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## NCERT Exemplar Problems Class 10 Maths Solutions Chapter 2 Polynomials

**Exercise 2.1 Multiple Choice Questions (MCQs)**

**Question 1:**

**Solution:**

(a) Given that, one of the zeroes of the quadratic polynomial say p(x) = (k- 1)x^{2} + kx + 1

**Question 2:
**A quadratic polynomial, whose zeroes are -3 and 4, is

(a) x

^{2}– x + 12 (b)x

^{2}+ x + 12 (c)\(\frac { { x }^{ 2 } }{ 2 } -\frac { x }{ 2 } -6\) (d)2x

^{2}+ 2x-24

**Solution:**

(c) Let ax

^{2}+ bx + c be a required polynomial whose zeroes are -3 and 4.

We know that, if we multiply/divide any polynomial by any constant, then the zeroes of polynomial do not change.

**Alternate Method**

Let the zeroes of a quadratic polynomial are α = – 3 and β = 4.

Then, sum of zeroes =α + β = -3+4=1 and product of zeroes = αβ = (-3) (4) = -12

**Question 3:
**If the zeroes of the quadratic polynomial x

^{z}+ (a +1)* + b are 2 and -3, then

(a) a = -7, b = -1 (b) a = 5,b = -1

(c) a=2, b = -6 (d)a=0,b = -6

**Solution:**

(d) Let p{x) =\({ x }^{ 2 }+\left( a+1 \right) x\)+ b

Given that, 2 and -3 are the zeroes of the quadratic polynomial p(x).

required values are a = 0 and b = – 6.

**Question 4:
**The number of polynomials having zeroes as -2 and 5 is

(a) 1 (b) 2 (c) 3 (d) more than 3

**Solution:**

(d) Let p (x) = ax

^{2}+ bx + c be the required polynomial whose zeroes are -2 and 5.

Hence, the required number of polynomials are infinite i.e., more than 3.

**Question 5:**

If one of the zeroes of the cubic polynomial ax^{3} + bx^{2} + cx + d is zero,

the product of then other two zeroes is

(a)\(\frac { -c }{ a }\) (b)\(\frac { c }{ a }\) (c)0 (d) \(\frac { -b }{ a }\)

**Solution:**

(b) Let p(x) =ax^{3} + bx^{2} + cx + d

Given that, one of the zeroes of the cubic polynomial p(x) is zero.

Let α, β and γ are the zeroes of cubic polynomial p(x), where a = 0.

We know that,

**Question 6:
**If one of the zeroes of the cubic polynomial x

^{3}+ ax

^{2}+ bx + c is -1, then the product of the other two zeroes is

(a) b – a +1 (b) b – a -1 (c) a – b +1 (d) a – b -1

**Solution:**

(a) Let p(x) = x

^{3}+ ax

^{2}+ bx + c

Let a, p and y be the zeroes of the given cubic polynomial p(x).

∴ α = -1 [given]

and p(−1) = 0

⇒ (-1)

^{3}+ a(-1)

^{2}+ b(-1) + c = 0

⇒ -1 + a- b + c = 0

⇒ c = 1 -a + b …(i)

We know that,

αβγ = -c

⇒ (-1)βγ = −c [∴α = -1]

⇒ βγ = c

⇒ βγ = 1 -a + b [from Eq. (i)]

Hence, product of the other two roots is 1 -a + b.

**Alternate Method**

Since, -1 is one of the zeroes of the cubic polynomial f(x) = x

^{2}+ ax

^{2}+ bx + c i.e., (x + 1) is a factor of f{x).

Now, using division algorithm,

⇒x

^{3}+ ax

^{2}+ bx +c = (x + 1) x {x

^{2}+ (a – 1)x + (b – a + 1)> + (c – b + a -1)

⇒x

^{3}+ ax

^{2}+ bx + (b – a + 1) = (x + 1) {x

^{2}+ (a – 1)x + (b -a+ 1}}

Let a and p be the other two zeroes of the given polynomial, then

**Question 7:
**The zeroes of the quadratic polynomial x

^{2}+ 99x + 127 are

(a) both positive (b) both negative

(c) one positive and one negative (d) both equal

**Solution:**

(b) Let given quadratic polynomial be p(x) =x

^{2}+ 99x + 127.

On comparing p(x) with ax

^{2}+ bx + c, we get

a = 1, b = 99 and c = 127

Hence, both zeroes of the given quadratic polynomial p(x) are negative.

Alternate Method

the above condition.

So, both zeroes of the given quadratic polynomial are negative.

**Question 8:
**The zeroes of the quadratic polynomial x

^{2}+ kx + k where k ≠ 0,

(a) cannot both be positive (b) cannot both be negative

(c) are always unequal (d) are always equal

**Solution:**

(a)Let p(x) = x

^{2}+ kx + k, k≠0

On comparing p(x) with ax

^{2}+ bx + c, we get

Here, we see that

k(k − 4)> 0

⇒ k ∈ (-∞, 0) u (4, ∞)

Now, we know that

In quadratic polynomial ax

^{2}+ bx + c

If a > 0, b> 0, c> 0 or a< 0, b< 0,c< 0,

then the polynomial has always all negative zeroes.

and if a > 0, c < 0 or a < 0, c > 0, then the polynomial has always zeroes of opposite sign

Case I If k∈ (-∞, 0) i.e., k<0

⇒ a = 1>0, b,c = k<0

So, both zeroes are of opposite sign.

Case II If k∈ (4, ∞)i.e., k≥4

=> a = 1> 0, b,c>4

So, both zeroes are negative.

Hence, in any case zeroes of the given quadratic polynomial cannot both be positive.

**Question 9:
**If the zeroes of the quadratic polynomial ax

^{2}+ bx+ c, where c≠0, are equal, then

(a) c and a have opposite signs (b) c and b have opposite signs

(c) c and a have same signs (d) c and b have the same signs

**Solution:**

(c) The zeroes of the given quadratic polynomial ax

^{2}+ bx + c, c ≠ 0 are equal. If coefficient of x

^{2}and constant term have the same sign i.e., c and a have the same sign. While b i.e., coefficient of x can be positive/negative but not zero.

which is only possible when a and c have the same signs.

**Question 10:
**If one of the zeroes of a quadratic polynomial of the form x

^{2}+ ax + b is the negative of the other, then it

(a) has no linear term and the constant term is negative

(b) has no linear term and the constant term is positive

(c) can have a linear term but the constant term is negative

(d) can have a linear term but the constant term is positive

**Solution:**

(a) Let p(x) = x

^{2}+ ax + b.

Put a = 0, then, p(x) = x

^{2}+ b = 0

⇒ x

^{2}= -b

⇒ x = ±\( \pm \sqrt { -b } \)

[∴b < 0]

Hence, if one of the zeroes of quadratic polynomial p(x) is the negative of the other, then it has no linear term i.e., a = O and the constant term is negative i.e., b< 0.

**Alternate Method**

Let f(x) = x

^{2}+ ax+ b

and by given condition the zeroes area and – α.

Sum of the zeroes = α- α = a

=>a = 0

f(x) = x

^{2}+ b, which cannot be linear,

and product of zeroes = α .(- α) = b

⇒ -α

^{2}= b

which is possible when, b < 0.

Hence, it has no linear term and the constant term is negative.

**Question 11:
**Which of the following is not the graph of a quadratic polynomial?

**Solution:**

(d) For any quadratic polynomial ax

^{2}+ bx + c, a≠0, the graph of the Corresponding equation y = ax

^{2}+ bx + c has one of the two shapes either open upwards like u or open downwards like ∩ depending on whether a > 0 or a < 0. These curves are called parabolas. So, option (d) cannot be possible.

Also, the curve of a quadratic polynomial crosses the X-axis on at most two points but in option (d) the curve crosses the X-axis on the three points, so it does not represent the quadratic polynomial.

**Exercise 2.2 Very Short Answer Type Questions
**

**Question 1:
**Answer the following and justify.

(i) Can x

^{2}-1 be the quotient on division of x

^{6}+2x

^{3}+x-l by a polynomial in x of degree 5?

(ii) What will the quotient and remainder be on division of ox

^{2}+ bx + c by px

^{3}+qx

^{2}+ rx+ s, p≠ 0 ?

(iii) If on division of a polynomial p(x) by a polynomial g(x),the quotient is zero, what is the relation between the degree of p(x)and g(x)l

(vi) If on division of a non-zero polynomial p(x)by a polynomial g(x),the remainder is zero, what is the relation between the degrees of p(x) and g(x)?

(v) Can the quadratic polynomial x

^{2}+ kx + k have equal zeroes for some odd integer k > 1?

**Solution:**

(i) No. because whenever we divide a polynomial x

^{6}+ 2x

^{3}+ x -1 by a polynomial in x of degree 5, then we get quotient always as in linear form i.e., polynomial in x of degree 1. Let divisor = a polynomial in x of degree 5

= ax

^{5}+ bx

^{4}+ cx

^{3}+ dx

^{2}+ ex + f

quotient = x

^{2}-1

and dividend = x

^{6}+ 2x

^{3}+ x -1

By division algorithm for polynomials,

Dividend = Divisor x Quotient + Remainder

= (ax

^{5}+ bx

^{4}+ cx

^{3}+ dx

^{2}+ ex + f)x(x

^{2}-1) + Remainder

= (a polynomial of degree 7) + Remainder

[in division algorithm, degree of divisor > degree of remainder]

= (a polynomial of degree 7)

But dividend = a polynomial of degree 6

So, division algorithm is not satisfied.

Hence, x

^{2}-1 is not a required quotient.

(ii) Given that, Divisor px

^{3}+ gx

^{2}+ rx + s, p≠0

and dividend = ax

^{2}+ bx + c

We see that,

Degree of divisor > Degree of dividend

So, by division algorithm,

quotient = 0 and remainder = ax

^{2}+ bx + c

If degree of dividend < degree of divisor, then quotient will be zero and remainder as same as dividend.

(iii) If division of a polynomial p(x) by a polynomial g(x), the quotient is zero, then relation between the degrees ofp(x) and g(x) is degree of p(x) < degree of g(x).

(iv) If division of a non-zero polynomial p(x) by a polynomial g(x), the remainder is zero, then g(x) is a factor of p(x) and has degree less than or equal to the degree of p(x). e., degree of g(x) < degree of p(x).

(v) No, let p(x) = x

^{2}+ kx + k

If p(x) has equal zeroes, then its discriminant should be zero.

D = B

^{2}-4AC = 0 ,..(j)

On comparing p(x) with Ax

^{2}+ Bx + C, we get

A =1 B = k and C = k

∴ (k)

^{2}-4(1)(k) = 0 [from Eq. (i)]

⇒ k(k- 4)=0

⇒ k =0, 4

So, the quadratic polynomial p(x) have equal zeroes only at k =0, 4.

**Question 2:
**Are the following statements True’ or ‘False’? Justify your answer.

(i) If the zeroes of a quadratic polynomial ax

^{2}+ bx + c are both positive, then a, b and c all have the same sign.

(ii) If the graph of a polynomial intersects the X-axis at only one point, it cannot be a quadratic polynomial.

(iii) If the graph of a polynomial intersects the X-axis at exactly two points, it need not be a quadratic polynomial.

(iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.

(v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign.

(vi) If all three zeroes of a cubic polynomial x

^{3}+ax

^{2}-bx + c are positive, then atleast one of a, b and c is non-negative.

(vii) The only value of k for which the quadratic polynomial kx

^{z}+ x + k has equal zeroes is \( \frac { 1 }{ 2 } \)

**Solution:**

(ii) True, if the graph of a polynomial intersects the X-axis at only one point, then it cannot be a quadratic polynomial because a quadratic polynomial may touch the X-axis at exactly one point or intersects X-axis at exactly two points or do not touch the X-axis.

(iii) True, if the graph of a polynomial intersects the X-axis at exactly two points, then it may or may not be a quadratic polynomial. As, a polynomial of degree more than z is possible which intersects the X-axis at exactly two points when it has two real roots and other imaginary roots.

(iv) True, let a, p and y be the zeroes of the cubic polynomial and given that two of the zeroes have value 0.

Let α =β = 0

and f(x) = (x- α)(x-β)(x-γ)

= (x-a)(x-0)(x-0)

= x

^{3}– ax

^{2 }which does not have linear and constant terms.

(v) True, if f(x) = ax

^{3}+ bx

^{2}+ cx + d. Then, for all negative roots, a, b, c and d must have same sign.

(vii) False, let f(x) = kx

^{2}+ x + k

For equal roots. Its discriminant should be zero i.e., D = b

^{2}– 4ac = 0

⇒ 1-4k.k = 0

=> k = ± \( \frac { 1 }{ 2 } \)

So, for two values of k, given quadratic polynomial has equal zeroes

**Exercise 2.3 Short Answer TypeQuestions**

**Question 1:
**Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials

**(i) 4x**

^{2}– 3x – 1.**Solution:**

**(ii) 3x ^{2} + 4x – 4.
Solution:
**

**(iii) 51 ^{2} + 12t + 7.
Solution:
**

**(iv) t ^{3} – 2t^{2} – 15t.**

Solution:

**(v) \(2{ x }^{ 2 }+\frac { 7 }{ 2 } x+\frac { 3 }{ 4 } \)
Solution:
**

**(vi) 4×2 +5√2x – 3.
Solution:
**

**(vii) 2s ^{2} -(1+2√2)s +√2
Solution:
**

**(viii) v ^{2} + 4√3v – 15.
Solution:
**

**(ix) y ^{2} + \(\frac { 3 }{ 2 } \)√5y – 5.
Solution:
**

**(x) \(7{ y }^{ 2 }-\frac { 11 }{ 3 } y-\frac { 2 }{ 3 } .\)
Solution:
**

**Exercise 2.4 Long Answer Type Questions
**

**Question 1:
**For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorisation.

**Solution:**

Using factorisation method, = 2√5 x

^{2}+ 5x – 2x – √5

= √5x (2x + √5)-1(2x + √5)

= (2x + √5) (√5x – 1)

Hence,the zeroes of f(x) are \(-\frac{\sqrt{5}}{2}\) and \(\frac { 1 }{ \sqrt { 5 } }\)

**Question 2:
**If the zeroes of the cubic polynomial x

^{3}– 6x

^{2}+ 3x + 10 are of the form a,a + b and a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

**Solution:**

Using factorisation method,

a

^{2}-5a+a-5 = 0 =3

⇒ a (a – 5) + 1 (a – 5) = 0

⇒ (a – 5) (a + 1) = 0

⇒ a = -1, 5

when a = -1, then b = 3

When a = 5, then b = – 3 [using Eq. (i)]

∴Required zeroes of f(x) are

When a = -1 and b = 3

then, a,(a+b),(a + 2) = -1, (-1+3), (-1+6) or -1,2, 5

When a = 5and b = -3 then

a, (a + b), (a + 2b) = 5, (5 -3), (5 -6) or 5,2,-1.

Hence, the required values of a and b are a = – 1 and d = 3ora = 5, b = -3 and the zeroes are -1,2 and 5.

**Question 3:
**If √2 is a zero of the cubic polynomial 6x

^{3}+ √2x

^{2}– 10x – 4√2, the find its other two zeroes.

**Solution:**

Let f(x) = 6x

^{3}+ √2x

^{2}-10x – 4√2 and given that. √2 is one of the zeroes of f(x) i.e.,(x – √2) is one of the factor of given cubic polynomial.

Now, using division algorithm,

**Question 4:
**Find k, so that x

^{2}+ 2x + k is a factor of 2x

^{4}+ x

^{3}– 14x

^{2}+ 5x + 6. Also, find all the zeroes of the two polynomials.

**Solution:**

Given that, x

^{2}+ 2x+ k is a factor of 2x

^{4}+ x

^{3}-14x

^{2}+ 5x+ 6, then we apply division algorithm,

**Question 5:
**If x – √5 is a factor of the cubic polynomial x

^{3}– 3√5x

^{2}+ 13x – 3√5, then find all the zeroes of the polynomial.

**Solution:**

Let f(x) = x

^{3}– 3√5x

^{2}+ 13x – 3√5 and given that, (x – √5) is a one of the factor of f(x). Now, using division algorithm,

**Question 6:
**For which values of a and b, the zeroes of q(x) = x

^{3}+ 2x

^{2}+ a are also the zeroes of the polynomial p(x) = x

^{5}– x

^{4}– 4x

^{3}+ 3x

^{2}+ 3x + b? Which zeroes of p(x) are not the zeroes of p(x)?

**Solution:**

Given that the zeroes of q(x) = x

^{3}+ 2x

^{2}+ a are also the zeroes of the polynomial p(x) = x

^{5}– x

^{4}– 4x

^{3}+ 3x

^{z}+ 3x + b i.e., q(x) is a factor of p(x). Then, we use a division algorithm.

If (x

^{3}+ 2x

^{2}+ a) is a factor of (x

^{5}– x

^{4}– 4x

^{3}+ 3x

^{2}+ 3x + b), then remainder should be zero.

i.e., – (1 + a) x

^{2}+ (3 + 3a) x + (b – 2a) = 0

= 0.x

^{2}+ 0.x+0

On comparing the coefficient of x, we get

a + 1 = 0

⇒a = -1

and b – 2a = 0

⇒b =2a

b = 2(-1) = -2 [va = -1]

For a = -1 and b = – 2, the zeroes ofq(x) are also the zeroes of the polynomial p(x).

∴ q(x) = x

^{3}+ 2x

^{2}-1

and p(x) = x

^{5}– x

^{4}– 4x

^{3}+ 3x

^{2}+ 3x – 2

Now, Divident = divisor xquotient + remainder

p(x) = (x

^{3}+2x

^{2}-1)(x

^{2}-3x + 2)+ 0

= (x

^{3}+ 2x

^{2}-1){x

^{2}-2x – x + 2}

= (x

^{3}+ 2x

^{2}– 1) (x – 2) (x – 1)

Hence, the zeroes of p(x) are land 2 which are not the zeroes of q(x).

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