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JEE Main & Advanced JEE Main Paper Phase-I Held on 07-1-2020 Morning

  • question_answer
    If f \[(a+b+1-x)=f(x),\] for all x, where a and b are fixed positive real numbers, then \[\frac{1}{a+b}\int_{a}^{b}{x(f(x)+f(x+1))dx}\] is equal to                    [JEE MAIN Held on 07-01-2020 Morning]

    A) \[\int_{a-1}^{b-1}{f(x)dx}\]

    B) \[\int_{a+1}^{b+1}{f(x)dx}\]

    C) \[\int_{a-1}^{b-1}{f(x+1)dx}\]

    D) \[\int_{a+1}^{b+1}{f(x+1)dx}\]

    Correct Answer: A , B , C , D

    Solution :

    (BONUS)
    \[\text{l}=\frac{1}{a+b}\int_{a}^{b}{x\left[ f(x)+f(x+1) \right]dx...(1)}\]
    Replace x by \[(a+b-x)\]
    \[\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(a+b-x)+f(a+b+1-x) \right]}dx\]
    As \[f(a+b+1-x)=f(x)\]
    \[\Rightarrow f(a+b+1-a-b+x)=f(a+b-x)\]
    \[\Rightarrow f(1+x)=f(a+b-x)\]
    \[\therefore \,\,\,\,\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(1+x)+f(x) \right]dx...(2)}\]
    Add (1) & (2)
    \[2l=\frac{1}{a+b}\int_{a}^{b}{(a+b)\left[ f(1+x)+f(x) \right]dx}\]
    \[\Rightarrow \frac{1}{2}\int_{a}^{b}{\left[ f(1+x)+f(x) \right]dx}\]
    \[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\]
    \[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+a+b-x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\]
    \[=\int_{a}^{b}{f(x)dx}\]
    Put \[x=t+1\]
    \[=\int_{a-1}^{b-1}{f(t+1)dt}\]
    Further, \[\text{l}=\int_{a-1}^{b-1}{f(a+b-1-t)dt=\int_{a-1}^{b-1}{f(t-2)dt}}\]
    \[=\int_{a+1}^{b+1}{f(x)dx}\]

    Solution :

    (BONUS)
    \[\text{l}=\frac{1}{a+b}\int_{a}^{b}{x\left[ f(x)+f(x+1) \right]dx...(1)}\]
    Replace x by \[(a+b-x)\]
    \[\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(a+b-x)+f(a+b+1-x) \right]}dx\]
    As \[f(a+b+1-x)=f(x)\]
    \[\Rightarrow f(a+b+1-a-b+x)=f(a+b-x)\]
    \[\Rightarrow f(1+x)=f(a+b-x)\]
    \[\therefore \,\,\,\,\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(1+x)+f(x) \right]dx...(2)}\]
    Add (1) & (2)
    \[2l=\frac{1}{a+b}\int_{a}^{b}{(a+b)\left[ f(1+x)+f(x) \right]dx}\]
    \[\Rightarrow \frac{1}{2}\int_{a}^{b}{\left[ f(1+x)+f(x) \right]dx}\]
    \[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\]
    \[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+a+b-x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\]
    \[=\int_{a}^{b}{f(x)dx}\]
    Put \[x=t+1\]
    \[=\int_{a-1}^{b-1}{f(t+1)dt}\]
    Further, \[\text{l}=\int_{a-1}^{b-1}{f(a+b-1-t)dt=\int_{a-1}^{b-1}{f(t-2)dt}}\]
    \[=\int_{a+1}^{b+1}{f(x)dx}\]

    Solution :

    (BONUS)
    \[\text{l}=\frac{1}{a+b}\int_{a}^{b}{x\left[ f(x)+f(x+1) \right]dx...(1)}\]
    Replace x by \[(a+b-x)\]
    \[\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(a+b-x)+f(a+b+1-x) \right]}dx\]
    As \[f(a+b+1-x)=f(x)\]
    \[\Rightarrow f(a+b+1-a-b+x)=f(a+b-x)\]
    \[\Rightarrow f(1+x)=f(a+b-x)\]
    \[\therefore \,\,\,\,\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(1+x)+f(x) \right]dx...(2)}\]
    Add (1) & (2)
    \[2l=\frac{1}{a+b}\int_{a}^{b}{(a+b)\left[ f(1+x)+f(x) \right]dx}\]
    \[\Rightarrow \frac{1}{2}\int_{a}^{b}{\left[ f(1+x)+f(x) \right]dx}\]
    \[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\]
    \[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+a+b-x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\]
    \[=\int_{a}^{b}{f(x)dx}\]
    Put \[x=t+1\]
    \[=\int_{a-1}^{b-1}{f(t+1)dt}\]
    Further, \[\text{l}=\int_{a-1}^{b-1}{f(a+b-1-t)dt=\int_{a-1}^{b-1}{f(t-2)dt}}\]
    \[=\int_{a+1}^{b+1}{f(x)dx}\]

    Solution :

    (BONUS)
    \[\text{l}=\frac{1}{a+b}\int_{a}^{b}{x\left[ f(x)+f(x+1) \right]dx...(1)}\]
    Replace x by \[(a+b-x)\]
    \[\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(a+b-x)+f(a+b+1-x) \right]}dx\]
    As \[f(a+b+1-x)=f(x)\]
    \[\Rightarrow f(a+b+1-a-b+x)=f(a+b-x)\]
    \[\Rightarrow f(1+x)=f(a+b-x)\]
    \[\therefore \,\,\,\,\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(1+x)+f(x) \right]dx...(2)}\]
    Add (1) & (2)
    \[2l=\frac{1}{a+b}\int_{a}^{b}{(a+b)\left[ f(1+x)+f(x) \right]dx}\]
    \[\Rightarrow \frac{1}{2}\int_{a}^{b}{\left[ f(1+x)+f(x) \right]dx}\]
    \[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\]
    \[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+a+b-x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\]
    \[=\int_{a}^{b}{f(x)dx}\]
    Put \[x=t+1\]
    \[=\int_{a-1}^{b-1}{f(t+1)dt}\]
    Further, \[\text{l}=\int_{a-1}^{b-1}{f(a+b-1-t)dt=\int_{a-1}^{b-1}{f(t-2)dt}}\]
    \[=\int_{a+1}^{b+1}{f(x)dx}\]


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