Let the function f : [-7, 0] → R be continuous on [-7, 0] and differentiable on (-7, 0) . If f(-7) = -3 and f'(x) ≤ 2 for all x (-7, 0) , then for all - Mathematics MCQ Mathematics Mock Test - 5 2026 | ExamTest.live

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Let the function f : [- 7, 0] \to R be continuous on [- 7, 0] and differentiable on (- 7, 0) . If f (- 7) = - 3 and f^{'} (x) \leq 2 for all x \in (- 7, 0), then for all such functions f, f (- 1) + f (0) lies in the interval:

Solution & Step-by-step Explanation

Using Lagrange Mean Value Theorem (LMVT) on [- 7, x] : \frac{f ( x ) - f ( - 7 )}{x - ( - 7 )} = f^{'} (c) \leq 2 f (x) - (- 3) \leq 2 (x + 7) f (x) + 3 \leq 2 x + 14 ⟹ f (x) \leq 2 x + 11 For x = - 1 : f (- 1) \leq 2 (- 1) + 11 = 9 For x = 0 : f (0) \leq 2 (0) + 11 = 11 Adding the two inequalities: f (- 1) + f (0) \leq 9 + 11 = 20 The range of values for the sum is (- \infty, 20] .

Try it yourself before checking the explanation above.

Let the function f : [- 7, 0] \to R be continuous on [- 7, 0] and differentiable on (- 7, 0) . If f (- 7) = - 3 and f^{'} (x) \leq 2 for all x \in (- 7, 0), then for all such functions f, f (- 1) + f (0) lies in the interval:

[- 6, 20]

(- \infty, 20]

(- \infty, 11]

[- 3, 11]

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