Prob 3
sin8 x ?? cos8 x = (sin4 x ?? cos4 x)(sin4 x + cos4 x) = (sin2 x ?? cos2 x)(sin4 x + cos4 x)
1 ?? 2 sin2 x cos2 x = (sin2 x + cos2 x)2 ?? 2 sin2 x cos2 x = (sin4 x + cos4 x)
Thus sin8 x??cos8 x
1??2 sin2 x cos2 x = sin2 x ?? cos2 x = ??cos 2x. Thus
R sin8 x??cos8 x
1??2 sin2 x cos2 xdx = ??sin 2x
2 + C
Prob 4
for the function f(x) = jxj
f0(x) =
(
??1 ??1 x < 0
1 0 < x 1
Thus @ c in (-1,1) where f0(c) = 0.
The function f(x) = jxj does not satis es hypotheses of Lagrange`s Mean Value Theorem because it is not
di erentiable in (-1,1).
Prob 5
Using the derivative of sin??1 x which is d
dx sin??1 x = p 1
1??x2 and chain rule we get the required derivative
which is x??1
x
p
1??x2 .
1
Prob 6
729 = 93 ()) log9 729 = 3 then log3 log3 log9 729 = log3 log3 3 = log3 1 = 0 and 3 logp
3 3 = 3 2 = 6.
Answer is 6.
Prob 7
Given function is f(x) = 1+2 sin x+3 cos2 x to nd maximum or minimum rst nd the derivative which is
f0(x) = 2 cos x??3 sin 2x = 2 cos x(1??3 sin x). Thus critical points are obtained by cos x = 0 and sin x = 1=3.
As the given interval is 0 x 2=3 ()) critical points are x = =2 ; x = sin??1(1=3) .
second derivative is f00(x) = ??2 sin x ?? 3 cos 2x.
f00(=2) = 1 > 0 hence x = =2 is point of minima and minimum value is f(=2) = 3.
f00(sin??1(1=3)) = ??2??4
p
2
3 < 0, thus x = sin??1(1=3) is point of local maxima and local maximum value is
f(sin??1(1=3) = 13
3 .
Prob 8
Given f(x) = x2??x+1
x2+x+1 has domain R and f0(x) = 2(x2??1)
(x2+x+1)2 . Thus critical points are x = 1.
f00(x) = ??4x3+12x+1
(x2+x+1)3
f00(1) = 7=27 > 0
f00(??1) = ??7 < 0
f(1) = 1=3 ; f(??1) = 3 are minimum and maximum values of the function respectively.
Prob 9
x2+1
x+1 ?? ax ?? b = (1??a)x2??x(a+b)+(1??b)
x+1
given limx!1
(1??a)x2??x(a+b)+(1??b)
x+1 = 0
thus degree(num)
raju , 12 Years ago
Grade 12
