f(x)=2x+tan−1x and g(x)=ln(√1+x2+x) and x∈[0,3] g′(x)=
1
√1+x2
Now, 0≤x≤3 0≤x2≤9 1≤1+x2≤10 So, 2+
1
10
≤f′(x)≤3
21
10
≤f′(x)≤3 and
1
√10
≤g′(x)≤1 option (4) is incorrect From above, g′(x)<f′(x)∀x∈[0,3] Option (1) is incorrect. f′(x)&g′(x) both positive so f(x)&g(x) both are increasing So, max(f(x). at x=3 is 6+tan−13 Max(g(x) at x=3 is ln(3+√10) And 6+tan−13>ln(3+√10) Option (2) is correct