Question

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

Answer 1

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

)