Class 9

MCQs of Chapter - 2 Polynomials

1.) If x²+kx+6=(x+2)(x+3) for all x, then the value of k is

(A) 1

(B) –1

(D) 3

Ans : (C) 5

2.) Which one of the following is a polynomial?

Ans : (C)

3.) √2 is a polynomial of degree

(A) 2
(B) 0
(C) 1
(D) 1/2

Ans : (B) 0

4.) Degree of the polynomial 4x⁴+0x³+0x⁵+5x+7 is
(A) 4
(B) 5
(C) 3
(D) 7

Ans : (A) 4

5.) Degree of the zero polynomial is

(A) 0
(B) 1
(C) Any natural number

(D) Not defined

Ans : (D) Not defined

6.) If p(x)=x²-2√2x+1, then p(2√2) is equal to

(A) 0

(B) 1

(D) 8√2+1

Ans : (B) 1

7.) The value of the polynomial 5x–4x²+3, when x=–1 is

(A)– 6

(B) 6

(D) –2

Ans : (A) – 6

8.) If p(x)=x+3, then p(x)+ p(–x) is equal to

(A) 3

(B) 2x

(D) 6

Ans : (D) 6

9. Zero of the zero polynomial is

(A) 0

(B) 1

(D) Not defined

Ans : (C) Any real number

10. Zero of the polynomial p(x)=2x+5 is

(A) -2/5

(B) -5/2

(D) 5/2

Ans : (B) -5/2

11. One of the zeroes of the polynomial 2x²+7x–4 is

(A) 2

(B) 1/2

(D) -2

Ans : (B) 1/2

12. If x^51+51 is divided by x+1, the remainder is

(A) 0

(B) 1

(D) 50

Ans : (D) 50

13. If x+1 is a factor of the polynomial 2x²+kx, then the value of k is

(A) –3

(B) 4

(D) –2

Ans : (C) 2

14. x+1 is a factor of the polynomial

(A) x³+x²–x+1

(B) x³+x²+x+1

(D) x⁴+3x3+3x2+x+1

Ans : (B) x³+x²+x+1

15. One of the factors of (25x²–1)+(1+5x)² is

(A) 5+x

(B) 5–x

(D) 10x

Ans : (D) 10x

16. The value of 249²–248² is

(A) 12

(B) 477

(D) 497

Ans : (D) 497

17. The factorisation of 4x²+8x+3 is

(A) (x+1)(x+3)

(B) (2x+1)(2x+3)

(D) (2x–1)(2x–3)

Ans : (B) (2x+1)(2x+3)

18. Which of the following is a factor of (x+y)³-(x³+y³)?

(A) x²+y²+2xy

(B) x²+y²-xy

(D) 3xy

Ans : (D) 3xy

19. The coefficient of x in the expansion of (x+3)³ is

(A) 1

(B) 9

(D) 27

Ans : (D) 27

20.

(A) 1

(B) -1

(D) 1/2

Ans : (C) 0

21.

(A) 0

(B) 1/√2

(D) 1/2

Ans : (C) 1/4

22. If a+b+c=0, then a³+b³+c³ is equal to

(A) 0

(B) abc

(D) 2abc

Ans : (C) 3abc

If x=root(a+2b) + root(a-2b) / root(a+2b) - root(a-2b), prove that bx²-ax+b=0

Question : If x={√[a+2b] + √[a-2b]} / {√[a+2b] - √[a-2b]}, prove that bx²-ax+b=0

Doubt by Veer

Solution :

Rationalising the denominator

2bx-a=√(a²-4b²)

SBS

(2bx-a)²=a²-4b²

4b²x²+a²-4abx=a²-4b²

4b²x²-4abx=-4b²

4b²x+4b²-4abx=0

4b(bx²+b-ax)=0

4b(bx²-ax+b)=0

4b=0 OR bx²-ax+b=0

b=0/4 OR bx²-ax+b=0

b=0 OR bx²-ax+b=0

but b≠0

Hence bx²-ax+b=0.

The area of the isosceles triangle is 5√11/4 cm² if the . . .

Question : Write true or false for the given statement.

"The area of the isosceles triangle is 5√11/4 cm² if the perimeter is 11 cm and the base is 5 cm."

Doubt by Purvi

Solution :

Let the equal sides of the isosceles triangle be
a and base be b. So the three sides of the isosceles triangle would be a, a and b.

b=5 cm (Given)
Perimeter (P) = 11 cm
a+a+b=11

2a+5=11
2a=11-5
2a=6
a=6/2
a=3 cm

So the three sides of the isosceles triangle would be a, a and b where a=3 cm and b=5 cm

Now,
Area of isoseceles triangle is given by

Area of isoseceles triangle is 5√11/4 cm². Hence, the above statement is absolutely correct.

If x3+ax3+10-bx is divisible by (x-2) . . .

Question : If x³+ax²+10-bx is divisible by (x-2) and (x-1) then find the values of a and b.

Doubt by Purvi

Solution :

p(x)=x³+ax²+10-bx

p(x)=x³+ax²-bx+10

Case I

g(x)=(x-2)
p(x) is divisible by g(x), it means g(x) must be a factor of p(x).

g(x)=0
x-2=0
x=2

p(2)=0
(2)³+a(2)²-b(2)+10=0
8+4a-2b+10=0
4a-2b+18=0
4a-2b=-18
2(2a-b)=-18
2a-b=-18/2
2a-b=-9 — (1)

Case II

g(x)=(x-1)
p(x) is divisible by g(x), it means g(x) must be a factor of p(x).

g(x)=0
x-1=0
x=1

p(1)=0
(1)³+a(1)²-b(1)+10=0
1+a-b+10=0
a-b+11=0
a-b=-11 — (2)

Subtracting equation (2) from equation (1)

2a-b-(a-b)=-9-(-11)
2a-b-a+b=-9+11
a+0=2
a=2
putting this value of a in equation (1)

2(2)-b=-9
4-b=-9
4+9=b
b=13

Hence, a=2 and b=13

In Fig, if AB || CD || EF, PQ || RS, ∠RQD = 25 . . .

Question : In Fig, if AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then ∠QRS is equal to

(A) 85°
(B) 135°
(C) 145°
(D) 110°

Doubt by veer

Solution :

AB || CD || EF (Given)

∠RQD = 25° and ∠CQP = 60° (Given)

∠QRS=?

∠ARQ=∠RQD =25° [Alternate Interior Angles]
∠ARQ=25° — (1)

∠CQP=∠SRB=60° [Alternate Exterior Angles]
∠SRB=60°

Now,
∠SRB + ∠SRA = 180° [Linear Pair]

60° + ∠SRA = 180°

∠SRA=180°-60°
∠SRA=120° — (2)

Adding equation (1) and (2)

∠ARQ+∠SRA=25°+120°
∠QRS = 145°
Hence, (C) 145°, would be the correct option.

The base of an isosceles triangle is of length 48 cm and its. . .

Question : The base of an isosceles triangle is of length 48 cm and its area is 240 cm². Use Heron's formula and hence find the perimeter of the triangle.

Doubt by Veer

Solution :

Here
a=48 cm

b=x cm

c=x cm

Semiperimeter (s)

s=[a+b+c]/2
s=[48+x+x]/2
s=[48+2x]/2
s=2[24+x]/2
s=[24+x] cm

Area = 240 cm² [Given]

√[s(s-a)(s-b)(s-c)] = 240
√[24+x(24+x-48)(24+x-x)(24+x-x)] = 240
√[24+x(x-24)(24)(24)] = 240
Squaring both sides
[(x+24)(x-24)24×24]=240×240
x²-(24)²=(240×240)/(24×24)
x²-576=100
x²=100+576
x²=676
x=√676
x=26

b=c=26 cm

So,
Perimeter of triangle
P=a+b+c
P=48+26+26
P=48+52
P=100 cm²

If x=1/7-4root3, show that x2-14x+1=0 and hence evaluate . . .

Question : If x=1/(7-4√3), show that x²-14x+1=0 and hence evaluate x³-9x²-69x+7.

Doubt by Purvi

Solution :

x=1/(7-4√3)
Rationising the denominator

x=1/(7-4√3)×(7+4√3)/(7+4√3)
x=(7+4√3)/[(7)²-(4√3)²]
x=(7+4√3)/(49-48)
x=(7+4√3)/1
x=7+4√3

x²-14x+1=0

LHS

x²-14x+1
=[7+4√3]²-14[7+4√3]+1
=(7)²+(4√3)²+2(7)(4√3)-14(7)-14(4)√3+1
=49+48+56√3-98-56√3+1
=50+48-98
=98-98
=0
= RHS

LHS = RHS

Hence Proved.

Now

x²-14x+1=0
x²=14x-1— (1)

Also
x³=x².x
x³=(14x-1)x [∵ Using Equation (1)]

x³=14x²-x — (2)

Now

x³-9x²-69x+7
=14x²-x-9(14x-1)-69x+7
[Using Equation (1) and (2)]

=14x²-x-126x+9-69x+7
=14x²-196x+16
=14x²-196x+14+2
=[14x²-196x+14]+2
=14[x²-14x+1]+2
=14[0]+2 [∵x²-14x+1=0]
=0+2
=2

Hence, x³-9x²-69x+7=2

If one angle of a parallelogram is 24° less than twice the . . .

Question : If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of a parallelogram is :

(a) 176°

(b) 68°

(d) 102°

Doubt by Noor

Solution :

Let the smallest angle of the parallelogram be x and the largest angle be y

We also know that in parallelogram opposite angles are equal to each other and adjacent angles are supplementary (180°).

ATQ
y=2x-24° — (1)

x+y=180° (Sum of adjacent angles of a parallelogram)

x+2x-24°=180° [From Equation (1)]
3x-24°=180°
3x=180°+24°
3x=204°
x=204°/3
x=68°

Substituting in equation (1)

y=2(68°)-24°
y=136°-24
y=112°

The largest angle of the parallelogram is 112°.

Hence, (c) 112°, would be the correct option.

If two equal chords of a circle . . .

Question : If two equal chords of a circle intersect within a circle. Prove that the segment of one chord are equal to the corresponding segment of another.

Doubt by Samridhi

Solution :

Given : A circle with centre O. AB and CD are two chords such that AB=CD.

To Prove :
(i) AP=DP
(ii) BP=CP

Construction : Draw OM⊥AB and ON⊥CD

Proof :
(i) In ∆OMP and ∆ ONP

∠OMP = ∠ONP (Each 90°)
OP = OP (Common)
OM = ON (Equal chords are equidistant from the centre of a circle)
∆OMP≅∆ONP (By RHS)
MP=NP — (1) (By CPCT)

AM=BM (Perpendicular drawn from the centre to the chord bisect the chord)
Similarly
DN=CN

Also
AB=CD — (2) (Given)
AM+BM=DN+CN
AM+AM=DN+DN
2AM=2DN
AM=DN — (3)

Adding equation (1) and (2)
AM+MP=DN+NP
AP=DP — (4)

(ii) Subtracting equation (4) from (2)

AB-AP=CD-DP

BP=CP

AD, BE, CF are equal altitudes of a a triangle ABC . . .

Question : AD, BE, CF are equal altitudes of a ΔABC. Prove that ΔABC is an equilateral triangle.

Doubt by Suraj

Solution :

Given : AD, BE and CF are altitudes of ΔABC.
AD=BE=CF
To Prove : ABC is an equilateral triangle.

Proof :
Area of ΔABC
= ½×Base×Height
= ½×BC×AD — (1)
Again,
Area of ΔABC
= ½×Base×Height
= ½×AB×CF — (2)

From equation (1) and (2)
½×BC×AD=½×AB×CF
BC×AD=AB×CF
BC×AD=AB×AD [∵AD=CF]
BC=AB — (3)

Similarly
BC=AC — (4)

From equation (3) and (4)

AB=BC=CA

All sides of triangle ABC are equal.

So, ABC is an equilateral triangle.

Hence Proved.

Note : The above question can also be proved by using the concept of congruency. Can you do that by yourself?

Similar Question : In given figure the altitudes AD, BE and CF, the altitudes of triangle ABC are equal. Prove that ABC is an equilateral triangle.

Similar Question : The altitudes of ΔABC, AD, BE and CF are equal. Prove that ΔABC is an equilateral triangle.

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