1[-pi/2,pi/2],,,,,am very weak in this plz.if am correct!
10 Answers
1f(x)=cosx + cos2x
=let cosx =t,then
e(t)=2t2+t-1
find the range of this expression.
39y = cosx + cos2x = cosx + 2cos²x - 1
Or y/2 = t² - t/2 - 1/2
=> y/2 = t² - t/2 + 1/16 - 1/16 - 1/2
=> y/2 = (t - 1/4)² - 9/16
=> (8y - 9)/16 = (t - 1/4)²
Now as RHS is always non-negative, same condition applies to LHS.
(8y - 9)/16 ≥ 0
=> y ≥ 9/8
So range is [9/8, ∞)
1f(x)=cosx + cos2x =let cosx =t,then e(t)=2t2+t-1 we know that range of expression f(x)=ax2+bx +c ki range hoti hai range=[-D/4a,∞) now the range can b found.
but the expression can have max value of 2
as
-1≤ cosx≥1
39Pranav...try yourself. Put t = cos(x). cos(x) oscillates from -1 to 1, so the RHS oscillates from -5/4 to 3/4.
Now take the extra values from the LHS to the RHS and see how y ranges.
1pritish we have to take into account that -1<=cos x<=1 .hence answer [-9/8,2]
39Yes I forgot to mention that the answer I gave is a subset of the range. I should've mentioned the testing part.
1can we find the inverse of function and then find domain.
its inverse is:-
y=cos^-^1(\frac{2.2^.^5\sqrt{x+9/8}-1}{4}})
take 2^.5 as √2 it 2√2