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Quadratic Equation and Inequalities question

2022 · 28 Jun · Shift 2 · Q23
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  5. /2022 · 28 Jun · Shift 2 · Q23

Quadratic Equation and Inequalities question

2022 · 28 Jun · Shift 2 · Q23

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let f(x) be a quadratic polynomial such that f(−-− 2) + f(3) = 0. If one of the roots of f(x) = 0 is −-− 1, then the sum of the roots of f(x) = 0 is equal to :
  1. A
    113{{11} \over 3}311​
  2. B
    73{{7} \over 3}37​
  3. C
    133{{13} \over 3}313​
  4. D
    143{{14} \over 3}314​
View written solutionFree

Correct answer: A

  1. Let the quadratic polynomial be f(x)=ax2+bx+c,f(x)=ax^2+bx+c,f(x)=ax2+bx+c, where a≠0a\neq 0a=0.

  2. Given: f(−2)+f(3)=0.f(-2)+f(3)=0.f(−2)+f(3)=0. Compute each term: f(−2)=4a−2b+c,f(-2)=4a-2b+c,f(−2)=4a−2b+c, f(3)=9a+3b+c.f(3)=9a+3b+c.f(3)=9a+3b+c. So, f(−2)+f(3)=(4a−2b+c)+(9a+3b+c)=13a+b+2c=0.f(-2)+f(3)=(4a-2b+c)+(9a+3b+c)=13a+b+2c=0.f(−2)+f(3)=(4a−2b+c)+(9a+3b+c)=13a+b+2c=0. Hence, 13a+b+2c=0.(1)13a+b+2c=0. \qquad (1)13a+b+2c=0.(1)

  3. One root of f(x)=0f(x)=0f(x)=0 is −1-1−1, so f(−1)=0.f(-1)=0.f(−1)=0. That gives a−b+c=0.(2)a-b+c=0. \qquad (2)a−b+c=0.(2)

  4. From (2), c=b−a.c=b-a.c=b−a. Substitute into (1): 13a+b+2(b−a)=013a+b+2(b-a)=013a+b+2(b−a)=0 13a+b+2b−2a=013a+b+2b-2a=013a+b+2b−2a=0 11a+3b=0.11a+3b=0.11a+3b=0. Therefore, b=−11a3.b=-\frac{11a}{3}.b=−311a​.

  5. For a quadratic ax2+bx+cax^2+bx+cax2+bx+c, the sum of roots is −ba.-\frac{b}{a}.−ab​. Hence, sum of roots=−ba=−−11a/3a=113.\text{sum of roots}=-\frac{b}{a}=-\frac{-11a/3}{a}=\frac{11}{3}.sum of roots=−ab​=−a−11a/3​=311​.

  6. Therefore, the correct option is 113.\boxed{\frac{11}{3}}.311​​.

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