TargetIIT

Q1. Let [im]http://codecogs.izyba.com/gif.latex?%5CLARGE%5C%21%5Cphi%20%28x%29%20%3D%20sinx%5Cint_%7B0%7D%5E%7Bx%7D%7Bcostdt%7D%20%2B%202%5Cint_%7B0%7D%5E%7Bx%7D%7Btdt%7D%20%20%2B%20cos%5E2x%20-%20x%5E2[/im].
If [im]http://codecogs.izyba.com/gif.latex?%5CLARGE%5C%21x%5E2%20-%20%202x%20%2B3%20%5Cgeq%20%5Cphi%20%28x%29%20%5Cvee%20x%5Cepsilon%20%20R%2C[/im] then greatest area bounded by xφ(x) and coordinate x=0 and x=5 is

(a) 16   (b) 25   (c) 8   (d) 35/2

Q2. Let [i]a[/i] is a complex number, satisfying[im]http://codecogs.izyba.com/gif.latex?%5CLARGE%5C%21ia%5E3%2Ba%5E2-a%2Bi%3D0[/im]
then maximum value of |a-3-4i| is

(a) 2   (b) 4   (c) 5    (d) 6

Q3. A/R
Stmnt 1: The area of the triangle on the argand plane formed by the complex nos. z,iz and z+iz is [frac]1[/]2[/frac]|z|[p]2[/p]

Stmnt 2: Multiplying any complex number by i rotates it by π/2 wrt itself in anticlockwise direction

(A or B???)

Q4. Let [im]http://codecogs.izyba.com/gif.latex?%5CLARGE%5C%21I_%7B1%7D%20%3D%20%5Cint_%7B0%7D%5E%7B1%7D%7B%5Cfrac%7B1%2Bx%5E8%7D%7B1%2Bx%5E4%7Ddx%7D%5C%3B%20and%20%5C%3B%20%20I_%7B2%7D%3D%5Cint_%7B0%7D%5E%7B1%7D%7B%5Cfrac%7B1%2Bx%5E9%7D%7B1%2Bx%5E2%7Ddx%7D[/im],
then

(a) 0<I1<I2<1
(b) 0 < I2 < I1 < 1

Q5. Normal of parabola y[p]2[/p]=4x at P and Q meets at R(x[ss]2[/ss],0) and tangents at P and Q meets at T(x[ss]1[/ss],0), If x[ss]2[/ss]=3, then the length of latus rectum plus tangent PT will be
(a) 3   (b) 6   (c) 1 (d) 8

Ans 3) C ??

Multiplication of Z with i then vector for Z rotates a right angle in the positive sense.

ok if it is pie/2, then ans shuld be (a)

why (a)??

How does it explain that?

Q1.

diff phi(x) ...
we get phi'(x)=0 for all x ...

so its const. fxn.

max val. of phi(x) = min val of that polynomial = 2 ..

so area under 2x frm 0 to 5 => 25...

min val of x^2-2x+3 is at x=1 ...

1-2+3 =2 ..

but then
just simplify the expression of phi(x)

∫costdt from 0 to x = sinx and 2∫tdt  from 0 to x = x[p]2[/p]

So, phi(x) = sin[p]2[/p]x + x[p]2[/p]+cos[p]2[/p]x-x[p]2[/p]
              = 1 for all x

Ya asish i was also getting φ(x) = 1..but was confused bcoz the second statement says that φ(x) ≤ 2..dunno how the equality is coming..

Q2

all i could think of is ... put a=x+iy

equate

a^3 =0

and

a^2-a+1=0

im gettin 2...???[doubtful]

oh!! asish i never thought of that...

for 2. ans given (D)

i was getting a=i, and a[p]2[/p] = -i (=> a=±[frac]1[/]√2[/frac](1-i))

Q3 ...the reason helps  find out that its a rt angled triangle so i think it shud be A

q4 is b na...

Q4. given (A) ... i too think B

q5 is 6 na.//

yeah but pls explain this

[i]tangent PT[/i]
this means length of PT or something else?

yeah length of pt ...

we have

2a + a(t^2 + t^2 -t^2) =2

so t =±1 ..

so pt of intersection of tangents become -1,0

PT = root(S1) = root(4) = 2

2+LR =2+4=6...

But then T≡(-1,0)
P≡(1,2)

But PT≠ 2  (as obtained by doing [sqrt]S1[/sqrt])

yeah ... i think its only for circles ..sry..... some error..

maybe the guy who set the question did the same error :P///

GUYZ...FOR Q5.  ANS IS  6.

BUT MY SOLN IS VERY LENGTHY  AND  RISKY......

I DID IT USING PARAMETRIC FORM ......

CAN ANY TELL ME OF A SHORTER METHOD

can anyone tell me which ques are still unsolved here ?????

@eure: all unsolved except 3 (no general consensus for others)

1. looks like options wrong (please verify)
2. again seems like options wrong
I was getting
a=i, and a[p]2[/p] = -i (=> a=±[frac]1[/]√2[/frac](1-i))

4. ans given A while i got B

5. length of tangent PT means length of PT naa?
If we calc. length of PT it does not come 2

while [sqrt]S1[/sqrt] = 2

2nd. is teh last term in the equation 1 or[b] i[/b]

more doubts added.

Q6. [im]http://codecogs.izyba.com/gif.latex?%5Cvec%7Ba%7D%20%5Ctextup%7B%20and%20%7D%5Cvec%7Bb%7D%5Ctextup%7B%20are%20non-zero%2C%20non-collinear%20vectors%20such%20that%7D[/im]
[im]http://codecogs.izyba.com/gif.latex?%7C%5Cvec%7Ba%7D%7C%20%3D2%2C%5C%3B%20%5Cvec%7Ba%7D.%5Cvec%7Bb%7D%3D1%5Ctextup%7B%20and%20the%20angle%20between%20%7D%20%5Cvec%7Ba%7D%5Ctextup%7B%20and%20%7D%5Cvec%7Bb%7D%5Ctextup%7B%20is%20%7D%5Cpi%20/3.[/im]
[im]http://codecogs.izyba.com/gif.latex?%5Ctextup%7BIf%20%7D%5Cvec%7Br%7D%5Ctextup%7B%20is%20any%20vector%20satisfying%20%7D%5Cvec%7Br%7D.%5Cvec%7Ba%7D%3D2%2C%5C%3B%20%5Cvec%7Br%7D.%5Cvec%7Bb%7D%3D8%2C%5C%3B%20%28%5Cvec%7Br%7D+2%5Cvec%7Ba%7D-10%5Cvec%7Bb%7D%29.%28%5Cvec%7Ba%7D%5Ctimes%20%5Cvec%7Bb%7D%29%20%3D%204%5Csqrt%7B3%7D[/im]
[im]http://codecogs.izyba.com/gif.latex?%5Ctextup%7Band%20is%20equal%20to%20%7D%5Cvec%7Br%7D+2%5Cvec%7Ba%7D-10%5Cvec%7Bb%7D%3D%5Clambda%20%28%5Cvec%7Ba%7D%5Ctimes%5Cvec%7Bb%7D%29%2C%20%5C%3B%20then%5C%3B%20%5Clambda%20%3D[/im]
(a) 0.5  (b) 2   (c) 0.25   (d) 4

Q7/8/9 PARAGRAPH
Vertices of a variable acute angled triangle ABC lies in a circle of radius R such that [im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7Bda%7D%7BdA%7D+%5Cfrac%7Bdb%7D%7BdB%7D+%5Cfrac%7Bdc%7D%7BdC%7D%3D6[/im].
Distance of orthocentre of triangle ABC from vertex A,B and C is [im]http://codecogs.izyba.com/gif.latex?x_%7B1%7D%2Cx_%7B2%7D%2Cx_%7B3%7D[/im] respectively

7. Inradius of triangle ABC
(a) 1  (b) 2   (c) 3  (d)  4

8. Maximum value of [im]http://codecogs.izyba.com/gif.latex?x_%7B1%7Dx_%7B2%7Dx_%7B3%7D[/im] is
(a) 4   (b) 6   (c) 8    (d) 10

9. [im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7Bdx_%7B1%7D%7D%7Bda%7D+%5Cfrac%7Bx_%7B2%7D%7D%7Bdb%7D+%5Cfrac%7Bdx_%7B3%7D%7D%7Bdc%7D[/im] is always ≤  
(a) -3[sqrt]3[/sqrt]   (b) 3[sqrt]3[/sqrt]   (c) 1    (d) 6

@che: last term 'i'.. edited

so for 2nd is not the ans [b] 6[/b]

@che: pls see #19

Q10. For the curve x+y+1=e[p]y[/p] (actually ques. is dy/dx = 1/(x+y) and curve passes thru origin), area bounded by curve and abscissa y=0 and y=1 is
im getting (e-5/2) while given (e-3/2)

Q11. Consider hyperbola xy=4. Tangent at any point P of the hyperbola intersects the coordinate axes at A and B.
Area of triangle OAB
(a) a const =16
(B) const =32
(c) const=64
(d) variable

I got const =8

so for 2nd wats the prob ?

u hav got [im]http://codecogs.izyba.com/gif.latex?\left|a%20\right|=1[/im]

so [im]http://codecogs.izyba.com/gif.latex?\\\left|a%20+(-3-4i)\right|\leq%20\left|a%20\right|+\left|-3-4i%20\right|\\%20\left|a%20+(-3-4i)\right|\leq%201+5\\%20\left|a%20+(-3-4i)\right|\leq%206\\[/im]

Whats the use of finding maximum value (as you have done) of that when u have already got the EXACT values?

Shouldnt we instead find the maximum value by substituting a= what we have obtained?

actually i guess u r rit....

the thing which i did in #32 is wen we dunno wat a is but we know mod a...wer any complex number mod equals 1.

in this case jus substitute the value of a which u got and find which is max.

so options r rong

thnks..

So finally.
1. options wrong
2. options wrong
3. A
4. need one more opinion (iitimcomin has already answerd and i agree with him) given A while we think B
5. still no conclusion

6/7/8/9 unsolved in #26

x[ss]1[/ss]=2RcosA
x[ss]2[/ss]=2RcosB
x[ss]3[/ss]=2RcosC

a=2RsinA
b=2RsinB
c=2RsinC

da/dA=2RcosA
db/dB=2RcosB
dc/dC=2RcosC

so 2R(cosA+cosB+cosC)=6

so cosA+cosB+cosC=3/R

from AM≥GM

(x[ss]1[/ss]+x[ss]2[/ss]+x[ss]3[/ss])/3 ≥(x[ss]1[/ss]x[ss]2[/ss]x[ss]3[/ss])[p]1/3[/p]

x[ss]1[/ss]+x[ss]2[/ss]+x[ss]3[/ss]=2R(cosA+cosB+cosC)=6

so 6/3≥(x[ss]1[/ss]x[ss]2[/ss]x[ss]3[/ss])[p]1/3[/p]

2[p]3[/p]≥x[ss]1[/ss]x[ss]2[/ss]x[ss]3[/ss]

8≥x[ss]1[/ss]x[ss]2[/ss]x[ss]3[/ss]

got anything about inradius though?

thanks for the others

can u just show this in a diag?
x1=2RcosA
x2=2RcosB
x3=2RcosC

well nothing much abt inradius

got jus one cond R+r=3 .......searching for one more

and regarding that diag jus check in Mlk or for that mateer ne jee maths book its given there...i cant drw all that

and in q 9 is it ≤ or [b]≥[/b]  

9. ≤ only

7/9/10/11 unsolved

Q11) without any doubt ans is 8

surely options r wrong

Q10) ur ans is correct

Confirm...

7) a
9) b (not too sure with this)....though I think chetan has a point in his last post....

7. given (C) .. but do post how u got A
9. will confirm tomorrow.

actually ans for 7 shud be 1 only

cosA +cosB+cosC=3/R

and cos A+ cosB+cosC=1+[frac]r[/]R[/frac]

so from this we get R+r=3

and we also know that the circumradius of a triangle is greater than or equal to twice its inradius. R≥2r

so R cant be less than r

if r=1 we get R=2 which is the limiting cond

[b]r cant be 3 bec than R=0 which is not possible[/b]

similarly r=2 and r=4 also caNCEL OUT ...with r=2 we get R=1 which is not possible and with r=4 we get R=-1 which is also not possible

but this is with help of options.....still din get a proper soln...maybe soumik will do.