\[ \large \lfloor x^2 \rfloor - 5\lfloor x \rfloor + 6 - \sin(x) = 0 \]

Find the sum of all values of \(x\) that satisfy the equation above.

\[ \large \lfloor x^2 \rfloor - 5\lfloor x \rfloor + 6 - \sin(x) = 0 \]

Find the sum of all values of \(x\) that satisfy the equation above.

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TopNewestWhat are you having doubts about? – Rahul Saha · 1 year, 3 months ago

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Dear @Sandeep Bhardwaj pls help – Radhesh Sarma · 1 year, 3 months ago

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– Sandeep Bhardwaj · 1 year, 3 months ago

I think \(\pi\) is the only value satisfying the given equation, when assumed that \([..]\) represents greatest integer function.Log in to reply

– Radhesh Sarma · 1 year, 3 months ago

yes sir, represents greatest integer function.Log in to reply

\([x] ^{2}-5[x]+6=sin(x) \)

Now we can see that LHS part is always integer. It means \(sin(x) \) can be equal to \(-1,0, 1\).

Further we can see that - 1 and 1 get rejected. This gives 2 solutions \([x]=2, 3\).

This gives \(2 \leq x <4\) In the following range \(sin(x) \) is 0 for only one value of x. That is \(x= \pi\). – Shubhendra Singh · 1 year, 3 months ago

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– Akhil Bansal · 1 year, 3 months ago

How do you take out exponent of x outside from G.I.FLog in to reply

– Shubhendra Singh · 1 year, 3 months ago

Yes you are right, I accept my mistake. Luckily the answer got matched. I'm really sorry:-[Log in to reply

– Radhesh Sarma · 1 year, 3 months ago

he has factorized itLog in to reply

– Shubhendra Singh · 1 year, 3 months ago

Sorry Radhesh, my solution is a bit incorrect. I took \([x]^{2}\) in place of \([x^{2}]\) . So now I think you must try and solve the equation using graph.Log in to reply

– Radhesh Sarma · 1 year, 3 months ago

koi baaat nahiLog in to reply

@shubhendra singh ,thank a lot – Radhesh Sarma · 1 year, 3 months ago

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– Shubhendra Singh · 1 year, 3 months ago

You're welcome 8-)Log in to reply