If f1(x) = sinx + tanx and f2(x) = 2x then:-
a) f1(x) > f2(x) for all x ∈ (0 , π/2)
b) f1(x) < f2(x) for all x ∈ (0 , π/2)
c) f1(x) - f2(x) = 0 has exactly one root for all x ∈ (0 , π/2)
d) None of these
Answers
Answered by
2
Answer:
et f(x)=f
1
(x)−f
2
(x)
f(x)=2x−3sinx+xcosx
⇒f
′
(x)=2−2cosx−xsinx
⇒f
′′
(x)=sinx−xcosx=cosx(tanx−x)
⇒f
′′
(x)>0∀x∈(0,
2
π
).
Thus f
′
(x) is increasing in (0,
2
π
);f
′
(0)=0
⇒f
′
(x)>0∀x∈(0,
2
π
).
⇒f(0)=0
⇒f(x)>0∀x∈(0,
2
π
).
⇒f
1
(x)>f
2
(x)
⇒f(x) is increasing in (0,
2
π
)
Similar questions