Question

if fx = tanx (0, 90) then the function is said to be what type of function

Answer 1

Step-by-step explanation:

f(x)=tanx−x

f(x)=tanx−xf

f(x)=tanx−xf ′

f(x)=tanx−xf ′ (x)=sec

f(x)=tanx−xf ′ (x)=sec 2

f(x)=tanx−xf ′ (x)=sec 2 x−1

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec 2

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec 2 x≥1∀∈R [By squaring on both sides]

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec 2 x≥1∀∈R [By squaring on both sides]f

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec 2 x≥1∀∈R [By squaring on both sides]f ′

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec 2 x≥1∀∈R [By squaring on both sides]f ′ (x)=sec

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec 2 x≥1∀∈R [By squaring on both sides]f ′ (x)=sec 2

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec 2 x≥1∀∈R [By squaring on both sides]f ′ (x)=sec 2 x−1≥0 [By sub. 1 on both sides]

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec 2 x≥1∀∈R [By squaring on both sides]f ′ (x)=sec 2 x−1≥0 [By sub. 1 on both sides]∴f

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec 2 x≥1∀∈R [By squaring on both sides]f ′ (x)=sec 2 x−1≥0 [By sub. 1 on both sides]∴f ′

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec 2 x≥1∀∈R [By squaring on both sides]f ′ (x)=sec 2 x−1≥0 [By sub. 1 on both sides]∴f ′ (x)≥0∀∈R

f(x)=tanx−xf ′ (x)=sec 2 x−1we know secx≥1∀x∈R (in domain)sec 2 x≥1∀∈R [By squaring on both sides]f ′ (x)=sec 2 x−1≥0 [By sub. 1 on both sides]∴f ′ (x)≥0∀∈R⇒f(x) is increasing

Answer 2

क) 'पण्डिता' 'सरस्वती' इति उपाधिभ्यां का विभूषिता? (ख) रमा कुतः संस्कृतशिक्षां ...