CBSE Class 10 Mathematics Polynomials MCQs Set 21

Answer: (d) -32

Explanation:
1. \( p(-3) = (-3)^3 - (-3)^2 + (-3) + 7 \)
2. \( = -27 - 9 - 3 + 7 = -32 \)
3. Hence the value is -32.

Teacher's Note:
1. Put the number in brackets so signs are not lost.
2. Powers of a negative number: odd power stays negative, even power becomes positive.

Answer: (c) 6

Explanation:
1. \( p(1) = 1 + 5 - 6 + 2 = 2 \)
2. \( q(-1) = -6 + 4 - 3 + 1 = -4 \)
3. \( p(1) - q(-1) = 2 - (-4) = 6 \)

Teacher's Note:
1. Find each value separately before subtracting.
2. Subtracting a negative number adds its size.

Answer: (b) a = -1

Explanation:
1. If \( (x - a) \) is a factor, then the polynomial is 0 at \( x = a \).
2. \( a^6 - a^6 + a^4 - a^4 + 3a - a + 2 = 0 \)
3. \( 2a + 2 = 0 \), so \( a = -1 \).

Teacher's Note:
1. Factor Theorem: (x - a) divides f(x) exactly when f(a) = 0.
2. Pairs of equal terms cancel here, which makes the work short.

Answer: (c) not defined

Explanation:
1. The zero polynomial has no non-zero term.
2. Without a non-zero term there is no highest power, so its degree is not defined.

Teacher's Note:
1. A non-zero constant such as 5 has degree 0, but the zero polynomial itself has no defined degree.

Answer: (b) -37

Explanation:
1. \( 2(-2)^3 - 3(-2)^2 + 7(-2) + 5 \)
2. \( = -16 - 12 - 14 + 5 = -37 \)

Teacher's Note:
1. Evaluate each term separately: 2 × (-8) = -16 and 3 × 4 = 12.
2. Add the four results at the end.

Answer: (c) \( \frac{61}{4} \)

Explanation:
1. \( \alpha + \beta = -\frac{7}{2} \) and \( \alpha\beta = -\frac{3}{2} \).
2. \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{49}{4} + 3 = \frac{61}{4} \)

Teacher's Note:
1. Use the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \) instead of finding the zeroes.
2. The product is negative, so -2αβ adds 3.

Answer: (b) a parabola opening downward

Explanation:
1. The graph of a quadratic polynomial is a parabola.
2. Its shape depends on the sign of a: it opens upward if a > 0 and downward if a < 0.
3. Since a < 0, the parabola opens downward.

Teacher's Note:
1. The variable y is a function of x, so a vertical line meets the graph at most once. Sideways parabolas and ovals are not graphs of polynomials.

Answer: (d) 4

Explanation:
1. Zeroes are the x-coordinates of the points where the graph meets the x-axis.
2. The curve crosses the x-axis at four points, so there are 4 zeroes.

Teacher's Note:
1. Count only the crossing or touching points on the x-axis, not the turning points.
2. A polynomial of degree n has at most n zeroes, so this polynomial has degree 4 or more.

Answer: (d) 2

Explanation:
1. Reading the graph, the curve meets the x-axis at x = -2, x = 1 and x = 3.
2. Sum = -2 + 1 + 3 = 2.

Teacher's Note:
1. Read the zeroes only where the curve actually meets the axis.
2. The marks at -1, 0 and 2 are only scale points; the curve does not cut the axis there.

Answer: (a) 1

Explanation:
1. The zeroes of this polynomial are -2, 1 and 3.
2. Between -2 and 2, excluding both, only x = 1 lies in the interval.

Teacher's Note:
1. Check carefully whether the end values are included.
2. Here -2 is a zero but the question excludes it.

Answer: (d) none of these

Explanation:
1. Dividing by \( x^2 - 3 \) gives quotient \( 3x^2 + 5x + 2 = (3x + 2)(x + 1) \).
2. So the polynomial is \( (x^2 - 3)(3x + 2)(x + 1) \).
3. Its zeroes are \( \pm\sqrt{3}, -1, -\frac{2}{3} \), which match none of the options.

Teacher's Note:
1. Divide first, then factorise the quotient.
2. Always list all four zeroes of a degree 4 polynomial.

Answer: (c) -13

Explanation:
1. \( f(x) = (x - 1)(x^2 - 4x + 4) + 5 \).
2. By the Remainder Theorem the remainder on division by \( x + 1 \) is \( f(-1) \).
3. \( f(-1) = (-2)(1 + 4 + 4) + 5 = -18 + 5 = -13 \).

Teacher's Note:
1. Rebuild f(x) from dividend = divisor × quotient + remainder.
2. Remainder on dividing by (x + 1) is the value at x = -1.

Answer: (d) none of these

Explanation:
1. Dividing by \( 2x^2 + 4 \) leaves remainder \( 8x + 5 \); dividing by \( 4 - 2x^2 \) leaves \( 8x + 5 \); dividing by \( 2x^2 + 5 \) leaves \( 8x + \frac{19}{2} \).
2. No remainder is zero, so f(x) is divisible by none of them.

Teacher's Note:
1. A polynomial g(x) divides f(x) only when the remainder is 0.
2. Compare degrees: dividing a degree 4 polynomial by a quadratic leaves a remainder of degree at most 1.

Answer: (d) -4

Explanation:
1. Sum of the three zeroes \( = -\frac{5}{1} = -5 \).
2. Let the third zero be t: \( 2 + (-3) + t = -5 \), so \( t = -4 \).
3. Check: product \( 2 \times (-3) \times (-4) = 24 = -\frac{d}{a} \).

Teacher's Note:
1. For a cubic, the zeroes add up to -b/a.
2. The product of the zeroes must equal -d/a as a check.

Answer: (b) -2, 8

Explanation:
1. Dividing, the quotient is \( 2x^2 - 5x + (2m + 12) \) and the remainder is \( (-11m - 18)x + (2m^2 + 12m + 24) \).
2. Comparing with \( 4x + n \): \( -11m - 18 = 4 \), so \( m = -2 \).
3. Then \( n = 2(4) + 12(-2) + 24 = 8 \).

Teacher's Note:
1. Match the x-coefficient of the remainder first to find m.
2. Then substitute m into the constant term to get n.

Answer: (c) x² + x - 2

Explanation:
1. Sum of zeroes = 1 + (-2) = -1; product = -2.
2. Polynomial = \( x^2 - (\text{sum})x + \text{product} = x^2 + x - 2 \).

Teacher's Note:
1. Remember the minus sign in front of the sum.
2. Verify by factorising: \( x^2 + x - 2 = (x - 1)(x + 2) \).

Answer: (a) -2

Explanation:
1. \( \alpha + \beta = -\frac{5}{2} \) and \( \alpha\beta = \frac{1}{2} \).
2. \( \alpha + \beta + \alpha\beta = -\frac{5}{2} + \frac{1}{2} = -2 \).

Teacher's Note:
1. Use sum = -b/a and product = c/a directly.
2. No need to find the zeroes themselves.

Answer: (c) x² + 5x + 3

Explanation:
1. Polynomial = \( x^2 - (\text{sum})x + \text{product} \).
2. = \( x^2 - (-5)x + 3 = x^2 + 5x + 3 \).

Teacher's Note:
1. A negative sum gives a positive middle term.
2. Check: sum of zeroes = -b/a = -5.

Answer: (d) x² - x + 1

Explanation:
1. Dividend \( = g(x) \times q(x) + r(x) \), so \( g(x)(x - 2) = x^3 - 3x^2 + x + 2 - (4 - 2x) = x^3 - 3x^2 + 3x - 2 \).
2. Dividing by \( x - 2 \) gives \( g(x) = x^2 - x + 1 \).

Teacher's Note:
1. First subtract the remainder, then divide by the quotient.
2. Multiply back \( (x - 2)(x^2 - x + 1) \) to check.

Answer: (d) x - 2

Explanation:
1. Dividend minus remainder \( = 3x^2 - x^3 - 3x + 2 \).
2. Divisor is \( -x^2 + x - 1 \). Dividing \( -x^3 + 3x^2 - 3x + 2 \) by \( -x^2 + x - 1 \) gives quotient \( x - 2 \).
3. Check: \( (-x^2 + x - 1)(x - 2) + 3 = -x^3 + 3x^2 - 3x + 5 \), the dividend.

Teacher's Note:
1. Write both polynomials in descending powers before dividing.
2. Always verify with dividend = divisor × quotient + remainder.

Answer: (d) 2x² - 4x - 1

Explanation:
1. \( f(x) = k\left(x^2 - 2x - \frac{1}{2}\right) \).
2. Taking \( k = 2 \) gives \( 2x^2 - 4x - 1 \).

Teacher's Note:
1. Clear fractions by choosing a suitable constant k.
2. Only the option with ratio 1 : -2 : -1/2 is correct.

Answer: (d) \( \frac{4}{5} \)

Explanation:
1. \( \alpha + \beta = -\frac{4}{16} = -\frac{1}{4} \) and \( \alpha\beta = -\frac{5}{16} \).
2. \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-1/4}{-5/16} = \frac{4}{5} \).

Teacher's Note:
1. Combine the fractions first: 1/α + 1/β = (α + β)/αβ.
2. Two negatives give a positive answer.

Answer: (a) a = 0, b = -6

Explanation:
1. Sum of zeroes \( = 2 - 3 = -1 = -(a + 1) \), so \( a = 0 \).
2. Product \( = -6 = b \).

Teacher's Note:
1. Compare sum with -(coefficient of x) and product with the constant term.
2. Check by substituting x = 2: 4 + 2(1) - 6 = 0.

Answer: (b) -13

Explanation:
1. \( \alpha + \beta = 5 \) and \( \alpha\beta = 6 \).
2. \( 5 - 3 \times 6 = 5 - 18 = -13 \).

Teacher's Note:
1. Use sum and product directly.
2. The factors are (x - 2)(x - 3), so the zeroes 2 and 3 confirm the result.

Answer: (b) \( \frac{\sqrt{3}}{4},\ -\frac{2}{\sqrt{3}} \)

Explanation:
1. \( 4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} \).
2. \( = 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = (\sqrt{3}x + 2)(4x - \sqrt{3}) \).
3. Zeroes: \( x = -\frac{2}{\sqrt{3}} \) and \( x = \frac{\sqrt{3}}{4} \).

Teacher's Note:
1. Split 5x into two terms whose product equals (4√3)(-2√3) = -24 x².
2. Check: the product of the zeroes is -2√3/(4√3) = -1/2.