CBSE Class 10 Mathematics Polynomials MCQs Set 09

Question. If a cubic polynomial with the sum of its zeroes, sum of the products and its zeroes taken two at a time and product of its zeroes as 2, \(-5\) and \(-11\) respectively, then the cubic polynomial is
(a) \(x^3 + 7x - 6\)
(b) \(x^3 + 7x + 6\)
(c) \(x^3 - 7x - 6\)
(d) \(x^3 - 7x + 6\)
Answer: (d) \(x^3 - 7x + 6\)
Let \(\alpha, \beta, \gamma\) be the zeros of the required polynomials
\(\alpha + \beta + \gamma = 0\)
\(\alpha\beta + \beta\gamma + \gamma\alpha = -7\)
\(\alpha\beta\gamma = -6\)
Required cubic polynomial is \(k[x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma]\) where \(k\) is non-zero constant
\(k[x^3 + (0)x^2 + (-7)x - (-6)] = x^3 - 7x + 6\) [consider, \(k = 1\)]

Question. If \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial \(f(x) = ax^2 + bx + c\), then the value of \(\alpha^4 + \beta^4\) is
(a) \(\frac{(b^2 - 2ac)^2 + a^2c^2}{a^4}\)
(b) \(\frac{(b^2 + 2ac)^2 - a^2c^2}{a^4}\)
(c) \(\frac{(b^2 - 2ac)^2 - 2a^2c^2}{a^4}\)
(d) \(\frac{(b^2 + 2ac)^2 + 2a^2c^2}{a^4}\)
Answer: (c) \(\frac{(b^2 - 2ac)^2 - 2a^2c^2}{a^4}\)
Given, \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(f(x) = ax^2 + bx + c\)
Sum of zeroes, \(\alpha + \beta = -\frac{b}{a}\)
and product of zeroes, \(\alpha\beta = \frac{c}{a}\)
Now, \(\alpha^4 + \beta^4 = (\alpha^2)^2 + (\beta^2)^2 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2\)
\(= [(\alpha + \beta)^2 - 2\alpha\beta]^2 - 2(\alpha\beta)^2\)
On substituting, \(\alpha + \beta = -\frac{b}{a}\) and \(\alpha\beta = \frac{c}{a}\) in the above equation, we get
\(\alpha^4 + \beta^4 = [(\frac{-b}{a})^2 - 2(\frac{c}{a})]^2 - 2(\frac{c}{a})^2\)
\(= [\frac{b^2}{a^2} - \frac{2c}{a}]^2 - \frac{2c^2}{a^2}\)
\(= [\frac{b^2 - 2ac}{a^2}]^2 - \frac{2c^2}{a^2}\)
\(= \frac{(b^2 - 2ac)^2}{a^4} - \frac{2a^2c^2}{a^4} = \frac{(b^2 - 2ac)^2 - 2a^2c^2}{a^4}\)

Question. The polynomial \(f(x) = ax^3 + bx - c\) is divisible by the polynomial \(g(x) = x^2 + bx + c\), \(c \neq 0\), if
(a) \(ab = 2\)
(b) \(ab = 1\)
(c) \(ac = 2\)
(d) \(c = 2b\)
Answer: (b) \(ab = 1\)
If \(ax^3 + bx - c\) is exactly divisible by \(x^2 + bx + c\), then the remainder should be zero.
On dividing, we get Quotient = \(ax - ab\) and,
remainder = \((ab^2 + b - ac)x + abc - c = 0\)
\((ab^2 + b - ac)x + abc - c = 0\)
Comparing the coefficient of \(x\) and constant term both sides, we get
\(ab^2 + b - ac = 0\) and \(abc - c = 0 \Rightarrow ab = 1\)

Question. If one of the zeroes of a quadratic polynomial of the form \(x^2 + ax + b\) is the negative of the other, then which of the following is correct?
(a) Polynomial has linear factors
(b) Constant term of polynomial is negative
(c) Both (a) and (b) are correct
(d) Neither (a) nor (b) is correct
Answer: (c) Both (a) and (b) are correct
Let given polynomial be, \(p(x) = x^2 + ax + b\)
Again, let \(\alpha\) and \(\beta\) be the zeroes of \(p(x)\).
Then, product of zeroes = \(\frac{Constant term}{Coefficient of x^2}\)
\(\alpha\beta = \frac{b}{1} \Rightarrow \alpha\beta = b\) ...(1)
Since, one of the zeroes of the quadratic polynomial \(p(x)\) is negative of the other.
\(\alpha\beta < 0 \Rightarrow b < 0\) [from Eq. (1)]
So, \(b\) should be negative.
As, \((x - \alpha)\) and \((x - \beta)\) are the factors of polynomial \(p(x)\).
Then, \(p(x) = (x - \alpha)(x - \beta) = (x - \alpha)(x + \alpha) = x^2 - \alpha^2 = x^2 - k\) [\(\beta = -\alpha\)] [where, \(\alpha^2 = k\) is positive term]
So, \(x^2 + ax + b = x^2 - k\)
Then, \(a = 0\) and \((x + \sqrt{k}), (x - \sqrt{k})\) are the linear factors of polynomial \(p(x)\).
Hence, if one of the zeroes of quadratic polynomial \(p(x)\) is the negative of the other, then it has linear factor and the constant term is negative, i.e. \(b < 0\).

Question. If \(\alpha, \beta\) and \(\gamma\) are the zeroes of the polynomial \(p(x) = ax^3 + 3bx^2 + 3cx + d\) and having relation \(2\beta = \alpha + \gamma\), then \(2b^3 - 3abc + a^2d\) is
(a) \(-1\)
(b) 1
(c) 0
(d) None of the above
Answer: (c) 0
Given, \(p(x) = ax^3 + 3bx^2 + 3cx + d\)
On comparing with \(Ax^3 + Bx^2 + Cx + D\), we get \(A = a, B = 3b, C = 3c\) and \(D = d\)
Then, sum of zeroes, \(\alpha + \beta + \gamma = -\frac{B}{A} = -\frac{3b}{a}\) ...(1)
Product of zeroes taken two at a time, \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{C}{A} = \frac{3c}{a}\) ...(2)
and product of all zeroes, \(\alpha\beta\gamma = -\frac{D}{A} = -\frac{d}{a}\) ...(3)
Also, \(2\beta = \alpha + \gamma\) [given]
\(2\beta = -\frac{3b}{a} - \beta\) [From Eq. (1)]
\(3\beta = -\frac{3b}{a} \Rightarrow \beta = -\frac{b}{a}\)
From eq. (3), \((\alpha\gamma)\beta = -\frac{d}{a} \Rightarrow \alpha\gamma = -\frac{d}{a\beta} = -\frac{d}{a(-b/a)} = \frac{d}{b}\)
From Eq. (2), \(\beta(\alpha + \gamma) + \gamma\alpha = \frac{3c}{a}\)
\(\beta \times 2\beta + \gamma\alpha = \frac{3c}{a}\) [\(\alpha + \gamma = 2\beta\)]
\(2\beta^2 + \frac{d}{b} = \frac{3c}{a}\) [\(\alpha\gamma = \frac{d}{b}\)]
\(2(-\frac{b}{a})^2 + \frac{d}{b} = \frac{3c}{a}\) [\(\beta = -\frac{b}{a}\)]
\(\frac{2b^2}{a^2} + \frac{d}{b} = \frac{3c}{a}\)
\(\frac{2b^3 + a^2d}{a^2b} = \frac{3c}{a}\)
\(2b^3 + a^2d = \frac{3a^2bc}{a} = 3abc\)
\(2b^3 - 3abc + a^2d = 0\) Hence proved.

Question. If the square of difference of the zeroes of the quadratic polynomial \(x^2 + px + 45\) is equal to 144, then the value of p is
(a) \(\pm 9\)
(b) \(\pm 12\)
(c) \(\pm 15\)
(d) \(\pm 18\)
Answer: (d) \(\pm 18\)
Given that, \(f(x) = x^2 + px + 45\)
Then, \(\alpha + \beta = -\frac{p}{1} = -p\)
and \(\alpha\beta = \frac{45}{1} = 45\)
According to given condition, \((\alpha - \beta)^2 = 144\)
\((\alpha + \beta)^2 - 4\alpha\beta = 144\)
\((-p)^2 - 4(45) = 144\)
\(p^2 = 144 + 180\)
\(p^2 = 324 \Rightarrow p = \pm 18\)

Question. If \(\alpha\) and \(\beta\) are zeroes and the quadratic polynomial \(p(S) = 3S^2 - 6S + 4\), then the value of \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2(\frac{1}{\alpha} + \frac{1}{\beta}) + 3\alpha\beta\) is
(a) 7
(b) 6
(c) 8
(d) 10
Answer: (c) 8
Since, \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(p(S) = 3S^2 - 6S + 4\).
\(\alpha + \beta = -\frac{(-6)}{3} = 2\) and \(\alpha\beta = \frac{4}{3}\)
We have, \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2(\frac{1}{\alpha} + \frac{1}{\beta}) + 3\alpha\beta\)
\(= \frac{\alpha^2 + \beta^2}{\alpha\beta} + 2(\frac{\alpha + \beta}{\alpha\beta}) + 3\alpha\beta\)
\(= \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} + \frac{2(\alpha + \beta)}{\alpha\beta} + 3\alpha\beta\)
\(= \frac{(2)^2 - 2 \times \frac{4}{3}}{\frac{4}{3}} + \frac{2 \times 2}{\frac{4}{3}} + 3 \times \frac{4}{3}\)
\(= \frac{4 - \frac{8}{3}}{\frac{4}{3}} + 3 + 4 = \frac{4}{4} + 7 = 1 + 7 = 8\)

Question. Find the zeroes of the quadratic polynomial \(y^2 - 3y + 2\) with the help of the graph.
(a) \(1, -2\)
(b) \(\frac{-1}{4}, \frac{3}{2}\)
(c) \(6, -1\)
(d) 1, 2
Answer: (d) 1, 2

Question. If the sum of the zeroes of the equation \(\frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{c}\) is zero, then the product of zeroes of the equation is?
(a) \(\frac{a^2 + b^2}{2}\)
(b) \(-\frac{(a^2 + b^2)}{2}\)
(c) \(\frac{ab}{2}\)
(d) \(\frac{(a + b)^2}{2}\)
Answer: (b) \(-\frac{(a^2 + b^2)}{2}\)
Given equation is \(\frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{c}\)
\(\frac{x+b+x+a}{(x+a)(x+b)} = \frac{1}{c}\)
\(c(2x+a+b) = (x+a)(x+b)\)
\(2cx + (a+b)c = x^2 + (a+b)x + ab\)
\(x^2 + (a+b-2c)x + ab - ac - bc = 0\)
Let the zeroes of the above equation be \(\alpha\) and \(\beta\).
Given, \(\alpha + \beta = 0 \Rightarrow \frac{-(a+b-2c)}{1} = 0 \Rightarrow a+b = 2c\) ...(1)
Now, product of zeroes, \(\alpha\beta = \frac{(ab - ac - bc)}{1} = ab - (a+b)c\)
Substituting \(c = \frac{a+b}{2}\) from (1):
\(\alpha\beta = ab - (a+b)(\frac{a+b}{2}) = \frac{2ab - (a+b)^2}{2} = \frac{2ab - (a^2 + b^2 + 2ab)}{2} = -\frac{(a^2 + b^2)}{2}\)
\( = ab - (a + b)\left(\frac{a + b}{2}\right) \)
(From Eq. (1)]
\( = \frac{2ab - (a + b)^2}{2} \)
\( = \frac{2ab - (a^2 + b^2 + 2ab)}{2} \)
\( = -\frac{(a^2 + b^2)}{2} \)

Question. Draw the graph of the polynomial \( -x^2 + x + 2 \) and find the maximum value of the polynomial.
(a) 2
(b) \( \frac{5}{2} \)
(c) \( \frac{9}{4} \)
(d) None of these
Answer: (c)

FILL IN THE BLANK

Question. A .......... polynomial is of degree one.
Answer: Linear

Question. A cubic polynomial is of degree..........
Answer: Three

Question. We get the original number if we multiply the .......... together.
Answer: Factors

Question. Degree of remainder is always .......... than degree of divisor.
Answer: Smaller/less

Question. .......... equation is valid for all values of its variables.
Answer: Identity

Question. Polynomials of degrees 1, 2 and 3 are called .......... , .......... and .......... polynomials respectively.
Answer: linear, quadratic, cubic

Question. .......... is not equal to zero when the divisor is not a factor of dividend.
Answer: Remainder

Question. The zeroes of a polynomial \( p(x) \) are precisely the x-coordinates of the points, where the graph of \( y = p(x) \) intersects the .......... axis.
Answer: x

Question. The algebraic expression in which the variable has non-negative integral exponents only is called ..........
Answer: Polynomial

TRUE/FALSE

Question. A polynomial cannot have more than one zero
Answer: False, a polynomial can have any number of zeroes. It depends upon the degree of the polynomial.

Question. Graph of a quadratic polynomial is an ellipse.
Answer: False

Question. The degree of the sum of two polynomials each of degree 5 is always 5.
Answer: False, \( x^5 + 1 \) and \( -x^5 + 2x + 3 \) are two polynomials of degree 5 but the degree of the sum of the two polynomials is 1.

Question. \( \frac{6\sqrt{x} + x^{3/2}}{\sqrt{x}} \) is a polynomial, \( x \neq 0 \).
Answer: True, because \( \frac{6\sqrt{x} + x^{3/2}}{\sqrt{x}} = 6 + x \), which is a polynomial.

Question. Every polynomial equation has at least one real root.
Answer: False

Question. Product of zeroes of quadratic polynomial \( = -\frac{\text{constant term}}{(\text{coefficient of } x^2)} \)
Answer: False

Question. If \( p(x) = ax + b \) then zero of \( p(x) \) is \( \frac{-b}{a} \).
Answer: True

Question. Zeroes of quadratic polynomial \( x^2 + 7x + 10 \) are 2 and -5
Answer: False

Question. Sum of zeroes of \( 2x^2 - 8x + 6 \) is -4
Answer: False

Question. Degree of a quadratic polynomial is less than or equal to two.
Answer: False

ASSERTION AND REASON

DIRECTION : In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion : If \( \alpha, \beta, \gamma \) are the zeroes of \( x^3 - 2x^2 + qx - r \) and \( \alpha + \beta = 0 \), then \( 2q = r \).
Reason : If \( \alpha, \beta, \gamma \) are the zeroes of \( ax^3 + bx^2 + cx + d \), then \( \alpha + \beta + \gamma = -\frac{b}{a} \), \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \), \( \alpha\beta\gamma = -\frac{d}{a} \).
(a) A
(b) B
(c) C
(d) D
Answer: (a)
Clearly, Reason is true. [Standard Result]. For Assertion: \( \alpha + \beta + \gamma = -(-2) = 2 \). Given \( \alpha + \beta = 0 \), so \( 0 + \gamma = 2 \implies \gamma = 2 \). Product of zeroes \( \alpha\beta\gamma = -(-r) = r \implies \alpha\beta(2) = r \implies \alpha\beta = \frac{r}{2} \). Also, \( \alpha\beta + \beta\gamma + \gamma\alpha = q \implies \frac{r}{2} + \gamma(\alpha + \beta) = q \implies \frac{r}{2} + \gamma(0) = q \implies r = 2q \). Assertion is true.

Question. Assertion : \( (2 - \sqrt{3}) \) is one zero of the quadratic polynomial then other zero will be \( (2 + \sqrt{3}) \).
Reason : Irrational zeros (roots) always occurs in pairs.
(a) A
(b) B
(c) C
(d) D
Answer: (a)
As irrational roots/zeros always occurs in pairs therefore, when one zero is \( (2 - \sqrt{3}) \) then other will be \( 2 + \sqrt{3} \). So, both A and R are correct and R explains A.

Question. Assertion : Zeroes of \( f(x) = x^2 - 4x - 5 \) are 5, -1
Reason : The polynomial whose zeroes are \( 2 + \sqrt{3}, 2 - \sqrt{3} \) is \( x^2 - 4x + 7 \).
(a) A
(b) B
(c) C
(d) D
Answer: (c)

Question. Assertion : \( x^2 + 4x + 5 \) has two zeroes.
Reason : A quadratic polynomial can have at the most two zeroes.
(a) A
(b) B
(c) C
(d) D
Answer: (d)

Question. Assertion : If one zero of poly-nominal \( p(x) = (k^2 + 4)x^2 + 13x + 4k \) is reciprocal of other, then \( k = 2 \).
Reason : If \( (x - \alpha) \) is a factor of \( p(x) \), then \( p(\alpha) = 0 \).
(a) A
(b) B
(c) C
(d) D
Answer: (b)
Let \( \alpha, \frac{1}{\alpha} \) be the zeroes of \( p(x) \), then \( \alpha \cdot \frac{1}{\alpha} = \frac{4k}{k^2 + 4} \implies 1 = \frac{4k}{k^2 + 4} \implies k^2 - 4k + 4 = 0 \implies (k - 2)^2 = 0 \implies k = 2 \). Assertion is true. Reason is true but not the correct explanation.

Question. Assertion : \( P(x) = 14x^3 - 2x^2 + 8x^4 + 7x - 8 \) is a polynomial of degree 3.
Reason : The highest power of x in the polynomial \( p(x) \) is the degree of the polynomial.
(a) A
(b) B
(c) C
(d) D
Answer: (d)
The highest power of x in \( p(x) \) is 4. Degree is 4. So, A is incorrect but R is correct.