# XAT Scan – 3 ## 7 thoughts on “XAT Scan – 3”

1. Saurabh |

Sir, the value of y =15 comes at 4 and 12.. after that I did not get how you got 7 solutions… please explain this point…

• Saurabh |

sir i think I have got it.. we have to look at points at which y = 4 and 12 cuts the graph… suppose if it was f(f(f(15))) then????

• catcracker |

I assume you mean f(f(fx))) = 15, in that case we would find the values for which f(x) = -8, f(x) = 1, f(x) = 7 and so on for all the seven values found in the previous question.

regards
J

2. catcracker |

We need f(x) to be 4 or 12. Looking at the graph, we see that there are 4 values of x for which f(x) = 4 (see the green line in the graph – approx at -8, 1, 7 and 10). Similarly there are 3 more where it is 12 (see the red line)

regards
J

3. Saurabh |

Hello Sir, Thanks for your reply… just a clarification on the second concept f(f(f(x))) = 15
at f(x) = -8, there will be zero values satisfying…
f(x)=1 , will have 4 values … etc… we will calculate for other points…
also sir,
please guide me to the place where i can find the theory regarding this… could not find in Arun Sharma…

• catcracker |

I don’t think there is any “theory” as such. Common sense apply karna hai. That is why most people mess up exams like this – they only know a formula or theorem and cannot solve a twisted question like this!

regards
J

4. Saurabh |

Oh my God…. now I got it… I was missing such a simple concept in the 3 para, 2nd line…. I really thought that this is a difficult problem.. thank you 😀