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Answer 1

Answer:

max =π²/4     min= π²/8

Step-by-step explanation:

to find max and min differentiate the eqn and equals to 0

after differentiating, u get

sin⁻¹x =cos⁻¹x

it happens only when x = 1/√2

sin⁻¹x =cos⁻¹x = π/4

put the value in eqn = π²/16 +π²/16 =π²/8

for maxx , sin⁻¹x or cos⁻¹x lies in domain from( -1 ,1)

if one get π/2 then other get 0

so max vaπlue =π²/4