Hey Nikhila and Ankeet,

This's simple yaar....

 

For any triangle,

 

" AREA  = Inradius*( Perimeter of triangle ) / 2 "

 

So : here, Area = 1/2*3*4 = 6

               ie, inradius = 2*area / perimeter

                                = 12 / (3 + 4 + 5)

                                = 1

 

Please let me know if there are any mistakes....

 

An easy question for you :

 

log₃₄5 lies between :

 

a) 1/5 and 1/2

b) 1/25 and 1/5

c) 1/3 and 1/2

d) 1/5 and 1/3

2)

 

a = (cos x + sin x),

b = (cos³x + sin³x) = (cos x + sin x)(1 - sin x cos x)

 

So, a³ - 3a + 2b = a(a² - 1) - (2a - b)

 

=(cos x +sin x)(2 cos x sin x) - (2 {cos x +sin x} - {cos x +sinx}{1 - sin x cosx} )

 

=                    "                    - (cos x + sin x)(1 + sin x cos x)

 

=(cos x + sin x) ( {2 cos x sin x} - {1 + sin x cos x} )

 

=(cos x + sin x)(cos x sin x - 1)

 

= - (cos³x + sin³x)                 = ( - b )

 

Please let me know if there are any mistakes... 

 

                                         " Thathwamasi "

1)

 

 (1 + sin - cos) / (1 + sin +cos)  +  (1 + sin + cos) / (1 + sin + cos)

 

={ (1 + sin - cos)²  + (1+ sin + cos)² } / (1 +sin + cos)(1 + sin - cos)

 

=(4 + 4sin) / (2sin +2sin²)

 

=4(1 + sin) / 2sin(1 + sin)

 

=2cosec

  

                                          " Thathwamasi "

Thanx rebel.
Now.......
Someone pleazzzzzzzzz answer my last question

 

And someone...pleazzz send good questions on " progressions ", including A.P., G.P., H.P., special series etc.. 

Someone answer pleazzz:

 

(1 + x)n = a₀ + a₁x + a₂x² +  .... + a_nxⁿ, then prove that :

 

( i )  a₀ - a₂ + a₄ - a₆ + ........ = 2^n/2 cos(npie / 4)

 

( ii ) a₁ - a₃ + a₅ - a₇ + ......... = 2^n/2 sin(npie / 4)   

 

I' m getting many blocks on my way to this one...Explain ok??

Thanx for that....and let me now kill that tricky question :

 

As per the qn, there are 6t terms in an A.P. S1 : sum of first 2t terms

                                                                S2 : sum of first 4t terms

                                                                S3 : sum of last 2t terms

 

I believe, you know this : given 3n terms of an A.P.,

Sum of 1st n terms, Sum of next n terms, Sum of last n terms are in A.P.

 

Then here S1, (S2 - S1), S3 are in A.P.  ---------- (1)

 

Also given, (S1*S2*S3 - S1^2*S3) = 4;

ie, S1*(S2 - S1)*S3 = 4

 

A.M. >= G.M.

 

{ S1 + (S2 - S1) + S3 } / 3  >=  cuberoot(S1*(S2 - S1)*S3)

                                        >=  cuberoot(4)

From eqn(1)  3(S2 - S1) / 3 >=  cuberoot(2^2)

                      (S2 - S1)     >=  2^2/3   

SO : (S2 - S1) not less than 2^2/3 .........      

You like it???

                                                        "Thathwamasi"

 

 

Hey rebel, it's not (a) for ' tricky no.1 ' now please don't go on confirming each option yaar....!! This is really a tricky question and.. you want the answer.... or you want to continue your effort to kill the question...?? You can do it, come on...

 

Oh... before this, someone pleazzzz answer this :

 

Simplify :  (1 + sin + icos)ⁿ / (1 + sin - cos)ⁿ     

 

A little starting trouble over here...

1) (x² + x + 12) / (x² + 3x + 3) < 2

    

    =>(x² + x + 12) < 2(x² + 3x + 3)

    =>(x² + x + 12) < (2x² + 6x + 6)

 

    =>(x² + 5x - 6) >0  ie, x = (-, -6) or x = (1, )

 

2) (x² + 3x + 4) / (x² + 2) < 1/3

 

    =>3(x² + 3x + 4) < (x² + 2)

    =>(3x² + 9x + 12) < (x² + 2)

 

    =>(2x² + 9x + 10) < 0 ie, x = (-5/2, -2)

    

If your question is such that, the same 'x' satisfies both the inequalities, then there is no possible real value for 'x'.

 

Please let me know, whether the answer is right or wrong.

 

                                               Thathwamasi   

projectile thrown horizontally down with a velocity 'u' from the top A of an inclined plane making an angle$ with the plane so that it reaches B on the inclined plane. (hope you got that inclined plane makes angle$ with the ground)  

 

Time taken : 2u tan$ / g

AB : 2u^2 tan$ / gcos$

Vertical distance covered : AB * cos$

 

projctile projected from the bottom of an inclined plane with a velocity 'u' making angle$ with the plane while another one is projected downwards from the top of the plane with velocity ' v ' making angle@ with the plane.

They meet if : usin$ = vsin@ 

 

                                         

Projectile projected with velocity 'u' making angle# with the plane from the bottom A of an inclined plane(making angle@ with the ground) so that it reaches B. (Let C be a point on the ground such that AC perp. to BC)

 

Time taken,t : 2u sin# / g cos@

 

AC : u cos(# + @) t

      : 2u^2 sin# cos(# + @) / g cos@

 

AB : AC / gcos@

 

 

 

Well, when the spring gets dissolved, the elastic potential energy is passed on to the acid that the internal energy of the acid increases and hence the temperature.

 

If the compressed spring gets completely dissolved, then the entire potential energy should be transferred on to the acid's internal energy. If not, the energy released may depend on the % of dissolution.