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  5. Definite Integrals

12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    Let \[g(x)=\int_{0}^{x}{f(t)\,dt}\] where \[\frac{1}{2}\le f(t)\le 1,\,t\in [0,\,1]\] and \[0\le f(t)\le \frac{1}{2}\]  for \[t\in (1,\,2]\], then [IIT Screening 2000]

    A) \[-\frac{3}{2}\le g(2)<\frac{1}{2}\]                                 

    B) \[0\le g(2)<2\]

    C) \[\frac{3}{2}<g(2)\le \frac{5}{2}\]                                   

    D) \[2<g(2)<4\]

    Correct Answer: B

    Solution :

    • \[g(2)=\int_{0}^{2}{f(t)dt=\int_{0}^{1}{f(t)dt+\int_{1}^{2}{f(t)dt}}}\]           
    • \[\text{As}\,\,\frac{1}{2}\le f(t)\le 1\,\text{for}\,\,0\,\le t\,\le 1,\]           
    • \[\int_{0}^{1}{\frac{1}{2}dt\,\,\le \int_{0}^{1}{f(t)dt\,\le \int_{0}^{1}{tdt}}}\] or \[\frac{1}{2}\le \int_{0}^{1}{f(t)\,dt\le 1}\]  ?..(i)           
    • \[\text{As}\,\,0\,\le f(t)\le \frac{1}{2}\text{for}\,\,1<t\le 2,\]\[\int_{1}^{2}{0dt\le \int_{1}^{2}{f(t)dt\le \int_{1}^{2}{\frac{1}{2}dt}}}\]           
    • \[\therefore 0\le \int_{1}^{2}{f(t)dt\le \frac{1}{2}}\]  ?..(ii)           
    • Adding (i) and (ii), \[1/2\le g(2)\le 3/2\]                   
    • \[\therefore g(2)\] satisfies the inequality \[0\le g(2)<2\].


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