Solutions of half-yearly maths question paper of class 10 CBSE 2020

 

Here you can study the Solutions of the half-yearly maths question paper of 10 class CBSE 2020, the class 10 maths CBSE question paper is taken from one of the reputed schools of Delhi i.e Sanjivani Senior Secondary School of Mohan Garden, New Delhi. Solutions of this question paper will help all the students of class 10 in their preparation for the CBSE annual exam. All questions are solved by an expert teacher of Maths.

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Sanjivani Senior Secondary School, Mohan Garden, New Delhi

MM -80

Time-3 hr

Subject-Maths

Class 10 Maths Question Paper CBSE 2022 Half yearly exam with solutions

General Instruction: Read the following instructions very carefully and strictly follow them

(i) This question paper comprises four sections – A, B, C, and D. This question paper carries 40 questions. All questions are compulsory.

(ii) Section A.  question number 1 to 20 comprises 20 questions of one mark each.

Section B. question number . 21 to 26 comprises 6 questions of two marks each.

Section C. question number 27 to 34 comprises 8 questions of three marks each.

Section D.question number 35 to 40 comprises 6 questions of four marks each.

SECTION-A

question number 1 to 10 are a multiple-choice questions. Select the correct option.

Q1.The graphs of polynomial x = p(y) are given in the figure. Then no of its zeroes is

(a) 3

(b) 1

(c) 2

(d) 4

Ans.(a)

Q2.Which of the following rational number have terminating decimal

Ans.(d)

Q3.The sum and the product of zeroes of a quadratic polynomial are 3 and -10 respectively.The quadratic polynomial is.

(a) x² – 3x + 10 

(b) x² + 3x – 10

(c)x² – 3x – 10

(d)x² + 3x + 10

Ans. Applyig the formula for quadratic polyomial

x² – (α + β)x + αβ

Where α and β are the zereoes of the quadratic polynomial given

We are given α + β = 3, αβ = -10

x² – 3x -10

Hence and is (c) x² – 3x – 10

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Q4.Write number of solutions of following pair of linear equations

x + 2y -8 = 0

2x + 4y = 16

(a) Unique solution

(b) No solutions

(c) Infinite many solutions

(d) None

Ans.Comparing the coefficients of given pair of linear equations with the coefficients of standard pair of linear equations  a₁x + b₁y + c₁= 0 and a₂x + b₂y + c₂= 0

x + 2y -8 = 0, 2x + 4y = 16(rewritten as 2x + 4y -16 = 0)

a₁=1 ,b₁=2, c₁= -8and a₂=2 , b₂=4 , c₂= -16

Since the relationship between the coefficients of both equations are as following

Therefore the given pair has infinite solutions

Q5.Which of the following are quadratic equation ?

(a) x³ + x² + 4 = 0

(b)2x² – 7x = 0

Ans. The equation (a) x³ + x² + 4 = 0 is cubic, the equation (b)2x² – 7x = 0 is quadratic equation, (c) is cubic and (d) is of power 4

Hence the answer is (b)2x² – 7x = 0

 

Q6. If b² – 4ac ≥ 0 then what would be the roots of the quadratic equation ax² + bx + c = 0

(d) None

Ans. (a)

Q7. Find the common difference of AP.

(a) 2

(b) -2

(c) -3

(d) None

Ans. The given AP is as following

The common difference (d) of the given AP is

Hence the answer is (b) -2

Q8.In figure DE ll BC, find the value of x.

(a) 3 cm

(b) 2 cm

(c) 4 cm

(d) 1 cm

Ans.

In ΔABC we are given DE ll BC, applying the BPT theorem

x = 2

Hence the answer is (b) 2 cm

Q9.Sides of triangles are given below. Determine which of the sides are of right triangles.If the right triangle then finds the length of the hypotenuse.

7 cm, 24 cm and 25 cm

(a) Yes, 24 cm

(b) Yes, 25 cm

(c) No

(d) Yes, 7 cm

Ans.The condition of a triangle being a right triangle is as follows

The square of longest side = sum of the squares of other sides

25² = 24² + 7²

625 = 576 + 49

625 = 625

Therefore the given triangle is right triangle and its hypotenuse is 25 cm

Hence the answer is (b) Yes, 25 cm

Q10. Find 10^th term of  AP 2, 7, 12……..

(a) 37

(b) 47

(c) 57

(d) None

Ans.

n^th term of AP is = a + (n-1)d

a = 2, d = 7-2 = 5 and n = 10

a₁₀ = 2 + (10-1)5

= 2 + 9×5 = 2 + 45 = 47

Therefore the answer is (b) 47

In question number 11 to 15 , fill in blank

Q11. If equation kx -2y = 3 and 3x + y = 5 represent two intersection lines at unique point then value of k is ….

Ans. The given equations are kx -2y = 3 and 3x + y = 5

Writting both equation in standard form

kx -2y – 3=0, 3x + y – 5 = 0

The condition of two lines a₁x + b₁y + c₁= 0 and a₂x + b₂y + c₂= 0 itersecting each other is as follows

Comparing the coefficients of both given lines

a₁=k ,b₁= -2, c₁= -3 and a₂=3 , b₂=1 , c₂= -5

k ≠ -6

The given line has unique solutions for all values of k except to -6

So, answer ,  k  is  not equal to -6

Q12.If quadratic equation has equal roots 3x² – 4x + k = 0 has equal roots then value of k is………

Ans. The given equation is 3x² – 4x + k = 0

The condition that the ax² + bx + c = 0 has equal roots is as follows

b² – 4ac = 0

Comparing the given equation with standard equation

a = 3, b = -4 and c = k

(-4)² – 4× 3 ×k = 0

12k = 16

Hence the answer is   

Q13. The HCF of two number is 27 and their LCM is 162. If one of the number is 54 . Find other number……..

Ans. We are given HCF of two number = 27 and LCM of two number = 162

One of the number = 54, let another number = x

The relationship between two numbers and their LCM and HCF is as follows

LCM × HCF = Product of both numbers

27× 162 = 54 × x

Therefore another number is 81

Q14. Sum of first n natural number is ……..

Ans. The natural numbers are 1,2,3,4…

First term, a = 1, common difference, d = 1, sum of n terms is = S_n

Putting the value of d = 1, a = 1

 

Q15. I fig. OP = 5 cm, OQ = 5√3 cm, ∠POQ = 90°, ∠RPQ = 90° and PR = 24 cm, find RQ……

 

Ans. In fig. right ΔPQR,we are given PR = 24 cm , In right ΔPOQ , OQ = 5√3 cm, OP = 5 cm

Applying Pythagoras theorem in triangle ΔPQR

RQ² = PQ² + PR²

RQ² = PQ² + 24²

RQ² = PQ² + 576…….(i)

Let’s find the value of PQ from  ΔPOQ

In right ΔPOQ

PQ² = OP² + OQ²

PQ² = 5² + (5√3)²

PQ² = 25 + 75

PQ² = 100

Putting the value of PQ in equation (i)

RQ² = 100 + 576

RQ² = 676

RQ = 26

The value of RQ = 26 cm

Question number 16- 20, answer the following

Q16.The n^th term of an AP is  (7 – 4n), then what is its common difference.

Ans. Let n^th is represented by T_n

We are given

T_n= 7 – 4n

T₁= 7 – 4×1 = 3

T₂= 7 – 4×2 = -1

……

The common difference, d = T₂– T₁= -1 -3 = -4

Q17. ΔABC is an isosceles triangle with AC = BC, if AB² = 2AC², then find the measure of ∠C.

Ans. We are given ΔABC is an isosceles

AC = BC

If AB² = 2AC²

AB² = AC² + AC²

Putting AC = BC

AB² = AC² + BC²

 

Therefore ΔABC is a right triangle in which AB is hypotenuse with sides AC and BC and ∠C

 

= 90°

Q18.The decimal expansion of       will terminates after how many places of decimal.

Ans. We are given the fraction     

Converting the denominator into the power of 10, for this multiplying the denominator and numerator by 5

Hence the decimal expansion of the given fraction will terminate after 2 places.

Q19.Form the quadratic polynomial whose zeroes are √2 + 3 and √2 – 3.

Ans. If zeroes of a quadratic polynomial are given as α and β then the quadratic polynomial is given as follows

x² – (α + β)x + αβ

Putting the values α = √2 + 3, β = √2 – 3

x² – (√2 + 3 + √2 – 3)x + (√2 + 3) (√2 – 3)

x² – 2√2 x + √2² – 3²

x² – 2√2 x + 2 – 9

x²- 2√2 x  – 7

Hence required quadratic polynomial is x²- 2√2 x  – 7

Q20. If -5 is a root of 2x² + px – 15 = 0, find value of p.

Ans. We are given -5 is the root of 2x² + px – 15 = 0

Therefore x = -5 will satisfy the equation 2x² + px – 15 = 0

2×(-5)² + p(-5) – 15 = 0

50 -5p -15 = 0

-5p + 35 = 0

-5p = -35

p = 7

Section B

Question number 21 -26 carry 2 marks each

Q21. Divide the polynomial (9x² + 12x + 10) by  (3x + 2).

Ans. Dividing (9x² + 12x + 10) by  (3x + 2)

 

 

 

 

 

Therefore the quotient is (3x + 2) and the reminder is 6

Q22.Find HCF and LCM of 540 and 72.

Ans. Factorizing the numbers

540 = 2²× 3³×5

72 = 3²× 2³

HCF = 2² ×3² = 36

LCM = 2³×3³× 5 = 1080

Q23. Solve for x and y 

x -4y = 1, 2x – 5y = 5

Ans.We are given the equations

x -4y = 1….(i) 2x – 5y = 5…..(ii)

Multiplyig the equation (i) by 2 , we get equation (iii)

2x -8y = 2…….(iii)

Subtractig the equation (iii) form (ii), we get

3y =  3

y = 1

Putting the value y = 1 in equation (i)

x -4× 1 = 1

x = 1 + 4 = 5

Hence,the value of x =5 and y = 1

Q24. Find roots by quadratic formula

4x² + 4√3 x + 3 = 0 OR 2x² + x -4 = 0

Ans. We are given the quadratic equation

4x² + 4√3 x + 3 = 0

Applying the quadratic equation formula

Putting the value a = 4, b= 4√3 and c = 3

Both of the roots of the given equation are -√3/2, -√3/2

Similarily the equation 2x² + x -4 = 0 can be solved by using the quadratic equation formula

Q25.In an AP find missing terms

Ans. We are given second term  13 and fourth term 3

Applying the n^th term formula of AP a_n = a + (n-1)d

a₂ = a + (2-1)d, a₄ = a + (4-1)d

a₂ = a + d, a₄ = a + 3d

13 =a + d…..(i) 3 = a + 3d…..(ii)

Subtracting equation (ii) from equation (i)

-2d = 10

d = -5

Putting the value of d = -5 in equation (i)

a -5 = 13

a = 18

Therefore the required term of AP are

18, 13, 8, 3

Q26. S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ∼ ΔRTS.

Ans.

 

GIVEN:  Δ PQR

S is a point on PR and T is the point on QR

∠P = ∠RTS

CONSTRUCTION: Joining the point T and S, we get a ΔRTS

TO PROVE: ΔRPQ ∼ ΔRTS

PROOF: In ΔPQR  and ΔRTS

∠P = ∠RTS (given)

∠R = ∠R (common)

ΔRPQ ∼ ΔRTS (AA rule)

Hence proved

Section C

Question number 27  to 34 carry 3 marks each

Q27. Prove that √3 is an irrational number.

Ans.

Let √3 is a rational number

Where a and b are co-prime number (i.e the numbers who has no common factor except to 1)

b√3 = a

Squaring both sides

3b² = a²……..(i)

It is clear from here that a² is divisible by 3

So, a will also be divisible by 3

Therefore putting a = 3c in equation (i) (where c is another positive integer)

3b² = (3c)² = 9c²

b²  = 3c²……(ii)

It is clear from here that b² is divisible by 3

So, b will also be divisible by 3

From equation (i) and equation (ii), we concluded to the point that 3 is a common factor between a and b, hence a and b can’t be co-prime number

Since our assumption that √3 is a rational number is wrong, therefore √3 is an irrational number.

OR

Find HCF of 378, 180, and 420 by the prime factorization method. Is HCF × LCM of three numbers equal to the product of three numbers.

Ans. The given numbers are 378, 180, and 420

Factorizing the numbers

378 = 2 ×3³×7

180 = 2²× 3²×5

420 = 2²× 5×3×7

HCF = 2×3 = 6

LCM = 2²×3³×7×5= 3780

HCF × LCM = 6×3780= 22680

Product of numbers = 378 × 180 ×420 =28576800

HCF × LCM ≠ Product of numbers

Therefore HCF × LCM of three numbers is not equal to the product of three numbers.

OR

Solve graphically  2x + 3y = 2, x – 2y = 8

Ans. We are given the equation  2x + 3y = 2, x – 2y = 8

The solutions of both equations is as following

Solutions of the equation 2x + 3y = 2

x	1	4	-2
y	0	-2	2

Solutions of the equation  x – 2y = 8

x	0	2	4
y	-4	-3	-2

Drawing the graph of the equation

 

 

Hence both of the lies intersect at(4,-2), so the solution of both equations is (4,-2)

 

Q28. Solve for x ad y

Ans.

We are given the equations

Let

5a + b = 2……..(iii)

6a  – 3b = 1……(iv)

Multiplying the equation (iv) by 5 and equation (iii) by 6 , we get equation (v) and equation (vi)

30 a + 6b = 12….(v) and 30a – 15b = 5…..(vi)

Subtractig equation (vi) from equation (v)

21b = 7

Putting the value of b in equation (iii)

As

x – 1 = 3,   y – 2 = 3

x = 4, y = 5

Q29.Find zeroes of quadratic polynomial 6x² + 2 + 7x  and verify the relationship between zeroes and coefficients.

Ans. We are given the quadratic polynomial 6x² + 2 + 7x

Rewriting the quadratic polynomial in standard form

6x² + 7x + 2

Factorizing the polynomial by the splitting up method

6x² + 4x + 3x+ 2

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

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

3x + 2 = 0, 2x + 1 = 0

Therefore zeroes of quadratic polynomial(α and β)

The relationship between zeroes and coefficients of a quadratic polynomial  ax² +bx + c are as follows

comparing the equation with standard quadratic equation

a = 6, b = 7, c= 2

Putting the values of a,b,c and zeroes of the polynomial in the relationship formula of zeroes and coefiicients.

Therefore the relationship of zeroes and coefficients is verified

OR

If α and β are zeroes of a quadratic polynomial 4x² + 4x + 1, then form a quadratic polynomial whose zeroes are 2α and 2β.

Ans. We are given the quadratic polynomial 4x² + 4x + 1

The relationship between zeroes(α and β) and coefficients of a quadratic polynomial  ax² +bx + c are as follows

Comparing the equation with the standard quadratic equation

a = 4, b = 4 and c = 1

Puttig the values in the relationship formula

α + β = -4/4  and αβ = 1/4

α + β = -1….(i)  αβ = 1/4…..(ii)

Let’s find out the quadratic polynomial whose zeroes are 2α and 2β by the following formula

x² -(sum of zeroes)x + product of zeroes

x² -(2α+ 2β)x + 2α×2β

x² -2(α+ β)x + 4 αβ

Putting the values of α+ β and αβ

x² -2(-1) x + 4 ×1/4

x² + 2x + 1

Hence the required quadratic polynomial is x² + 2x + 1

Q30.Find the roots of the equations

Ans. The given equation is

Simplifying it to get the standard quadratic equation

11x² -33x – 308 = -330

11x² -33x – 308 + 330 = 0

11x² -33x + 22 = 0

Factorisig it by splitting method

11x² -22x -11x +  22 = 0

11x(x – 2) – 11(x – 2) = 0

(x – 2) (11x -11) = 0

x = 2, x = 1

Hence the solution of the given equation is x = 2, x = 1

Q31.Which term of AP 3,15,27,39…….. is 132 more than its 54^th term.

Ans. The given AP is 3,15,27,39……..

The first term, a of AP is = 3, common difference, d = 15-3 = 12

Let n^th term of given AP is 132 more than its 54^th term.

a_n= a₅₄+ 132

a + (n-1)d = a + 53d + 132

(n-1)d – 53d = 132

d(n –1 -53) = 132

12(n – 54) = 132

n – 54 = 132/12

n = 54 + 11 = 65

Hence 65^th term is 132 more than the 54^th term of the given AP

OR

In an AP,  given a = 3, n = 8, S = 192, Find d

Ans. We are given a = 3, n = 8, S = 192

Applying the formula for sum of an AP

4(6 +7d) = 192

24 + 28d = 192

28d = 192 – 24 = 168

d = 4

Hence the value of d is 4

Q32. If a line is drawn parallel to one side of a triangle to intersect other two sides in district points .Prove that the other two sides are divided in the same ratio. 

Ans.

 

 

GIVEN: ΔABC in which DE ll BC

CONSTRUCTION: Drawing EF ⊥ BD and DG ⊥ AC

TO PROVE: 

PROOF:  Triangles ΔDBE  and  ΔDEC made on the same base and between the same parallel lines

∴arΔDBE = ar ΔDCE……(i)

arΔADE = 1/2 AD×EF

arΔDBE = 1/2 BD×EF

arΔADE = 1/2 DG×AE

arΔDCE = 1/2 CE×DG

Dividing equation (ii) by equation (iii)

From equation (i) arΔDBE = ar ΔDCE

AD × CE = BD × AE

Hence proved

Q33. BL and CM are median of a triangle ABC, right-angled at A.

Prove that 4(BL² + CM²) = 5 BC².

Ans.

 

GIVEN: ΔABC in which

∠A = 90°

BL and CM are median of a Δ ABC

TO PROVE: 4(BL² + CM²) = 5 BC²

PROOF: Applying the Pythagoras theorem in right ΔABC, ΔAMC and ΔABL

BC² = AB² + AC²…….(i)

CM² = AM² + AC²….(ii)

BL² = AB² + AL²…..(iii)

Adding equation number (ii) and (iii)

CM² + BL²= AM² + AC² + AB² + AL²

From equation number (i)  AC²= BC² -AB²

CM² + BL²= AM² + BC² -AB² + AB² + AL²

CM² + BL²= AM² + BC²  + AL²

AM = AB/2 (CM is the median of the ΔABC  on BC)

AL = AC/2 (BL is the median of ΔABC on AC)

CM² + BL² = AB²/4 + BC² + AC² /4

Multiplying both sides by 4

4 (CM² + BL²) = AB² + 4BC² + AC²

4 (CM² + BL²) =  4BC² + AC²+ AB²

Substituting from equation (i) BC² = AB² + AC²

4 (CM² + BL²) = 4BC² + BC²

4 (CM² + BL²) = 5BC² , Hence proved

Q34. Solve for x

Ans. We are given the equation

(3x + 4)(x + 4) = 4(x² + 3x + 2)

3x² + 12x + 4x + 16 = 4x² + 12x + 8

4x² -3x² – 4x -8 = 0

x² -4x  -8 = 0

Factorising it by quadratic formula

x = 2 ±2√3

Hence x = 2 + 2√3, 2 – 2√3 are the required solutions of the given equation

OR

Yash scored 40 marks in a test getting 3 marks for each right answer and losing 1 mark for each wrong answer . Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer ,then Yash would have scored 50 marks . How many questions were there in the test.

Ans. Let Yash answered the questions correctly = x and answered the questions incorrectly = y

According to first condition of the question

3x – y = 40……(i)

According to second condition of the question

4x -2y = 50….(ii)

Multiplying equation (i) by 2, we get equation (iii)

6x -2y = 80…..(iii)

Subtracting equation (iii) from equation (ii)

-2x = -30

x = 15

Putting x = 15 in equation (i)

3×15- y = 40

45-y = 40

y = 5

The solutions of the equation is x = 15 and y = 5

Number of questions Yash answered correctly = 15 and answered incorrectly = 5

Hence total number of questions in the test were = x + y = 15 + 5 = 20

Section D

Question numbers 35 to 40 carry 4 marks each.

Q35. If 4 times the 4^th term of an AP is equal to 18 times the 18^th term ,then find 22 nd term.

Ans. n^th term  of an AP is given by

a_n = a + (n -1) d

Where a is first term, d is the common difference of an AP

We are given in the question

4 times the 4^th term of an AP is =  18 times the 18^th term

4[a + (4-1)d ]=18[a + (18 -1)d]

4a + 12d = 18a +306d

294d = -14a

21d = -a

a = -21d

Now,22_nd term  = a + 21d

Putting the value a = -21d

22_nd term  =  -21d+ 21d = 0

Hence the 22_nd term of the AP is 0.

OR

How many terms of AP 24, 21, 18…..must be taken so that their sum is 78.

Ans. The given AP is 24, 21, 18….

Common difference of the given AP,d is = 21 -24 = -3 and first term, a = 24

Let the number of terms whose sum is 78 are = n

Applying the following formula for the sum(S) of an AP

We are given S = 78

(51 – 3n)n =156

-3n² + 51n = 156

3n² – 51n + 156 = 0

n² – 17n + 52= 0

Factorising it by splitting up method

n² – 13n -4n+ 52= 0

n(n -13) – 4(n – 13) = 0

(n – 13)(n -4) = 0

n = 13, n = 4

Therefore the number of terms are either 13 or 4 whose sum is 78.

Q36.In an obtuse ΔABC, ∠B is obtuse AD is perpendicular to CB produced,then prove that AC² = AB² + BC² + 2BC.BD

Ans.

GIVEN: In an obtuse ΔABC, ∠B is obtuse

AD ⊥CB

TO PROVE:AC² = AB² + BC² + 2BC.BD

PROOF: In ΔADB and ΔADC, applying the Pythagoras theorem

AC² = DC² + AD²…..(i)

AB² = AD² + BD²…..(ii)

Subtracting equation (ii) from equation (i)

AC² – AB² = DC²  – BD²

In figure, we have DC = BD + BC

AC² – AB² = (BD + BC)²  – BD²

AC² – AB²= BD² + BC² + 2BC.BD   – BD²

AC² – AB² = BC² + 2BC.BD

AC² = AB² +  BC² + 2BC.BD, Hence proved

OR

The perpendicular from A on side BC of a ΔABC intersect BC at D, such that DB = 3CD,prove that 2AB² = 2AC² + BC²

Class 10 Maths NCERT Solutions from chapter 1-15

Q37.Solve for x

 

Ans. The given equation is

ax + bx +x² = -ab

x² + ax + bx + ab = 0

x(x + a) + b( x +a) = 0

(x + a)(x + b) = 0

x = -a, -b

Hence the solutions of the equation is x = -a and x = -b

Q38.A boat takes 4 hr to go 44 km downstream and it can go 20 km upward in the same time . Find speed of stream and that of the boat in still water.

Ans.Let the speed of boat in still water = x and speed of stream = y

The speed of the boat in downward stream = speed of boat + speed of stream=x+y

The speed of the boat in downward stream = Distace covered in downwardstream/Time taken = 44/4 = 11 km/h

∴ x + y = 11……(i)

The speed of the boat in upward stream = speed of boat -speed of stream=x-y

The speed of the boat in downward stream = Distance covered in upward stream/Time taken = 20/4 = 5 km/h

∴ x – y = 5……(ii)

Adding equation (i) and (ii)

2x = 16

x = 8

Putting the value of x in equation (i)

8 + y = 11

y = 3

Hence the speed of boat in still water is 8 km/h and speed of stream  is 3 km/h

Q39. Find all zeroes of 2x⁴– 3x³-3x²+ 6x -2, if two of its zeroes are √2 and -√2.

Ans. We are given the polynomial 2x⁴– 3x³-3x²+ 6x -2

If two of zeroes are √2 and -√2 then (x -√2)(x +√2) =x² -2 will be one of its factor.

Therefore further factorizing the polynomial by the reminder’s theorem

2x⁴– 3x³-3x²+ 6x -2

2x²(x² -2) +4x²- 3x(x²-2)-6x-3x²+ 6x -2

2x²(x² -2) – 3x(x²-2)-3x²  +4x²-3x²- 2

2x²(x² -2) – 3x(x²-2) + (x² – 2)

(x² -2)(2x² – 3x +1)

(x² -2)[2x² – 2x-x +1]

(x² -2)[2x(x -1) -1(x -1)]

(x² -2)(x -1)(2x -1)

(x² -2) = 0⇒x =±√2, x =1 and x = 1/2

Hence other two zeroes of the given polynomial are 1 and 1/2

Q40. Prove that 1/√2 is an irrational number.

Ans. Let  1/√2 is a rational number

So, 1/√2 = a/b (Where a and b are co-prime numbers)

a√2  = b

Squaring both sides

2a² = b²……(i)

b² is divisible by 2

∴ b is also divisible by 2

b is also written in the form of 2c, where c is another positive integer

b = 2c

Putting b = 2c in equation (i)

2a² =(2c)²= 4c²

a² = 2c²……(ii)

a² is  divisible by 2

∴ a is also divisible by 2

Equation (i) and equation (ii) implies that 2 is a common factor between a and b,so our assumption is wrong that a and b are co-prime number

Therefore 1/√2 can’t be a rational number, so 1/√2 is an irrational number

What are the physical and chemical properties of metals?

Class 10 Science NCERT Solutions from Chapter 1 to 16

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NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science 

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution