Let f:[-1, 2] → R be differentiable such that 0 ≤ f’(t) ≤ 1 for t ∈ [-1, 0] and -1 ≤ f’(t) ≤ 0 for t ∈ [0, 2]. Then options -2 ≤ f(2) – f(-1) ≤ 1 1 ≤ f(2) – f(-1) ≤ 2 - 3 ≤ f(2) – f(-1) ≤ 0 - 2 ≤ f(2) – f(-1) ≤ 0

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Y RAJYALAKSHMI · Answer

Property: If m & M are the minimum and maximum values of f(x) in [a, b], then ∫ f( x) between the limits a & b lies between m(b – a ) & M(b – a) 0 ≤ f ’ (t) ≤ 1 for –1 ≤ t ≤ 0 =>f (t) = ∫ f ‘ (t) between the limits -1 & 0 = f (0) – f (–1) By using above property we have 0(0 + 1) ≤ f (0) – f (–1) ≤ 1( 0 + 1) => 0 ≤ f (0) – f (–1) ≤ 1 -------------- (1) – 1 ≤ f ’ (t) ≤ 0 for 0 ≤ t ≤ 2 =>f (t) = ∫ f ‘ (t) between the limits 0 & 2 = f (2) – f (0) By using above property we have –1(2 – 0) ≤ f (2) – f (0) ≤ 0( 2 – 0) => –2 ≤ f (2) – f (0) ≤ 0 -------------- (1) Adding (1) & (2), we get –2 ≤ f (2) – f (– 1) ≤ 1 Ans: –2 ≤ f (2) – f (– 1) ≤ 1 – Option (1)

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Let f:[-1, 2] → R be differentiable such that 0 ≤ f’(t) ≤ 1 for t ∈ [-1, 0] and -1 ≤ f’(t) ≤ 0 for t ∈ [0, 2]. Then options -2 ≤ f(2) – f(-1) ≤ 1 1 ≤ f(2) – f(-1) ≤ 2 - 3 ≤ f(2) – f(-1) ≤ 0 - 2 ≤ f(2) – f(-1) ≤ 0

Let f:[-1, 2] → R be differentiable such that 0 ≤ f’(t) ≤ 1 for t ∈ [-1, 0] and -1 ≤ f’(t) ≤ 0 for t ∈ [0, 2]. Then
options

-2 ≤ f(2) – f(-1) ≤ 1

1 ≤ f(2) – f(-1) ≤ 2

- 3 ≤ f(2) – f(-1) ≤ 0

- 2 ≤ f(2) – f(-1) ≤ 0

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