Ques1) Find the point of inflexon of (x-5)55 (x-6)66.
Ques2) Show that the extremum value of the function f(x) = (Sin-1 x) 3 + (cos-1 x)3 (-1<x<1) is \pi 2/ 32.
Ques3) Find the least value of the expression
2log10x - log x(0.1) for x>1.
Ques4) Show that the function f(x) = x / (1+xtanx) has one point of minimum in the interval (0, \pi/2).
Ques5) If f(X) = \sum_{n=1}^{m}{} (x-n) 2; has minimum at X 0 then show that X 0 is (m+1)/2.
Qus6) Let f(x) = 4 tanx-tan2x+tan3xwhere x\neq n\pi+( \pi/2). Then show that f(x) has neither local minima nor maxima.
393) let log10x=t
then we have the given expression as 2t+1/t
taking the derivative we have 2-1/t2=0
this implies t=1/√2
means x=101/√2
taking the second derivative of 2t+1/t and substituting t=1/√2 we get the expression to be positive, this means we have a minimum at this point.
now substitute this value to get the minimum value of the given expression as need.
396)
let tanx=t
then the given expression can be rewritten as 4t-t2+t3
the derivative of which can be written as
(1+t2)(3t2-2t+4)
this can never be zero as 1+t2 can never be zero
and also 3t2-2t+4 also can never be zero since discriminant is <0.
395)
f(x)=(x-1)2+(x-2)2+.............(x-m)2
f`(x)=2[mx+m(m+1)/2]
f`(x)=0 implies x=(m+1)/2
also f``(x)=2m which is always positive
so the given expression has a minimum at x=m(m+1)/2 =X0
394) i think this u should be trying out yourself aas on differentiating u will get
( 1-x2sec2x)/(1+xtanx)2
for this to be zero we have
x2sec2x=1
which has a solution in (0,pi/2) and its easy to prove that it is point of minimum of the given expression in that particular interval.
Anonymous how do we confirm that it really has a solution in (0,pi/2) ??391) i am not sure if that is in the syllabus.
2) do it yourself (simple one)
