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JEE Main & Advanced JEE Main Paper (Held On 19 May 2012)

  • question_answer
    Consider a quadratic equation \[\text{a}{{\text{x}}^{\text{2}}}+\text{bx}+\text{c}=0,\]where \[\text{2a}+\text{3b}+\text{6c}=0\] and let\[g\left( x \right)=a\frac{{{x}^{3}}}{3}+b\frac{{{x}^{2}}}{2}+cx.\] Statement 1: The quadratic equation has at least one root in the interval (0, 1). Statement 2: The Rolle's theorem is applicable to function g(x) on the interval [0, 1].     JEE Main  Online Paper (Held On 19  May  2012)

    A) Statement 1 is false, Statement 2 is true.

    B) Statement 1 is true. Statement 2 is false. Statement 1 is true, Statement 2 is true,

    C) Statement 2 is not a correct explanation for Statement 1.

    D)                 Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

    Correct Answer: D

    Solution :

    Let \[g(g)=\frac{a{{x}^{3}}}{3}+b.\frac{{{x}^{2}}}{2}+cx\] \[g'(x)=a{{x}^{2}}+bx+c\] Given: \[a{{x}^{2}}+bx+c=0\]and\[2a+3b+6c=0\] Statement-2: (i) g(0) = 0 and g (1) \[=\frac{a}{3}+\frac{b}{2}+c=\frac{2a+3b+6c}{6}\] \[=\frac{0}{6}=0\]\[\Rightarrow \]\[g(0)=g(1)\] (ii) g is continuous on [0,1] and differentiable on (0,1) \[\therefore \]By Rolle's theorem \[\exists \,k\in (0,1)\] such that) = 0 This holds the statement 2. Also, from statement-2,we can say \[\text{a}{{\text{x}}^{\text{2}}}+\text{bx}+\text{c}=0\]has at least one root in (0,1). Thus Statement-1 and 2 both are true and Statement-2 is a correct explanation for    Statement-1.


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