Here you get the CBSE Class 10 Mathematics chapter 2, Polynomials: NCERT Exemplar Problems and Solutions (Part-II). This part of the chapter includes solutions for Exercise 2.2 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Polynomials. This exercise comprises of only the Very Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

**NCERT Exemplar Solution for CBSE Class 10 Mathematics Chapter: Polynomials (Part-I)**

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

**CBSE Class 10 Mathematics Syllabus 2017-2018**

**Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Polynomials:**

**Exercise 2.2**

**Very Short Answer Type Questions**

**Q. 1 **Answer the following and justify.** **

(i) Can *x*^{2} – 1 be the quotient on division of *x*^{6} + 2*x*^{3} + *x* − 1 by a polynomial in *x* of degree 5?

(ii) What will the quotient and remainder be on division of *ax*^{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*)?

(iv) 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?

**Sol. **

**(i)** No.

Let, the divisor polynomial in *x* of degree 5 be g (x) = *ax*^{5} + *bx*^{4} + *cx*^{3} + *dx*^{2} + *ex* + *f*

Dividend is, *f* (*x*) = *x*^{6} + 2*x*^{3} + *x* − 1

And quotient, *q* (*x*) = *x*^{2} − 1

Now, by division algorithm for polynomials,

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

⟹ [deg *p*(*x*) is 6] = [deg *g*(*x*) is 5] [deg *q*(*x*) is 2] + deg *r*(*x*) is less than 5

Polynomial *p*(*x*) of degree 6 = (a polynomial of degree 7) + Remainder

So, division algorithm is not satisfied.

Hence, *x*^{2} − 1 is not a required quotient.

(ii) Given, divisor, *f* (*x*) = *px*^{3} + *qx*^{2} + *rx* + *s*, *p* ≠ 0

And dividend, *g* (*x*) = *ax*^{2} + *bx* + *c* by

As we see that, Degree of divisor > Degree of dividend so, by division algorithm, when we divide g(x) by *f* (*x*), quotient will be zero and remainder will be *g* (*x*).

(iii) If on division of a polynomial *p* (*x*) by a polynomial *g* (*x*), the quotient is zero, then relation between the degrees of *p *(*x*) and *g *(*x*) will be:

degree of *p *(*x*) < degree of *g *(*x*).

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

Or degree of ≤ degree of .

(v) No.

Let the given polynomial be, *f* (*x*) = *x*^{2} + *kx* + *k*

For *f* (*x*) to have equal roots, its discriminant must be zero.

⟹ *b*^{2} – 4*ac* = 0

⟹ (*k*)^{2} – 4(1) (*k*) = 0

⟹ *k*^{2} – 4*k* = 0

⟹ *k* (*k* – 4) = 0

⟹ *k* = 0 or *k* = 4

But *k* > 1 so *k* = 4

So, the quadratic polynomial *f* (*x*) has equal zeroes only at* k* = 4.

**Q. 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*^{2} + *x* + *k* has equal zeroes is 1/2.

**Sol. **

**(i)** False.

Let *α* and *β* be the roots of the quadratic polynomial. If both *α* and *β* are positive, then from

*α* + *β* = −b/ a we conclude that −b/ a is negative. But sum of two positive numbers (α and β here) ,must be positive, i.e., either *b *or *a* must be negative. So, a, b and c will have different signs.

(ii) False.

The given statement is false, because when two zeroes of a quadratic polynomial are equals, then two intersecting points coincide to become one point.

(iii) True.

If a polynomial of degree more than two has two real roots and other roots are imaginary, then graph of the polynomial will intersect at two points on x-axis.

(iv) True.

Let *α*, *β* and *γ* be the zeroes of the cubic polynomial and given that two of the zeroes have value zero.

which does not have linear and constant terms.

**(v)** True.

Let given polynomial be *f *(*x*) = *ax*^{3} + *bx*^{2} – *cx *+ *d* with all its three roots say, *α*, *β* and *γ* being negative. Then,

(vi) False.

Let *α*, *β* and *γ* be the three zeroes of cubic polynomial* x*^{3} + *ax*^{2} – *bx *+ *c* hen, product of zeroes

Then,

So, the cubic polynomial has all three zeroes which are positive only when all constants a, b and c are negative.

(vii) False.

Let *f *(*x*) = *kx*^{2} + *x* + *k*

For equal roots. Its discriminant should be zero

⟹ *b*^{2} – 4*ac* = 0

⟹ 1 – 4*k *. *k *= 0

⟹ 1 – 4*k*^{2} = 0

⟹ *k* = ±1 / 2

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

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