CBSE Class 10 Polynomials Sure Shot Questions Set 06

Question. Let \( \alpha, \beta \) be the zeros of the polynomial \( x^2 – px + r \) and \( \frac{\alpha}{2}, 2\beta \) be the zeros of \( x^2 – qx + r \). Then the value of r is –
(a) \( \frac{2}{9}(p – q)(2q – p) \)
(b) \( \frac{2}{9}(q – p)(2p – q) \)
(c) \( \frac{2}{9}(q – 2p)(2q – p) \)
(d) \( \frac{2}{9}(2p – q)(2q – p) \)
Answer: (d) \( \frac{2}{9}(2p – q)(2q – p) \)

Question. When \( x^{200} + 1 \) is divided by \( x^2 + 1 \), the remainder is equal to –
(a) \( x + 2 \)
(b) \( 2x – 1 \)
(c) 2
(d) – 1
Answer: (c) 2

Question. If \( a(p + q)^2 + 2bpq + c = 0 \) and also \( a(q + r)^2 + 2bqr + c = 0 \) then pr is equal to –
(a) \( p^2 + \frac{a}{c} \)
(b) \( q^2 + \frac{c}{a} \)
(c) \( p^2 + \frac{a}{b} \)
(d) \( q^2 + \frac{a}{c} \)
Answer: (b) \( q^2 + \frac{c}{a} \)

Question. If \( a,b \) and \( c \) are not all equal and \( \alpha \) and \( \beta \) be the zeros of the polynomial \( ax^2 + bx + c \), then value of \( (1 + \alpha + \alpha^2) (1 + \beta + \beta^2) \) is :
(a) 0
(b) positive
(c) negative
(d) non-negative
Answer: (d) non-negative

Question. If 2 and 3 are the zeros of \( f(x) = 2x^3 + mx^2 – 13x + n \), then the values of m and n are respectively –
(a) –5, – 30
(b) –5, 30
(c) 5, 30
(d) 5, – 30
Answer: (b) –5, 30

Question. If \( \alpha, \beta \) are the zeros of the polynomial \( 6x^2 + 6px + p^2 \), then the polynomial whose zeros are \( (\alpha + \beta)^2 \) and \( (\alpha – \beta)^2 \) is –
(a) \( 3x^2 + 4p^2x + p^4 \)
(b) \( 3x^2 + 4p^2x – p^4 \)
(c) \( 3x^2 – 4p^2x + p^4 \)
(d) None of these
Answer: (c) \( 3x^2 – 4p^2x + p^4 \)

Question. If \( c, d \) are zeros of \( x^2 – 10ax – 11b \) and \( a, b \) are zeros of \( x^2 – 10cx – 11d \), then value of \( a + b + c + d \) is
(a) 1210
(b) – 1
(c) 2530
(d) – 11
Answer: (a) 1210

Question. If the ratio of the roots of polynomial \( x^2 + bx + c \) is the same as that of the ratio of the roots of \( x^2 + qx + r \), then
(a) \( br^2 = qc^2 \)
(b) \( cq^2 = rb^2 \)
(c) \( q^2c^2 = b^2r^2 \)
(d) \( bq = rc \)
Answer: (b) \( cq^2 = rb^2 \)

Question. The quadratic polynomial whose zeros are twice the zeros of \( 2x^2 – 5x + 2 = 0 \) is –
(a) \( 8x^2 – 10x + 2 \)
(b) \( x^2 – 5x + 4 \)
(c) \( 2x^2 – 5x + 2 \)
(d) \( x^2 – 10x + 6 \)
Answer: (b) \( x^2 – 5x + 4 \)

Question. If \( \alpha, \beta, \gamma \) are the zeros of the polynomial \( x^3 – 3x + 11 \), then the polynomial whose zeros are \( (\alpha+\beta), (\beta+\gamma) \) and \( (\gamma+\alpha) \) is –
(a) \( x^3 + 3x + 11 \)
(b) \( x^3 – 3x + 11 \)
(c) \( x^3 + 3x – 11 \)
(d) \( x^3 – 3x – 11 \)
Answer: (d) \( x^3 – 3x – 11 \)

Question. If \( \alpha, \beta, \gamma \) are such that \( \alpha + \beta + \gamma = 2, \alpha^2 + \beta^2 + \gamma^2 = 6, \alpha^3 + \beta^3 + \gamma^3 = 8 \), then \( \alpha^4 + \beta^4 + \gamma^4 \) is equal to
(a) 10
(b) 12
(c) 18
(d) None
Answer: (c) 18

Question. If \( \alpha, \beta \) are the roots of \( ax^2 + bx + c \) and \( \alpha + k, \beta + k \) are the roots of \( px^2 + qx + r \), then k =
(a) \( \frac{1}{2} \left[ \frac{a}{b} - \frac{p}{q} \right] \)
(b) \( \left[ \frac{a}{b} - \frac{p}{q} \right] \)
(c) \( \frac{1}{2} \left[ \frac{b}{a} - \frac{q}{p} \right] \)
(d) \( (ab – pq) \)
Answer: (c) \( \frac{1}{2} \left[ \frac{b}{a} - \frac{q}{p} \right] \)

Question. The condition that \( x^3 – ax^2 + bx – c = 0 \) may have two of the roots equal to each other but of opposite signs is :
(a) \( ab = c \)
(b) \( \frac{2}{3}a = bc \)
(c) \( a^2b = c \)
(d) None
Answer: (a) \( ab = c \)

Question. If one zero of the polynomial \( ax^2 + bx + c \) is positive and the other negative then \( (a,b,c \in R, a \neq 0) \)
(a) a and b are of opposite signs.
(b) a and c are of opposite signs.
(c) b and c are of opposite signs.
(d) a,b,c are all of the same sign.
Answer: (b) a and c are of opposite signs.

Question. If \( \alpha, \beta \) are the zeros of the polynomial \( x^2 – px + q \), then \( \frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2} \) is equal to –
(a) \( \frac{p^4}{q^2} + 2 - \frac{4p^2}{q} \)
(b) \( \frac{p^4}{q^2} - 2 + \frac{4p^2}{q} \)
(c) \( \frac{p^4}{q^2} + 2q^2 - \frac{4p^2}{q} \)
(d) None of these
Answer: (a) \( \frac{p^4}{q^2} + 2 - \frac{4p^2}{q} \)

Question. If \( \alpha, \beta \) are the zeros of the polynomial \( x^2 – px + 36 \) and \( \alpha^2 + \beta^2 = 9 \), then p =
(a) \( \pm 6 \)
(b) \( \pm 3 \)
(c) \( \pm 8 \)
(d) \( \pm 9 \)
Answer: (d) \( \pm 9 \)

Question. If \( \alpha, \beta \) are zeros of \( ax^2 + bx + c \), \( ac \neq 0 \), then zeros of \( cx^2 + bx + a \) are –
(a) \( – \alpha, – \beta \)
(b) \( \frac{1}{\alpha}, \beta \)
(c) \( \frac{1}{\beta}, \alpha \)
(d) \( \frac{1}{\alpha}, \frac{1}{\beta} \)
Answer: (d) \( \frac{1}{\alpha}, \frac{1}{\beta} \)

Question. A real number is said to be algebraic if it satisfies a polynomial equation with integral coefficients. Which of the following numbers is not algebraic :
(a) 2/3
(b) \( \sqrt{2} \)
(c) 0
(d) \( \pi \)
Answer: (d) \( \pi \)

Question. The cubic polynomials whose zeros are 4, \( \frac{3}{2} \) and –2 is :
(a) \( 2x^3 + 7x^2 + 10x – 24 \)
(b) \( 2x^3 + 7x^2 – 10x – 24 \)
(c) \( 2x^3 – 7x^2 – 10x + 24 \)
(d) None of these
Answer: (c) \( 2x^3 – 7x^2 – 10x + 24 \)

Question. If the sum of zeros of the polynomial \( p(x) = kx^3 – 5x^2 – 11x – 3 \) is 2, then k is equal to
(a) \( k = – \frac{5}{2} \)
(b) \( k = \frac{2}{5} \)
(c) \( k = 10 \)
(d) \( k = \frac{5}{2} \)
Answer: (d) \( k = \frac{5}{2} \)

Question. If \( f(x) = 4x^3 – 6x^2 + 5x – 1 \) and \( \alpha, \beta \) and \( \gamma \) are its zeros, then \( \alpha\beta\gamma = \)
(a) 3/2
(b) 5/4
(c) – 3/2
(d) 1/4
Answer: (d) 1/4

Question. Consider \( f(x) = 8x^4 – 2x^2 + 6x – 5 \) and \( \alpha, \beta, \gamma, \delta \) are it's zeros then \( \alpha + \beta + \gamma + \delta = \)
(a) 1/4
(b) – 1/4
(c) – 3/2
(d) None
Answer: (d) None

Question. If \( x^2 – ax + b = 0 \) and \( x^2 – px + q = 0 \) have a root in common and the second equation has equal roots, then
(a) \( b + q = 2ap \)
(b) \( b + q = \frac{ap}{2} \)
(c) \( b + q = ap \)
(d) None of these
Answer: (b) \( b + q = \frac{ap}{2} \)