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NCERT Exemplar · Class 9 Mathematics Polynomials

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EXERCISE 2.1 11–21 (part 2 of 8)

  1. Write the correct answer in each of the following :

    Exercise 11

    If x51+51\displaystyle x^{51}+51 is divided by x+1\displaystyle x+1, the remainder is (A) 0\displaystyle 0 (B) 1\displaystyle 1 (C) 49\displaystyle 49 (D) 50\displaystyle 50

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 50\). \[\text{Remainder} = p(-1) \quad \text{(Remainder Theorem)} \] \[= (-1)^{51}+51 = -1+51 = 50 \]

  2. Exercise 12

    If x+1\displaystyle x+1 is a factor of the polynomial 2x2+kx\displaystyle 2 x^{2}+k x, then the value of k\displaystyle k is (A) -3\displaystyle 3 (B) 4\displaystyle 4 (C) 2\displaystyle 2 (D) -2\displaystyle 2

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 2\). \[x+1 \text{ a factor} \implies p(-1)=0 \quad \text{(Factor Theorem)} \] \[2(-1)^{2}+k(-1) = 2-k = 0 \implies k=2 \]

  3. Exercise 13

    x+1\displaystyle x+1 is a factor of the polynomial (A) x3+x2−x+1\displaystyle x^{3}+x^{2}-x+1 (B) x3+x2+x+1\displaystyle x^{3}+x^{2}+x+1 (C) x4+x3+x2+1\displaystyle x^{4}+x^{3}+x^{2}+1 (D) x4+3x3+3x2+x+1\displaystyle x^{4}+3 x^{3}+3 x^{2}+x+1

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    NCERT’s answer
    (B)
    (B) \(\displaystyle x^{3}+x^{2}+x+1\). By the factor theorem, \(\displaystyle x+1\) is a factor of \(\displaystyle p(x)\) exactly when \(\displaystyle p(-1)=0\). \[p(-1) = (-1)^{3}+(-1)^{2}+(-1)+1 = -1+1-1+1 = 0 \] \[x^{3}+x^{2}+x+1 = x^{2}(x+1)+(x+1) = (x+1)(x^{2}+1) \]

  4. Exercise 14

    One of the factors of (25x2−1)+(1+5x)2\displaystyle \left(25 x^{2}-1\right)+(1+5 x)^{2} is (A) 5+x\displaystyle 5+x (B) 5−x\displaystyle 5-x (C) 5x−1\displaystyle 5 x-1 (D) 10x\displaystyle 10 x

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 10x\). Expand the square and collect like terms. \[\left(25x^{2}-1\right)+(1+5x)^{2} = 25x^{2}-1+1+10x+25x^{2} \] \[= 50x^{2}+10x = 10x(5x+1) \]

  5. Exercise 15

    The value of 2492−2482\displaystyle 249^{2}-248^{2} is (A) 12\displaystyle 1^{2} (B) 477\displaystyle 477 (C) 487\displaystyle 487 (D) 497\displaystyle 497

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    NCERT’s answer
    (D)
    (D) 497. Use the difference-of-squares identity. \[249^{2}-248^{2} = (249-248)(249+248) \] \[= 1\times 497 = 497 \]

  6. Exercise 16

    The factorisation of 4x2+8x+3\displaystyle 4 x^{2}+8 x+3 is (A) (x+1)(x+3)\displaystyle (x+1)(x+3) (B) (2x+1)(2x+3)\displaystyle (2 x+1)(2 x+3) (C) (2x+2)(2x+5)\displaystyle (2 x+2)(2 x+5) (D) (2x−1)(2x−3)\displaystyle (2 x-1)(2 x-3)

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    NCERT’s answer
    (B)
    (B) \(\displaystyle (2x+1)(2x+3)\). Split the middle term. \[4x^{2}+8x+3 = 4x^{2}+6x+2x+3 \] \[= 2x(2x+3)+1(2x+3) = (2x+1)(2x+3) \]

  7. Exercise 17

    Which of the following is a factor of (x+y)3−(x3+y3)\displaystyle (x+y)^{3}-\left(x^{3}+y^{3}\right) ? (A) x2+y2+2xy\displaystyle x^{2}+y^{2}+2 x y (B) x2+y2−xy\displaystyle x^{2}+y^{2}-x y (C) xy2\displaystyle x y^{2} (D) 3xy\displaystyle 3 x y

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 3xy\). Expand the cube and cancel the cubes. \[(x+y)^{3}-(x^{3}+y^{3}) = x^{3}+3x^{2}y+3xy^{2}+y^{3}-x^{3}-y^{3} \] \[= 3x^{2}y+3xy^{2} = 3xy(x+y) \]

  8. Exercise 18

    The coefficient of x\displaystyle x in the expansion of (x+3)3\displaystyle (x+3)^{3} is (A) 1\displaystyle 1 (B) 9\displaystyle 9 (C) 18\displaystyle 18 (D) 27\displaystyle 27

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    NCERT’s answer
    (D)
    (D) 27. Expand the cube by the binomial identity. \[(x+3)^{3} = x^{3}+3(3)x^{2}+3(3)^{2}x+3^{3} \] \[= x^{3}+9x^{2}+27x+27 \]

  9. Exercise 19

    If xy+yx=−1(x,y≠0)\displaystyle \frac{x}{y}+\frac{y}{x}=-1(x, y \neq 0), the value of x3−y3\displaystyle x^{3}-y^{3} is (A) 1\displaystyle 1 (B) -1\displaystyle 1 (C) 0\displaystyle 0 (D) 12\displaystyle \frac{1}{2}

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    NCERT’s answer
    (C)
    (C) 0. Clear the denominators first. \[\frac{x}{y}+\frac{y}{x}=-1 \implies x^{2}+y^{2}=-xy \] \[x^{3}-y^{3} = (x-y)\left(x^{2}+xy+y^{2}\right) = (x-y)(0) = 0 \]

  10. Exercise 20

    If 49x2−b=(7x+12)(7x−12)\displaystyle 49 x^{2}-b=\left(7 x+\frac{1}{2}\right)\left(7 x-\frac{1}{2}\right), then the value of b\displaystyle b is (A) 0\displaystyle 0 (B) 12\displaystyle \frac{1}{\sqrt{2}} (C) 14\displaystyle \frac{1}{4} (D) 12\displaystyle \frac{1}{2}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle \frac{1}{4}\). Apply the difference-of-squares identity on the right. \[\left(7x+\frac{1}{2}\right)\left(7x-\frac{1}{2}\right) = 49x^{2}-\frac{1}{4} \] \[49x^{2}-b = 49x^{2}-\frac{1}{4} \implies b=\frac{1}{4} \]

  11. Exercise 21

    If a+b+c=0\displaystyle a+b+c=0, then a3+b3+c3\displaystyle a^{3}+b^{3}+c^{3} is equal to (A) 0\displaystyle 0 (B) abc\displaystyle a b c (C) 3abc\displaystyle 3 a b c (D) 2abc\displaystyle 2 a b c

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 3abc\). Use the standard identity for a sum of cubes. \[a^{3}+b^{3}+c^{3}-3abc = (a+b+c)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right) \] \[a+b+c=0 \implies a^{3}+b^{3}+c^{3}=3abc \]