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(d) For any quadratic polynomial ax² + bx + c, a≠0, the graph of the Corresponding equation y = ax2 + 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.

📝 Solution:

(i) No. because whenever we divide a polynomial x⁶ + 2x³ + 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

= ax⁵ + bx⁴ + cx³ + dx² + ex + f

quotient = x² -1

and dividend = x⁶ + 2x³ + x -1

By division algorithm for polynomials,

Dividend = Divisor x Quotient + Remainder

= (ax⁵ + bx⁴ + cx³ + dx² + ex + f)x(x² -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, x² -1 is not a required quotient.

📝 Solution:

📝 Solution:

📝 Solution:

True, if f(x) = ax³ + bx² + cx + d. Then, for all negative roots, a, b, c and d must have same sign.

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📝 Solution:

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📝 Solution:

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)

= x³ – ax²

which does not have linear and constant terms.

📝 Solution:

📝 Solution:

📝 Solution:

📝 Solution:

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.

📝 Solution:

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.

📝 Solution:

Given that, one of the zeroes of the quadratic polynomial say p(x) = (k – 1) x² + kx + 1 is -3,

then p(- 3) = 0

(k – 1) (- 3)² + k(- 3) + 1 = 0

9(k – 1) – 3k + 1 = 0

9k – 9 – 3k + 1 = 0

6k – 8 = 0

k = 4/3

📝 Solution:

False, let f(x) = kx² + x + k

For equal roots. Its discriminant should be zero i.e., D = b² – 4ac = 0

⇒ 1-4k.k = 0

=> k = ± 1/2

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

📝 Solution:

📝 Solution:

📝 Solution:

📝 Solution:

(ii) Given that, Divisor px³ + gx² + rx + s, p≠0 and dividend = ax² + bx + c

We see that,

Degree of divisor > Degree of dividend

So, by division algorithm,

quotient = 0 and remainder = ax2 + bx + c

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