2. Sample Papers
  3. KVPY

KVPY Sample Paper KVPY Stream-SX Model Paper-5

  • question_answer
    Let \[f\,(x)=\int_{0}^{x}{\sin ({{t}^{2}}-t+x)dt,}\] then the value of \[f''\left( \frac{-1}{2} \right)+f\left( \frac{-1}{2} \right)\] is

    A) \[0\]                 

    B) \[1\]

    C) \[\frac{1}{2}\]                          

    D) \[-\frac{1}{2}\]

    Correct Answer: A

    Solution :

    [a]
    We have, \[f(x)=\int_{0}^{x}{\sin ({{t}^{2}}-t+x)dt}\]
    \[\Rightarrow \]\[f'(x)=\sin ({{x}^{2}}-x+2)+\int_{0}^{x}{\cos ({{t}^{2}}-t+x)dt}\]
    \[\Rightarrow \]\[f''(x)=2x\cos {{x}^{2}}+\cos ({{x}^{2}})-\int_{0}^{x}{\sin ({{t}^{2}}-t+x)dt}\]
    \[\Rightarrow \]\[f''(x)+f(x)=(2x+1){{\cos }^{2}}x\]
    \[\Rightarrow \]\[f''\left( -\frac{1}{2} \right)+f\left( \frac{-1}{2} \right)=\left( 2\left( \frac{-1}{2} \right)+1 \right)\cos \left( \frac{1}{4} \right)=0\]


You need to login to perform this action.
You will be redirected in 3 sec spinner