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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    On the interval \[\left[ \frac{5\pi }{3},\,\,\frac{7\pi }{4} \right],\] the greatest value of the function \[f(x)=\int_{5\pi /3}^{x}{(6\cos t-2\sin t)\,dt=}\]

    A) \[3\sqrt{3}+2\sqrt{2}+1\]       

    B) \[3\sqrt{3}-2\sqrt{2}-1\]

    C) Does not exist                     

    D) None of these

    Correct Answer: B

    Solution :

    • \[f'(x)=(6\cos x-2\sin x)1-0\]                           
    • \[=2[3\cos x-\sin x]>0\]in \[\left[ \frac{5\pi }{3},\frac{7\pi }{4} \right]\]                   
    • \ \[f(x)\] is an increasing function, hence \[f(x)\] has greatest value at \[x=\frac{7\pi }{4}\]                   
    • \ Greatest \[f(x)=f\left( \frac{7\pi }{4} \right)=[6\sin t+2\cos t]_{5\pi /3}^{7\pi /4}\]                   
    • \[=-6\frac{1}{\sqrt{2}}+\frac{2}{\sqrt{2}}+6\frac{\sqrt{3}}{2}-1=3\sqrt{3}-2\sqrt{2}-1\].


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