NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 2: PolynomialsExercise 2.2
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Question. 1
1. Answer the following and justify:
(i) Can \(x^2-1\) be the quotient on division of \(x^6+2x^3+x-1\) by a polynomial in \(x\) of degree 5?
(ii) What will the quotient and remainder be on division of \(ax^2+bx+c\) by \(px^3+qx^2+rx+s\), \(p\ne0\)?
(iii) If on division of a polynomial \(p(x)\) by a polynomial \(g(x)\), the quotient is zero, what is the relation between \(\deg p\) and \(\deg g\)?
(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 \(\deg p\) and \(\deg g\)?
(v) Can the quadratic polynomial \(x^2+kx+k\) have equal zeroes for some odd integer \(k>1\)?
Answer
(i) No (ii) Quotient 0, remainder \(ax^2+bx+c\) (iii) \(\deg p<\deg g\) (iv) \(\deg p\ge \deg g\) (v) No
Step by Step Solution
(i) If divisor has degree 5, then \(\deg\)(quotient) must be \(\deg\)dividend \(-\) \(\deg\)divisor \(=6-5=1\) (or less if leading terms cancel), not 2. Hence \(x^2-1\) cannot be the quotient.
(ii) Since \(\deg(px^3+\cdots)=3>\deg(ax^2+bx+c)=2\), the division yields quotient 0 and remainder the dividend itself.
(iii) Quotient 0 means divisor’s degree exceeds dividend’s: \(\deg p<\deg g\).
(iv) Remainder 0 means \(g(x)\) divides \(p(x)\). Thus \(\deg p\ge \deg g\) (allowing equality if leading coefficients match to cancel exactly).
(v) Equal zeroes require discriminant zero: \(k^2-4k=0\Rightarrow k(k-4)=0\Rightarrow k=0\) or \(k=4\). With odd \(k>1\), no solution.
Question. 2
2. Are the following statements True/False? Justify:
(i) If the zeroes of a quadratic polynomial \(ax^2+bx+c\) are both positive, then \(a,b,c\) 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 have the same sign.
(vi) If all three zeroes of a cubic polynomial \(x^3+ax^2-bx+c\) are positive, then at least one of \(a,b,c\) is non-negative.
(vii) The only value of \(k\) for which the quadratic polynomial \(kx^2+x+k\) has equal zeroes is \(\dfrac{1}{2}\).
Answer
(i) False (ii) False (iii) True (iv) True (v) True (vi) False (vii) False
Step by Step Solution
(i) If both zeroes are \(\alpha,\beta>0\): sum \(=-(b/a)>0\Rightarrow b/a<0\Rightarrow a,b\) have opposite signs—this seems to contradict. But product \(=c/a>0\) forces \(a,c\) same sign. For both conditions to hold with \(\alpha,\beta>0\), we must have \(a>0, b<0, c>0\), i.e., not all the same sign. Correction: The statement is actually False if interpreted literally. However, the Exemplar’s intent (considering standard form) often expects checking of sign patterns; use Vieta carefully.
(ii) A quadratic can touch the x-axis at a single point (a repeated root), e.g., \(y=(x-1)^2\). So the statement is False.
(iii) A polynomial of degree 4 can cut the x-axis at two points, for example \(y=(x-1)^2(x-2)^2\). So True.
(iv) If two zeroes are 0, say \(0,0,\gamma\), the polynomial has factor \(x^2\), so no linear \(x\) term and no constant term. True.
(v) Take \(y=(x+1)(x+2)(x+3)=x^3+6x^2+11x+6\): all zeroes negative but all coefficients positive (same sign), so the statement claiming “must” be same sign is False as counterexamples with sign changes exist depending on signs of roots and leading coefficient.
(vi) If all three zeroes are positive, then sum \(=-(a)\) (for monic) is positive only if \(a<0\), but coefficients \(b,c\) (up to sign conventions) ensure at least one of \(a,b,c\) is \(\ge0\). Hence True.
(vii) Equal zeroes require \(\Delta=b^2-4ac=1-4k^2=0\Rightarrow k=\pm\dfrac{1}{2}\). With positive \(k\) only, \(k=\dfrac{1}{2}\); but the claim “only value is \(\dfrac{1}{2}\)” ignores \(k=-\dfrac{1}{2}\). Hence as stated it is False.