CBSE Class 11-science Answered

Qn. # (26)
Let C(3,-1) be centre of circle . AB is chord of length 6 cm and its equation is 2 x+ 5y +18 = 0
Let CD be perpendicular to AB ,
Since CD is perpendicular to AB , equation of CD is 5x -2y = k , where k is constant to be determined.
Line CD is passing through C (3, -1) . Hence we get equation of line as
5(3) - 2 (-1) = k or k = 17
Point of intersection is determined by solving the line equations as given below
2 x + 5 y = -18
5x - 2y = 17
By solving above equations , we get
x = 49/29 ; y = -124/29 ;
Point of intersection is D ( 49/29 , -124/29 )
Length of CD is
Radius R is calculated as
cm
If (h, k ) is centre of circle and R is radius of circle , then equation of circle is
(x-h)2 + ( y-k)2 = R2
Hence equaion of circle with centre ( 3, -1) and radius 4.63 cm is
(x-3)2 + (y+1)2 = (4.63)2 = 21.4
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Qn. # 28
y = √x
.........................(1)
If Δx → 0 , then
......................(2)
In above approximation, higher powers of Δx are neglected in the series expansion
..................(3)
Using eqn.(3) , we rewrite eqn.(2) as
Hence we rewrite eqn.(1) as
...............................(4)
if Δx → 0 , then we get from eqn.(3) , θ → 0 and cosθ → 1
From eqn.(3) , we can write Δx = 2 √x θ
Hence eqn.(4) becomes