16)b,c
see cos-13 is not possible.as cos can take values in interval [-1,1]
1.. √1-x2 + √1-y2 = a(x-y) then dy/dx is
a) (1-x2)/(1-y2) b) (1-y2)/ (1-x2) c) (√1-x2)/(1-y2) d) √1-y2)/ (1-x2)
2. If y= cosec x (cot–1x) thendy/dx
a. x/√1-x2 b) -x/√1+x2
c) x/√1+x2 d) None
3. The tangents to the circle x2 + y2 = 25 which is parallel to line 2x-y+1 = 0 are
a) y=2x+4√5 b) y=2x-4√5 c) y=2x+5√5 d) y=2x-5√5
4. ∫1/sin2x+cos2xdx =
a) -2cot2x+c b) tanx-cotx+c c) 9+(sec2x)/2 d) 3-tan2x/2 e) (3-sec2x)/2 f) (tan2x/2)-5
5. lim 0 to pi/4∫tann xdx , n belongs to I (more than one choice)
a) In+In-2=1/n-1
b) In+In-2=1/n
c) I1 = I3 + 2I5
d) None
6. At x=3/2 the value is real for (more tahn one choice)
(a) cos–1 2x (b) cosec–1 x (c) tan–1 x (d) None
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5 Answers
29For the third one just use the formula for the tanget with slope m
y = mx ± a√1+ m2 ...
1let the eqn be y=2x+c...
putting thge eqn..in the eqn of the circle....
we get,
5x2+4cx+(c2-25)=0
as tangent touches the circle at 1 pt.... D=0
ie.c=5√5....
hence aans is ..(c)
39For the first one substitute x = sinα and y = sinβ
\sqrt {1-sin^2\alpha} + \sqrt {1-sin^2\beta} = a(sin\alpha - sin\beta)
=> cos\alpha + cos\beta = 2asin(\frac{\alpha-\beta}{2})cos(\frac{\alpha+\beta}{2})
=> 2cos(\frac{\alpha-\beta}{2})cos(\frac{\alpha+\beta}{2}) = 2asin(\frac{\alpha-\beta}{2})cos(\frac{\alpha+\beta}{2})
=>
cot(\frac{\alpha-\beta}{2}) = a
=>\frac{\alpha-\beta}{2} = arccot(a)
=> arcsin(x) - arcsin(y) = 2arccot(a)
Now differentiate. It's much easier. Differentiation of cot-1a is zero.
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