠2, using an Euclid's axiom.

Ans:

and ∠2 - ∠4 ...(ii)

Adding (i) and (ii), we have

∠1 + ∠2 - ∠3 + ∠4

=> ∠A = ∠C

[∴ Euclid’ s second axiom says, if equals be added to the equals, the wholes are equal]

Or

Here, ∠3 - ∠4 , ∠1 = ∠3 and ∠2 = ∠4. Euclid’ s first axiom says, the things which are equal to equal thing are equal to one another. So ∠l = ∠2.

Here, line l is parallel to m and q is a transversal line. While drawing this figure, the teacher have no scale for measuring this length, but they know the side which is opposite to the smallest angle, is smaller and the side which is opposite to the largest angle, is larger. In this game, the teacher invite the two students Vicky and Vishal and said to them that especially Vicky stands on point A and Vishal stands on point B, respectively (assume that both have same space of walking).

Ans:

∠ABD + ∠DBE = 180° [by linear pair axiom]

⇒ ∠ABD + 110° =180° [∴ ∠DBE = 110°]

⇒ ∠ABD = 180°-110° =70° ...(i)

In AABD,

∠ABD + ∠DAB + ∠BDA = 180°

⇒ θ

θ

As I||m, so sum of the two interior angle is 180°

∴ θ

⇒ θ

⇒ θ

As I||m, ∠FDA = ∠DAB [alternate interior angles]

⇒ θ

Also, ∠HDC = ∠FDA [vertically opposite angles]

⇒ θ

Hence, angles are θ =80°, θ

Or

Let A be the set of first twelve natural numbers and let R be a relation on A defined by (x,y) ∈ R ⇔ x + 2y = 12, i.e, R = {(x, y) : x ∈ A ,y ∈ A and x + 2y = 12}.

Express R and R

Ans:

where A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

This shows that 1 is not related to any element of A. Similarly, we can see that 3,5,7,9,11 and 12 are not related to any element of A under the defined relation.

Thus R = {(2, 5), (4, 4), (6,3), (8,2), (10,1)}

⇒ R

Clearly, Domain (R) = {2,4,6,8,10} = Range (R-1) and, Range (R) = {5,4,3,2,1} = Domain (R

Ans:

**SECTION - C**

**Questions 27 to 34 carry 3 marks each**

**Q.27. Prove by using the principle of mathematical induction for all n ∈ N that:Ans:** Let P(n) be the given statement, i.e.,

L.H.S. of P(1)= 1

Thus, P(n) is true for n = 1.

Ans:

So the coordinate of the point will be

Therefore, is a solution of the equation

2x + 5y = 19.

So, Abscissa of the point = 2

So, Co-ordinate of the point is (2,3)

Or

Two circles intersect at two points B and C, see given figure. Through B, two line-segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively. Prove that : ∠ACP = ∠QCD

Ans:

Therefore, CD > AB. In other words, AB is the smallest of all chords passing through P

Or

Since angles in the same segments of a circle are equal

∴ ∠ACP = ∠ABP ...(i)

Similarly, ∠QCD = ∠QBD ...(ii)

Also, ∠ABP = ∠QBD ...(iii) [vert. opp. ∠s]

From (i), (ii)a nd (iii), we obtain

∠ACP = ∠QCD

Ans:

Let 1cm = 1 unit.

Mark a points on Y-axis at a 6 units distance in positive direction of Y-axis and mark a point C on X-axis at a 3 units distance in negative direction of X-axis. Now, draw a perpendicular line from C and A, which meets at a point B.

Thus, we get the rectangle OABC whose vertices are O(0,0), A(0, 6), B(-3, 6) and C(-3, 0).

∴ Area of rectangle = Length x Breadth

∴ Area of rectangle OABC = 6 x 3 = 18 sq units

(i) Write this relation in Roster form.

(ii) Draw arrow diagram for the above relation.

(iii) What is its domain, range and codomain ?

Ans:

R = {(9, 3), (9, -3), (4, 2), (4, -2), (25, 5), (25, -5)}

(ii) Relation R = {(9, 3), (9, -3), (4, 2), (4, -2), (25, 5), (25, -5)} from set P = {9, 4, 25} to set Q = {1,2, 3, 5, -2, -3, -5} can be represented by the following arrow diagram :

(iii) The domain of R is {9, 4, 25}

The range of R is {-5, -3, - 2, 2, 3, 5}

The co-domain of if is {1, 2, 3, 5, -2, -3, -5}

OR

Prove that the quadrilateral formed by the internal angle bisectors of any quadrilateral is cyclic.

Ans:

To Prove : PQ ⊥ AB and AC = BC

Construction : Join AQ, BQ, AP and BP

Proof: ΔAQP and ΔBPQ

PQ = PQ (common side)

AQ=BQ (radius of circle)

AP=BP (radius of circle)

In ΔAQC and ΔBQC

∠AQP = ∠BQP (proved)

AQ = BQ (radius of circle)

QC = QC (common side)

Hence, PQ bisect the common chord AB

∴ AC - BC (c.p.c.t.).. .(i)

[∴ Hence common chord bisect by the line segment PQ.]

Also, ∠ACQ+∠BCQ =180° (Linear pair)

2∠ACQ - 180° ∠ACQ = 90°

∴ PQ⊥AB ...(ii)

Thus line through their centre is the perpendicular bisector of the common chord.

Let ABCD be a quadrilateral in which the angle bisectors AH, BF, CF and DH of internal angles of A,B,C and D respectively form a quadrilateral EFGH

To prove : EFGH is a cyclic quadrilateral i.e

[∴ Angles A and B are bisector angles] Similarly ∠FGH = ∠CGD

[∴ Angles C and D are bisector angles]

On adding Eqs (i) and (ii), we get

It implies that sum of opposite angles of a quadrilateral is 180°

Hence, it shows that quadrilateral EFGH is cyclic quadrilateral.

Or

If 2x

Ans:

Clearly, constant term is 6

The factors of 6 are ±1, ±2, ±3, and ±6

By Trial method, we find that

f(-1) = (-1)

= -1+6-11+6=0

∴ x + 1 is a factor of f(x)

Now, divide f(x) by x + 1, we have

∴ f(x) = (x+1) (x2+5x+6)

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

= (x+1) [x(x+2) + 3(x+2)]

= (x+1) (x+2) (x+3)

Or

Consider f(x) = 2x

As (x-1) is a factor of f(x). So by Factor Theorem,

f(1) = 0

∴ ⇒ f(1) = 2(1)

= 2+a+b-6 = 0

⇒ a+b-4= 0 ...(i)

Also, it leaves remainder ‘2’ when divided by (x - 2). So by Remainder Theorem, f(2) = 2.

∴ f(2) = 2(2)

⇒ 2 x 8+a x 4+b x 2-6=2

⇒ 16 + 4a + 2b - 6 =2

⇒ 4a+2b+8 =0

⇒ 2a + b + 4 =0 ...(ii)

From (i) and (ii) we have

2a - a + 8 =0

⇒ a=-8

and -8 + b - 4 =0

⇒ b=12

Hence, the values of a and b are - 8 and 12.

Ans.

Then, R = 12.5cm and r = 12 cm

Now,

Area of outer surface = 2πR

Area of the inner surface = 2πr

Area of the ring at the top = πR

Total area to be painted = (2πR

∴ Cost of painting = Rs (1925.78 x 0.05)=Rs 96.28

**SECTION - D**

**Questions 35 to 40 carry 4 marks each**

**Q.35.**** In a survey of 25 students, it was found that 15 had taken Mathematics, 12 had taken Physics and 11 had taken Chemistry, 5 had taken Mathematics and Chemistry, 9 had taken Mathematics and Physics, 4 had taken Physics and Chemistry and 3 had taken all the subjects. Find the number of students that had taken(i) only Chemistry(ii) only Mathematics(iii) only Physics(iv) Physics and Chemistry but not Mathematics(v) Mathematics and Physics but not Chemistry(vi) only one of the subjects.**

(i) Number of students who had taken only Chemistry

(ii) Number of students who had taken only Mathematics

(iii) Number of students who had taken only Physics

(iv) Number of students who had taken Physics and Chemistry but not Mathematics

(v) Number of students who had taken Mathematics and Physics but not Chemistry

(vi) Number of students who had taken only one of the subjects

or

If x =1

Then p(1) = 6(1)

= 6 - 5 - 1 3 + 12

= 18 -18 = 0

∴ (x - 1) is a factor of given polynomial p(x) 1

6x

= (x -1) {(6x

= (x-l){3x(2x + 3)-4(2x + 3)}

= (x -1) (2x + 3) (3x - 4)

∴ 6x

Or

**Q.37. In the given figure, ABCD is a square. ΔDEC is an equilateral triangle. Prove that : (ii) AE = BEΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see fig.). If AD is extended to intersect BC at Pf show that :(iii) AP bisects ∠A as well as ∠D(iv) AP is the perpendicular bisector of BC.Ans:** (i) Here, ABCD is a square and ΔDEC is an equilateral triangle. In square ABCD, AB = BC = CD = AD and each angle is 90°. In equilateral triangle DEC, DE = DC = EC, each angle is 60°

Consider ΔADE and ΔBCE, we have

[by c.p.c.t.]

Or

Given : ΔABC and ΔDBC are isosceles triangles having common base BC, such that

AB = AC and DB = DC.

To Prove :(i) ΔABD ≅ ΔACD

(ii) ΔABP ≅ ΔACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC.

Proof : (i) Consider ΔABD and ΔACD, we have

AB = AC [given]

BD - CD [given]

AD - AD [common]

∴ By SSS congruence axiom, we have

In ΔABP and ΔACR we have

AB = AC [given]

∠1 = ∠2 [already proved above]

and AP = AP [common]

∴ By SAS congruence axiom, we have

∴ ΔABP ≅ ΔACP

(iii) ∠1 = ∠2

⇒ AP bisects ∠A ...(i)

⇒ AP bisects ∠D. [by c.p.c.t.]

(iv) As BP = CP. [by c.p.c.t.]

and ∠3 = ∠4

But ∠3 + ∠4 = 180° [ ∴ linear pair]

Hence, AP is the perpendicular bisector of BC.

Then, the amount contributed by Swati be (x)

According to the question,

x(x)

⇒ x

⇒ x

Now

Hence, Sapna's contribution is Rs 3125 and Swati’s contribution is Rs 125.

Let T

This term, will be independent of x, if the power of x is zero,

i.e., 12 - 3r = 0

⇒ r = 4

Substituting r = 4 in (1), we get

Hence, 5th term is independent of x and is given by 5/12.

Or

n

Ans:

∴ The coefficients of a

It is given that:

∴ C

⇒

⇒

Ans:

Here minimum class size=10-0=10

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