If the function f x = - x , x 1 a + cos - 1 ⁡ x + b , 1 ≤ x ≤ 2 is differentiable at x = 1 , then a b is equal to

Question

If the function f x = - x , x 1 a + cos - 1 ⁡ x + b , 1 ≤ x ≤ 2 is differentiable at x = 1 , then a b is equal to, π + 2 2, π - 2 2, - π - 2 2, - 1 - cos - 1 ⁡ ( 2 )

MARKS App · Accepted Answer

f x = - x x 1 a + cos - 1 ⁡ x + b 1 ≤ x ≤ 2 f x is continuous at x = 1 as f x is differentiable at x = 1 . ⇒ lim x → 1 f ( x ) = lim x → 1 + ( a + cos − 1 ( x + b ) ) = f ( 1 ) ⇒ - 1 = a + cos - 1 ⁡ 1 +b cos - 1 ⁡ 1 + b = - 1 - a ........ i f x is differentiable at x = 1 . ⇒ LHD = RHD ⇒ - 1 = - 1 1 - 1 + b 2 ⇒ 1 - 1 + b 2 = 1 ⇒ b = - 1 ...... ii From i ⇒ cos - 1 ⁡ ( 0 ) = - 1 - a ∴ - 1 - a = π 2 a = - 1 - π 2 a = - π - 2 2 ...... iii ∴ a b = π + 2 2

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