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# My function doubts!

• $$If\quad f\left( x+3 \right) \quad =\quad { x }^{ 2 }+x-6,\quad then\quad find\quad f(x).\quad$$

• $$If\quad f(x+3)\quad =\quad { x }^{ 2 }-7x+2,\quad find\quad the\quad remainder\quad when\quad f(x)\quad is\quad divided\quad by\quad x+1.$$

Note by Swapnil Das
1 year, 7 months ago

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They are very easy.

1.$$\text{Let g(x) = f(x), Not let y = x+3, we get} \\ g(y) = (y-3)^2 + (y-3) - 6 \\ f(x) = x^2 - 5x$$

2.$$\text{We know that the remainder will be f(-1). We have to find f(-1) } \\ \text{We know that } f(x+3) = x^2 -7x +2 \\ \text{Now if we let x = -4 we'll get f(-1) hence our answer.} \\ \text{Hence f(-1) will be} f(-1) = (-4)^2 - 7(4) + 2 = 46$$ · 1 year, 7 months ago

@Swapnil Das @Sravanth Chebrolu I think you must revise the simple remainder- theorem for the 2nd question. · 1 year, 7 months ago

Yup, I have. I just got a last doubt in the chapter, so I asked for help. Well I was now solving problems of Newton's Sums. @Nihar Mahajan · 1 year, 7 months ago

let, x+3 =y......

make a new function.....

and solve.... · 1 year, 7 months ago

i think i am little slow in typing.......:p · 1 year, 7 months ago

Why do you think so? · 1 year, 7 months ago

when i started posting the solution there were no comments but as soon as i finished i found 6 comments hovering around....:) · 1 year, 7 months ago

Whoah! That happens to me too! But, don't consider that as your inability. · 1 year, 7 months ago

ya.........

i am working on it......

always used to work with a pen and paper so.................. · 1 year, 7 months ago

Oh yes! The classic method of pen and paper . . . :P · 1 year, 7 months ago

can someone tell me what is multi_step decryption.....? · 1 year, 7 months ago

• First question: the answer is $$x^{2}-5x$$

• Second question: the answer is $$46$$.

Am I right? If I am I'll post the solution. Please reply @Swapnil Das · 1 year, 7 months ago

Ur answer to 2nd one is wrong. For a level 5 like U it should Child's play. Anyways u can see my solution. Cheers! · 1 year, 7 months ago

Sorry, I got it wrong as I got meddled up with the signs, I posted it correctly on Swapnil's message board though, cheers! xD · 1 year, 7 months ago

Yeah! · 1 year, 7 months ago

Hey! I posted a question on logic just now, as you wanted me to post such questions I did it, here's the link! · 1 year, 7 months ago

See who have solved... · 1 year, 7 months ago

Omg! That was qiuck! Hope you liked it. . . · 1 year, 7 months ago

Hey Sravanth, I don't know how I am getting some other remainder. Can you post the 2nd solution please? · 1 year, 7 months ago

Sure. Here it is.

As you know how to make expression, you'll get $$f(x):x^2-13x-10$$

Dividing it by $$x+1$$, we get $$4$$. . . · 1 year, 7 months ago

Can you please tell me how it became -5y instead of -5x? · 1 year, 7 months ago

BTW, I liked the new look of Brilliant! But I didn't find the explore icon anywhere. . . · 1 year, 7 months ago

community............. · 1 year, 7 months ago

Not the community, the was an icon explore, I didn't find that... · 1 year, 7 months ago

u can check community ....................

i think explore renamed to community................

latest posts are there which before were given under explore................ · 1 year, 7 months ago

Yes. I saw that. Thank you very much! · 1 year, 7 months ago

welcome...............

do u know what is multi step decryption or i hv to search on net.....? · 1 year, 7 months ago

I did hear about that in a novel Digital Fortress, but I don't know much about it. . . :( · 1 year, 7 months ago

ok then........................

thnx for giving a thought to it.................. · 1 year, 7 months ago

Welcome! · 1 year, 7 months ago

multi step decryption...............? · 1 year, 7 months ago

May I help U? · 1 year, 7 months ago

anyone.....?

multi step decryption....? · 1 year, 7 months ago

Sorry, that was a typo there! It's edited now. · 1 year, 7 months ago

I think you are right. Please post the solution. · 1 year, 7 months ago

But I think it's already posted! :P · 1 year, 7 months ago

Well Rajdeep also did the same, BTW thank you very much. Just a request, if you have time anytime, please post problems of this format, I would surely like them! · 1 year, 7 months ago

We'll take that into consideration. :P

Surely! I'll keep posting suchquestions! · 1 year, 7 months ago

Go to Community-All topics-Problems · 1 year, 7 months ago

Okay. So that's nothing but explore? · 1 year, 7 months ago

yeah! · 1 year, 7 months ago

Thank you very much! BTW, how did you like the new look? · 1 year, 7 months ago