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  5. Definite Integrals

12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    Let a, b, c be non-zero real numbers such that \[\int_{0}^{1}{(1+{{\cos }^{8}}x)(a{{x}^{2}}+bx+c)\,dx}=\int_{0}^{2}{(1+{{\cos }^{8}}x)(a{{x}^{2}}+bx+c)\,dx}\]Then the quadratic equation \[a{{x}^{2}}+bx+c=0\] has [IIT 1981; CEE 1993]

    A) No root in (0, 2)                  

    B) At least one root in (0, 2)

    C) A double root in (0, 2)      

    D) None of these

    Correct Answer: B

    Solution :

    • We have \[\int_{0}^{2}{f(x)dx=\int_{0}^{1}{f(x)dx+\int_{1}^{2}{f(x)dx}}}\],           
    • where \[f(x)=(a{{x}^{2}}+bx+c)(1+{{\cos }^{8}}x)\]                   
    • If \[f(x)>0(<0)\,x\in (1,\,2)\] then \[\int_{1}^{2}{f(x)dx>0(<0)}\].
    • Thus \[f(x)=(1+{{\cos }^{8}}x)(a{{x}^{2}}+bx+c)\] must be positive for some value of x in [1, 2] and must be negative for some value of x in [1, 2].
    • As \[(1+{{\cos }^{8}}x)\ge 1\] it follows that if \[g(x)=a{{x}^{2}}+bx+c,\]then there exist some \[\alpha ,\beta \in (1,\,2)\] such that \[g(\alpha )>0\]and \[g(\beta )<0\]. Since g is continuous on R, therefore there exist some c between \[\alpha \]and \[\beta \]such that \[g(c)=0\]. Thus \[a{{x}^{2}}+bx+c\]=0 has at least one root in (1, 2) and hence in (0, 2).


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