Doubt from algebra

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@Bhuvanitha_2020 @Sneha_2021 @Shwetanshu_2018 sir @Prajwal_2020_1 and anyone

Graph question jee 2011
Is the book tomato (isi)??

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Sorry i dont know the book ,it was asked by my friend
I searched the ques in jee 2011 paper but i did not find it there

Q17) 10
Q30) 2

For Q17,
1 must be a1
Proof : assume 1 is not, then the sum has maximum value 1/2 + 1/3 + 1/4 + 1/5 which is < 11/6

So now we need to solve 5/6 = 1/a2 + 1/a3 + 1/a4
By the same logic again, 1/2 must be present = a2

So now we need to solve 1/a3 + 1/a4 = 1/3
This is easy as we know that 1/n - 1/(n+1) is always reciprocal of an integer, hence a3 can be 4, and a4 hence 12.

Can you prove this is the only quartet solution set? It's just a little extra work.

Q30 is just a question of drawing graphs, noticing that x² grows faster than xsinx + cosx by showing xsinx + cosx < x+1 for x > 1.618, and that after pi/2 x² has already over taken xsinx + cosx and they will never cross again.

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f(x)=x^2-xsinx-cosx
Applying ivt theoram
f(0)f(\pi) <0
Implies 1 root in b/W [0,\pi]
Its even function one root in negative x- axis also hence 2 real root.
Its for sure jee problem

Check at 27 minute

f(x) = cos(1001x)
Although f(0)*f(pi) = - 1
f(x) does not have one root in 0 to pi.

IVT implies atleast 1 root

I got it @Amay_2018 sir

Sir but its strictly increaing function then their will be one root for sure?
@Amay_2018 sir

Sir can you elaborate me this please

From Diagram, If we take two points as A_{4}A_{10}. Then we have

10 choices for third vertices.

Her we have 6 such pairs of chords .. like A_{3}A_{9} ect..

So we have total no. of Right angle triangles =6\times 10 =60.

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Sir ur approach for first @Jagdish_Singh sir

For (30)

Let we take f(x)=x^2-x\sin x-\cos x.

Then we have f'(x)=2x-x\cos x-\sin x+\sin x

\Longrightarrow f'(x)=x\underbrace{(2-\cos x)}_{>0\;\forall\; x\;\in \mathbb{R}}.

So Here f'(x)>0 for x>0 and f'(x)<0 for x<0.

and at x=0, we have f(0)=-1. and

For x\rightarrow \pm\infty, f(x)\rightarrow \infty

So our graph is Concave upward cut x axis at exactly 2 points.

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i am getting perfect square at n=17 ,answer is given 1 , but my ques is how we can prove that there will be only one

Anyone, @Sneha_2021

Ur ans is correct its asking \textbf{number of integer}
Hence only 1 integer u r geeting as p^2