23btw tnx for q3
Q1] If ax2 -bx + 1 = 0 have 2 real roots (a,b are real) , then the greatest value of 2a-b is ?
a] -1/2 (b] 0 (c] 1/2 (d] none
Q2] Let the equation of a curve , x = a(θ+sinθ) ; y=a(1 - cosθ)
If θ changes at a const rate k , then the rate of change of slope of the tangent to the curve at θ = pi / 3 is
a] 2k/ √3 (b] k/√3 (c] k (d] None
(ans shud be 2k/3 na ? ans key gives ans as A , pls explain )
Q3] The largest distance btwn the point (a,0 ) frm the curve 2x2+y2 -2x = 0 is
a] √1-2a+a2 b] √1+2a+a2 c] √1+2a-a2 d] √1-2a+2a2
Q4] x =3rcosθ ;y=3+rsinθ ; z = -r(3cosθ+sinθ)
max value of xyz is ? options : 1,2,3 or 4
do we hav n e shorted method ? for Q4 ?
Q5] minimum value of sinA+sinB+sinC where A+B+C = pi is
(A,B,C are real )
a] 0 b] +ve c] -ve d]none
btw n e oder method than trigo ? pls post ur trigo method also
thnx for helping
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17 Answers
1a\left [ \left ( \frac{2ax-b}{2a} \right )^2 +\left ( \frac{a^2-b^2}{4a^2} \right ) \right ] > 0
\left [ \frac{(2ax-b)^2}{4a} +\left ( \frac{4a^2-b^2}{4a} \right ) \right ] > 0
\left (2ax-b)^2 + 4a-b^2 > 0
settin x= 1 we have
\left (2a-b)^2 > b^2- 4a^
(2a-b)^2 > 0 \rightarrow 2a-b >0
therefore any value > than 0 will hold ...
d)
2. qwerty said k = dθ / dx
dy / dx = k / sinθ
d / dx ( dy / dx ) = d / dx ( kcosec θ)
rate of chane of slope = 2k /3
ur crct
1is my answer is right?
i only apply the concept of r ange. we do this type of questions in function.
23jangra in post 12 i think u r wrong , bcz
b2 - 2b -2y ≥ 0
here when u r using D ≤ 0 , such that quadratic is in b ,
value of y depends linearly on b , i.e y = 2a-b ,
so , it isnt a constant like C in Ax2+Bx+C
1let (h,k) be the points which are closest to point P(a,0)
let z be the distance between them
z2 = (h - a)2 + (k)2 ......................... (1)
as point (h,k) also satisfy 2x2 + y2 - 2x = 0 ............(2)
from both the equations u can get the equation
z2 = -x2 + a2 + 2x(1-a)
now apply maxima and minima rule
u will get the answer (d) √1-2a+2a2
1let take 2a - b = k
ax2 - bx +1 = 0
now b2 - 4a ≥ 0 ...................................................... (1)
now put a = k+b / 2 in equation (1)
we will get
b2 - 2b -2y ≥ 0
here D ≤ 0
so 22 - 4(-2y) ≤ 0
hence we get y ≤ - 1/2
so greatest value of 2a - b is equal to - 1/2
23ans for 1 is given as A ,
but i hav dis doubt , take a = 4 , b = 5 ,
roots are 1/4 and 1
then 2a - b = 3 ???????? [2]
one more thing ,
wat is \frac{d}{dx^{n}}f^{n}(f(x))
62Q1] If ax2 -bx + 1 = 0 have 2 real roots (a,b are real) , then the greatest value of 2a-b is ?
a] -1/2 (b] 0 (c] 1/2 (d] none
2 real roots are given .. so D≥0
So b2>4a
We have to maximize 2a-b
2a-b< b2 /2-b=b2/2-b+1/2-1/2 = 1/2(b-1)2-1/2
Now, I think the problem setter has made a mistake...
For Q5 OOps.. my wrong.. the answer will be -ve.. and that is what i did show :P
62Q4] x =3rcosθ ;y=3+rsinθ ; z = -r(3cosθ+sinθ)
max value of xyz is ? options : 1,2,3 or 4
do we hav n e shorted method ? for Q4 ?
What is the value of r???
It depends on r completely!
If your question is over all r, then the answer will be zero.. so I am a bit confused... (even -ve!)
62Q5] minimum value of sinA+sinB+sinC where A+B+C = pi is
(A,B,C are real )
a] 0 b] +ve c] -ve d]none
Try to see jenson's inequality.. That just kills this question...
You will get ( sin a + sin b + sin c )/3 ≤ sin (a+b+c)/3 = sin (pi/3)
Hence the answer will be 3√3/2
(This will be true only if a, b and c are given to be +ve....
With the options at disposal.. it is clear that the minimum value will be none of these bcos it si not known if a, b and c are +ve..
we can chose a = -pi/2, b=-pi/2 and c= 2pi to get a value of -2 as well... so it is clear for this question that the answer will be d..
62Q3] The largest distance btwn the point (a,0 ) frm the curve 2x2+y2 -2x = 0 is
a] √1-2a+a2 b] √1+2a+a2 c] √1+2a-a2 d] √1-2a+2a2
the given equation is 2(x-1/2)2+y2=1/2
You have to find the minimum distance from the point (a,0)
The curve which is the given ellipse cuts the x axis at 0,1
You can either try parametric form as wlll for this question...