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 x² – 1 be the quotient on division of x⁶ + 2x³ + x − 1 by a polynomial in x of degree 5?

(ii) What will the quotient and remainder be on division  of ax² + bx + c by px³ + qx² + 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 x² + 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) = ax⁵ + bx⁴ + cx³ + dx² + ex + f

Dividend is,  f (x) = x⁶ + 2x³ + x − 1

And quotient, q (x) = x² − 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, x² − 1 is not a required quotient.

(ii) Given, divisor,  f (x) = px³ + qx² + rx + s, p ≠ 0

And dividend, g (x) = ax² + 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) egree 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) = x² + kx + k

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

⟹       b² – 4ac = 0    

⟹       (k)² – 4(1) (k) = 0

⟹       k² – 4k = 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 ax² + 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 x³ + ax² – 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 kx² + 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) = ax³ + bx² – 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 x³ + ax² – 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) = kx² + x + k

For equal roots. Its discriminant should be zero

⟹  b² – 4ac = 0

⟹  1 – 4k . k = 0

⟹   1 – 4k² = 0          

⟹    k = ±1 / 2

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