email: help@eenadupratibha.netøŒEî¦ô¢Ù 28 è…šúÙñô¢ª 2019

1 Mark Questions2

1. If , are the zeroes of the polynomial p(x)

1 1= 4x2 + 3x + 7 then find the value of + .

3

A: 7

2. If one zero of the polynomial p(x) = 5x2 + 13x + k is reciprocal of theother, then the value of ‘k’?

A: 5

3. If , and are the zeroes of the polynomialp(x) = 2x3 + 6x2 – 4x + 9 then

i) + + = a) – 2

ii) + + = b) – 3

9iii) = c)

2

A) i a, ii b, iii c B) i b, ii a, iii c

C) i b, ii c, iii a D) i a, ii c, iii b

A: B

4. Statement (i): p(x) = 2x + 3 is a linear polynomial.

Statement (ii): Polynomial of degree 1 iscalled linear polynomial.

A) Statement (i) is only correct.B) Statement (ii) is only correct.

C) Statement (i) is correct and Statement (ii)is correct explanation of Statement (i).

D) Statement (i) is correct and Statement (ii)is not correct explanation of statement (i).

A: C

5. If f(x) = ax2 + bx + c has no real zeroes anda + b + c < 0, then the value of ‘c’ is ........

A: c < 0

6. The zeroes of x2 – 1 are ........

A: 1

7. Write the general form of a cubic polynomialin one variable x?

A: p(x) = ax3 + bx2 + cx + d

8. Find the number of zeroes of z3.

A: 3

9. Find a quadratic polynomial, the sum andproduct of whose zeroes are –3 and 2respectively?

A: x2 – (–3x) + 2 = 0 x2 + 3x + 2 = 0

10. Give an example for quadratic polynomialwhose sum of the zeroes is zero.

A: x2 + 5 or 3x3 – 8 or any relevant example.

11. Find the quadratic polynomial whose sumand product of the zeroes is

2 and 13 .

1A: x2 – 2 x + = 0

3 3x2 – 3

2 x + 1= 0

12. xn + yn is divisible by x + y if ‘n’ is ........

A: odd

13. A point which satisfies y = x3 is ........

A: (–1, –1) or (2, 8) or (–2, –8) or any relevantanswer.

14. Write the condition that ax2 + bx + c is aquadratic polynomial.

A: a 0

15. (x3 – 8) (x – 2) = ........

A: x2 + 2x + 4

16. Degree of constant polynomial is ........

A: 0

1. Find the zeroes of the polynomial p(x) = x3 – 12x2 + 39x – 28, if it is given that thezeroes are in A.P.

Sol: Let = a – d, = a, = a + d be the zeroesof the polynomial p(x) = x3 – 12x2 + 39x – 28

–12 –28= –() = 12 and = –() = 281 1

(a – d) + a + (a + d) = 12 and (a – d). a. (a + d) = 28

3a = 12 and a(a2 – d2) = 28a = 4 and 4(42 – d2) = 28a = 4 and 16 – d2 = 7a = 4 and d2 = 9a = 4 and d = 3

If d = 3, then = a – d = 4 – 3 = 1

= a = 4 and = a + d = 4 + 3 = 7

If d = – 3, then

= a – d = 4 – (–3) = 4 + 3 = 7,

= a = 4 and = a + d = 4 + (–3) = 1

Hence the zeroes of the given polynomialare 1, 4, 7.

12. Verify that , 1, –2 are the zeroes of the cubic2

polynomial p(x) = 2x3 + x2 – 5x + 2 and then

verify the relationship between the zeroesand the coefficients.

Sol: p(x) = 2x3 + x2 – 5x + 2

Comparing this with ax3 + bx2 + cx + d,

we get a = 2, b = 1, c = –5, d = 2

1 1 3 1 2 1 p() = 2() + () – 5() + 2

2 2 2 2

1 1 5 1+1 – 10 + 8 10 – 10= + – + 2 = = = 04 4 2 4 4

p(1) = 2 13 + 12 – 5 1 + 2

= 2 + 1 – 5 + 2 = 5 – 5 = 0

p(–2) = 2(–2)3 +(–2)2 – 5(–2) + 2

= –16 + 4 + 10 + 2 = –16 + 16 = 0

1, 1, –2 are the zeroes of 2x3 + x2 – 5x + 22

Let = 12 , = 1, = –21 1 –1 b

+ + = + 1 + (–2) = 1 = = – –2 2 2 a

1 1 + + = (1) + (1)(–2) + (–2)()2 2

1 1 4 2 5 c= – 2 – 1 = = = 2 2 2 a

1 2 d = (1)(–2) = –1 = – (– ) = – 2 2 a

3. On dividing x3 – 3x2 + x + 2 by a polynomialg(x) we get the quotient and remainder are(x – 2) and –2x + 4 respectively. Find g(x).

Sol: Let p(x) = x3 – 3x2 + x + 2q(x) = x – 2r(x) = –2x + 4 and g(x) is the divisor.

By division algorithmp(x) = g(x). q(x) + r(x)

p(x) r(x)g(x) = q(x)

(x3 3x2 + x + 2) (2x + 4)= x 2

x3 3x2 + 3x 2=

x 2x2 – x + 1

x – 2) x3 – 3x2 + 3x – 2x3 – 2x2

– x2 + 3x– x2 + 2xx – 2

x – 20

g(x) = x2 – x + 1

4. Find all zeroes of 2x4 – 3x3 – 3x2 + 6x –2, ifyou know that two of its zeroes are

2 and

–2 .

Sol: Since 2 and –

2 are two zeroes

(x – 2)(x +

2)

= x2 – 2 is a factor of the polynomial.

x2 – 2)2x4 – 3x3 – 3x2 + 6x –2(2x2 – 3x + 1

2x4 – 4x2

– 3x3 + x2 + 6x

– 3x3 + 6x+ x2 – 2

+ x2 – 2

0

2x4 – 3x3 – 3x2 + 6x – 2

= (x2 – 2)(2x2 – 3x + 1)

Consider 2x2 – 3x + 1

= 2x2 – x – 2x + 1

= (2x – 1)(x – 1)

So the zeroes are 12 and 1.

1. p(x) = 8x – 6x5 + 4x3 + 2x7. Find degreeof p(x) and coefficient of x5.

Sol: p(x) = 8x – 6x5 + 4x3 + 2x7

Degree of p(x) = 7

Coefficient of x5 = –6

2. Check whether –3 and 3 are zeroes of x3 – 27.

Sol: Let p(x) = x3 – 27

p(–3) = (–3)3 – 27 = –27 – 27 = –54 0

p(3) = (3)3 – 27 = 27 – 27 = 0

–3 is not a zero of x3 – 27

and 3 is a zero of x3 – 27.

3. Find the sum and product of the zeroes ofthe polynomial 4x3 + 3x2 + 2x.

Sol: Let p(x) = 4x3 + 3x2 + 2x

Comparing this with ax3 + bx2 + cx + d,

a = 4, b = 3, c = 2, d = 0

b 3Sum of the zeroes = = a 4

d 0Product of the zeroes = = = 0a 4

4. If a – b, a, a + b are zeroes of the polynomial f(x) = 2x3 – 6x2 + 5x – 7, findthe value of the a.

Sol: Let are the zeroes of cubic polynomial ax3 + bx2 + cx + d

bthen = a6a + b + a + a – b = 2

3a = 3

a = 1

5. Sum of the zeroes of a polynomial 4x2 + kx + 5 is 5. Find the value of k.

Sol: Let be the zeroes of 4x2 + kx + 5

+ = – k4 k = –4(+ ) = –4 5 = –20

Degree of constant polynomial is?

l

Polynomials

One Mark Questions1. Check whether 14 and –1 are zeroes of the

polynomial p(x) = 4x2 + 3x – 1.

Sol: Given, p(x) = 4x2 + 3x – 1

1 1 2 1 p () = 4() + 3() – 1

4 4 4

1 3 1 + 3 – 4= 4 × + – 1 = = 016 4 4

P(–1) = 4(–1)2 + 3(–1) – 1

= 4 – 3 – 1 = 0

1 and –1 are zeroes of p(x)42. Find the zeroes of the polynomial x2 – 3 and

verify the relation between the zeroes andcoefficients.

Sol: x2 – 3 = (x + 3 )(x –

3 )

[... a2 – b2 = (a + b)(a – b)]

So the value of x2 – 3 is zero when

x = –3 and x =

3

Zeroes of x2 – 3 are –3 and

3

0Sum of the zeroes = –3 +

3 = 0 =

1–(Coefficient of x)

= (Coefficient of x2)

–3Product of zeroes = (–3 )(

3 ) = –3 =

1 Constant term

= Coefficient of x2

3. Divide p(x) = x4 – 3x2 + 4x + 5 by q(x) = x2 – 2and find the quotient and remainder.

Sol: p(x) = x4 + 0x3 – 3x2 + 4x + 5, q(x) = x2 – 2

x2 – 1x2 – 2) x4 + 0x3 – 3x2 + 4x + 5

x4 – 2x2

– x2 + 4x + 5– x2 + 2

4x + 3 Quotient = x2 – 1.

Remainder = 4x + 3

Two Marks Questions

Four Marks Questions

Dr. TSVS Suryanarayana Murty

Subject Expert

Writer

Target-2020

TenthMathematics Paper-1

100100

email: help@eenadupratibha.netøŒEî¦ô¢Ù 28 è…šúÙñô¢ª 2019

1. 23, 21, 19, .... 5 in the Arithematicprogression from last to 7th term.A) 19 B) 17 C) 15 D) 13

12. 3, , 2, ....... a11 =2

A) 28 B) 221C) 38 D) 48 2

3. 1 + 2 + 3 + 4 + ..... + 1000 =A) 50500 B) 50050 C) 500500 D) 505500

4. If x, y, z are in G.P. thenA) y x = z y B) y = x + z

x + zC) = y D) y2 = xz

25. There are 'n' arithmetic means between

a, b then d =

n + 1 b + aA) B)

a + b n + 1n + 1 b a

C) D) ab n + 1

6. The common ratio of the series x, x2, x3,x4, ..... is A) x B) 2x

xC) D) x + 12

7. 3, 6, 9, 12, ..... Arithmatic progressionthen an = A) 3n B) 3n C) 3 + n D) n3

8. The sum of p terms of an AP is the sameas the sum of q terms, then sum of(p + q) terms is equal to A) 1 B) 0 C) pq D) 2

Answers

1-B 2-B 3-C 4-D 5-D 6-A 7-B 8-B

Four Marks Questions1. Find the sum of all 3 digit natural numbers,

which are divisible by 7.Sol. The smallest and the largest number of

three digits, which are divisible by 7 are105, 994 respectiely. The sequence of 3 digit numbers

which are divisible by 7 are 105, 112,119, ......., 994.

Clearly it is an Arithmetic progression. a = 105, d = 112 105 = 7an = 994

994 = 105 + (n 1)7994 105 = (n 1)7

889 = (n 1)7889n 1 = = 1277

n = 127 + 1 = 128

Now, required sum

128= [105 + 994]2

= 64 1099 = 70,336

2. The sum of the series 25 + 22 + 19 + 16 +13 + ..... is 116. Find the last term and the

number of terms.Sol: a = 25

d = 22 25 = 3, Sn = 116nSn = [2a + (n 1)d]2n116 = [2(25) + (n 1)(3)]2

232 = n[50 3n + 3]232 = 50n 3n2 + 3n3n2 53n + 232 = 0

53 (53)2 4(3)(232)

= 2 3

53 2809 2784

= 6

53 25 53 + 5 53 5= = or

6 6 658 48 29= or or 86 6 3

Since, 'n' cannot be fractional29so rejecting n = 83

Hence the last term a8 = a + 7d= 25 + 7(3) = 25 21= 4

3. For what value of n, the nth term of the A.P.9, 7, 5 .... and 15, 12, 9, .... is the same?

Sol. 9, 7, 5, .......... in A.P.a = 9, d = 7 9 = 2

= 9 + (n 1)(2)an = 9 2n + 2 = 11 2n ........... (1)15, 12, 9, ............. in A.P.

a = 15, d = 12 15 = 3an = 15 + (n 1)(3)

= 15 3n + 3 = 18 3n ......... (2)As per question11 2n = 18 3n3n 2n = 18 11

n = 7Hence, for n = 7, the nth term of the givenseries are equal.

4. A teacher distributed money in geometricalprogression to his students Rashmi, Kalyani,Priyaka and Rajitha. Rashmi and Kalyanireceived total amount Rs.162. Priyanka &Rajitha received total amount Rs.4050. Findthe individuals received amount.

Sol: Here, the amounts are in G.P.Let it a, ar, ar2, ar3

an = a + (n 1)d

b2 4ac

n = b 2a

nSn = [a + l]2

an = a + (n 1)d

P. Venu GopalSubject Expert

Writer

Target-2020

TenthMathematics Paper-I

100100

1. If the 9th term of an A.P. is 99 and 99th

term is 9. Then find a100.Sol: a9 = a + 8d = 99 ........... (1)

a99 = a + 98d = 9 ............ (2)-----------------------------

9090d = 90 d = = 1 90

d = 1Substitute in (1)a + 8(1) = 99

a = 99 + 8 = 107 a = 107

a100 = a + 99d = 107 + 99(1)

= 107 99 = 82. Find the sum of n terms of A.P. whose 6th

term is 26 and 13th term is 54.Sol: a6 = 26

a + 5d = 26 ......... (1)a13 = a + 12d = 54 ......... (2)

(2) (1)a + 12d = 54a + 5d = 26 ------------------ 287d = 28 d = = 4

7Substitute in (1) we geta + 5 (4) = 26

a = 26 20 = 6

nSn = [2a + (n 1)d]2

n= [2(6) + (n 1)4]2

n nSn = [12 + 4n 4] = [8 + 4n]2 2

= n(4 + 2n) = 2n(n + 2)3. x + 1, x + 3, x + 7 are in geometric

progression. Then find x = ?Sol: x + 1, x + 3, x + 7 are in G.P.

x + 3 x + 7 =

x + 1 x + 3(x + 3)2 = (x + 1)(x + 7)x2 + 6x + 9 = x2 + 7x + x + 7x2 + 6x + 9 x2 8x 7 = 02x + 2 = 0

2x = 2 x = 14. Find the 15th term of G.P. whose 9th term

is 768 and a = 3.Sol: a9 = ar8 = 768

3(r8) = 768768r8 = = 256 = 28 r = 23

a15 = ar14 = 3(2)14

5. 101, 99, 97, ..... are in A.P. number ofterms is 28. By using

n Sn = [a + l] find the sum of 28 terms.

2

Sol: 101, 99, 97, ......

a = 101, d = 99 101 = 2

l = a28 = a + 27d

= 101 + 27(2) = 101 54 = 47nSn = [a + l]2

28= [101 + 47] = 14 148 = 2072

2

Progressions

Common ratio of the series is? a + ar = 162a(1 + r) = 162 .......... (1)ar2 + ar3 = 4050ar2 (1 + r) = 4050 ........ (2)(2) (1)ar2(1 + r) 4050 = = 25

a(1 + r) 162r2 = 25 r = 25 = 5Substitute in (1) a(1 + 5) = 162

162a = = 27

6Amount Rs.27, 27 5, 27 52, 27 53

Rs.27, Rs.135, Rs.675, Rs.3375 5. If the geometric progressions 13, 19, 127, ....

1 1and , .... have their nth term equal.2187 729

Find the value of n.Sol: The given geometric progressions are

13, 19, 127 .......... (1)

1 1 , .......... (2)2187 729The first term of G.P. (1) is a = 13,

19 1 3 1Common ratio is = = = r13 9 1 3

an = 13 ( 13 )n 1= ( 13 )

n 1 + 1= ( 13 )

n........ (3)

1The first term of G.P. (2) is a = ,

21871 1 r =

729 2187

1 2187= = 3729 1

1 1an = (3)n 1 (3)n 1

2187 37

= 3n 1 7 = 3n 8 .......... (4)By problem, (3) & (4) are equal

( 13 )n

= 3n 8

3n = 3n 8 n = n 88n + n = 8 n = = 42

an = arn 1

Two Marks Questions

1. Write example of Arithmatic progressionwhose common diference is 1.

Sol: 10, 9, 8, 7, 6, ..........d = 9 10 = 1

2. Sabitha said a d, a, a + d are in A.P.Justify your answer.

Sol: Common difference = a (a d) = a a + d = d

and(a + d) (a) = a + d a = dCommon difference is same

Hence Sabitha's answer is justified3. an = 12 5n then find a5 + a6 = ?Sol: a5 12 5(5) = 12 25 = 13

a6 = 12 5(6) = 12 30 = 18a5 + a6 = 13 18 = 31

4. Is 18

, 14

, 12

are in G.P.?

14 1 812 1 4

Sol: r = = = 2 and r = = = 218 4 1 14

2 1

Common ratio is same.

Hence 18

, 14

, 12

are in G.P.

One Mark Questions

12 Mark Questions

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