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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
  1.  -2 ≤ f(2) – f(-1) ≤ 1
  2. 1 ≤ f(2) – f(-1) ≤ 2
  3. - 3 ≤ f(2) – f(-1) ≤ 0
  4. - 2 ≤ f(2) – f(-1) ≤ 0

Grade:10

1 Answers

Y RAJYALAKSHMI
45 Points
7 years ago
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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