Q1. Let \LARGE\!\phi (x) = sinx\int_{0}^{x}{costdt} + 2\int_{0}^{x}{tdt} + cos^2x - x^2.
If \LARGE\!x^2 - 2x +3 \geq \phi (x) \vee x\epsilon R, 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 a is a complex number, satisfying\LARGE\!ia^3+a^2-a+i=0
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 12|z|2
Stmnt 2: Multiplying any complex number by i rotates it by π/2 wrt itself in anticlockwise direction
(A or B???)
Q4. Let \LARGE\!I_{1} = \int_{0}^{1}{\frac{1+x^8}{1+x^4}dx}\; and \; I_{2}=\int_{0}^{1}{\frac{1+x^9}{1+x^2}dx},
then
(a) 0<I1<I2<1
(b) 0 < I2 < I1 < 1
Q5. Normal of parabola y2=4x at P and Q meets at R(x2,0) and tangents at P and Q meets at T(x1,0), If x2=3, then the length of latus rectum plus tangent PT will be
(a) 3 (b) 6 (c) 1 (d) 8
1actually 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
106@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 a2 = -i (=> a=±1√2(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 √S1 = 2
106more doubts added.
Q6. \vec{a} \textup{ and }\vec{b}\textup{ are non-zero, non-collinear vectors such that}
|\vec{a}| =2,\; \vec{a}.\vec{b}=1\textup{ and the angle between } \vec{a}\textup{ and }\vec{b}\textup{ is }\pi /3.
\textup{If }\vec{r}\textup{ is any vector satisfying }\vec{r}.\vec{a}=2,\; \vec{r}.\vec{b}=8,\; (\vec{r}+2\vec{a}-10\vec{b}).(\vec{a}\times \vec{b}) = 4\sqrt{3}
\textup{and is equal to }\vec{r}+2\vec{a}-10\vec{b}=\lambda (\vec{a}\times\vec{b}), \; then\; \lambda =
(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 \frac{da}{dA}+\frac{db}{dB}+\frac{dc}{dC}=6.
Distance of orthocentre of triangle ABC from vertex A,B and C is x_{1},x_{2},x_{3} respectively
7. Inradius of triangle ABC
(a) 1 (b) 2 (c) 3 (d) 4
8. Maximum value of x_{1}x_{2}x_{3} is
(a) 4 (b) 6 (c) 8 (d) 10
9. \frac{dx_{1}}{da}+\frac{x_{2}}{db}+\frac{dx_{3}}{dc} is always ≤
(a) -3√3 (b) 3√3 (c) 1 (d) 6
106Q10. For the curve x+y+1=ey (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
1so for 2nd wats the prob ?
u hav got \left|a \right|=1
so \\\left|a +(-3-4i)\right|\leq \left|a \right|+\left|-3-4i \right|\\ \left|a +(-3-4i)\right|\leq 1+5\\ \left|a +(-3-4i)\right|\leq 6\\
106Whats 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?
1GUYZ...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
106thnks..
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
1x1=2RcosA
x2=2RcosB
x3=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
(x1+x2+x3)/3 ≥(x1x2x3)1/3
x1+x2+x3=2R(cosA+cosB+cosC)=6
so 6/3≥(x1x2x3)1/3
23≥x1x2x3
8≥x1x2x3
106got anything about inradius though?
thanks for the others
can u just show this in a diag?
x1=2RcosA
x2=2RcosB
x3=2RcosC
1well 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 ≥
1Q11) without any doubt ans is 8
surely options r wrong
Q10) ur ans is correct
11Confirm...
7) a
9) b (not too sure with this)....though I think chetan has a point in his last post....
1067. given (C) .. but do post how u got A
9. will confirm tomorrow.
1actually ans for 7 shud be 1 only
cosA +cosB+cosC=3/R
and cos A+ cosB+cosC=1+rR
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
r cant be 3 bec than R=0 which is not possible
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.
106for 2. ans given (D)
i was getting a=i, and a2 = -i (=> a=±1√2(1-i))
11Multiplication of Z with i then vector for Z rotates a right angle in the positive sense.
3Q1.
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...
106but then
just simplify the expression of phi(x)
∫costdt from 0 to x = sinx and 2∫tdt from 0 to x = x2
So, phi(x) = sin2x + x2+cos2x-x2
= 1 for all x
29Ya 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..
3Q2
all i could think of is ... put a=x+iy
equate
a^3 =0
and
a^2-a+1=0
im gettin 2...???[doubtful]
