Solved papers for JEE Main & Advanced JEE Main Paper (Held On 11 April 2015)

done JEE Main Paper (Held On 11 April 2015) Total Questions - 2

• question_answer1) The equation of a normal to the curve,\[\sin y=x\sin \left( \frac{\pi }{3}+y \right)\] at \[x=0,\]is: [JEE Main Online Paper (Held On 11 April 2015)]  
  A)
  \[2x+\sqrt{3}y=0\]
  done clear
  B)
  \[2x-\sqrt{3}x=0\]
  done clear
  C)
  \[2y+\sqrt{3}x=0\]
  done clear
  D)
  \[2x-\sqrt{3}y=0\] \[x=0\Rightarrow y=0\]
  done clear
  View Answer play_arrow

• question_answer2) Let k and K be the minimum and the maximum values of the function\[f(x)=\frac{{{(1+x)}^{0.6}}}{1+{{x}^{0.6}}}\]in \[[0,1]\]respectively, then the ordered pair (k, K) is equal to : [JEE Main Online Paper (Held On 11 April 2015)]  
  A)
  \[(1,{{2}^{0.6}})\]
  done clear
  B)
  \[({{2}^{-0.4}}1,{{2}^{0.6}})\]
  done clear
  C)
  \[({{2}^{-0.6}},1)\]
  done clear
  D)
  \[({{2}^{-0.4}},1)\]
  done clear
  View Answer play_arrow