3q5 is 6 na.//
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
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41 Answers
3Q3 ...the reason helps find out that its a rt angled triangle so i think it shud be A
106yeah but pls explain this
tangent PT
this means length of PT or something else?
3yeah 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...
106But then T≡(-1,0)
P≡(1,2)
But PT≠2 (as obtained by doing √S1)