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Mathematics Test 32

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Mathematics Test 32
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  • Question 1
    4 / -1

    Consider the function f(x) = |x – 2| + |x – 5|, x ∈ R. 
    Statement 1: f’(4) = 0 
    Statement 2: f is continuous in [2, 5], differentiable in (2, 5) and f(2) = f(5)

    Solution

    f(x) = 7 – 2x; x < 2 
    = 3; 2≤ x ≤ 5 
    = 2x – 7; x > 5 
    f(x) is constant function in [2, 5] 
    f is continuous in [2, 5] and differentiable in (2, 5) and f(2) = f(5) 
    by Rolle’s theorem f’(4) = 0 
    ∴ Statement 2 and statement 1 both are true and statement 2 is correct explanation for statement 1. 
    Hence, option B is correct.

  • Question 2
    4 / -1

    If the two circles (x − 1)2 + (y − 3)2 = r2 and x2 + y2 − 8x + 2y + 8 = 0 intersect at two distinct points, then

    Solution

    (x − 1)2 + (y − 3)2 = r2 (centre at 1,3 and radius of r)
    (x − 4)2 + (y + 1)2 − 16 − 1 + 8 = 0 
    (x − 4)2 + (y + 1)2 = 9 (centre at 4,-1 and radius of 3)
    Distance between the centres = 5
    Therefore, we know that 3+r > 5 or r>2
    Also, we know that r cannot be more than 5+3 (otherwise it will completely enclose the other circle and there will be no intersection)
    Hence, 2<r<8

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