Let x^2 f(a^2+7a+3) + y^2 f(3a+15) = 1 represent an ellipse with major axis along y-axis, where f is a strictly decreasing positive function on R . If the set of all possible values of a is R - [ , ], then ^2+ ^2 is equal to:

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Correct answer: B. 40

For the given equation to represent an ellipse with its major axis along the y-axis, the denominator of y^2 must be strictly greater than the denominator of x^2. f(3a+15) > f(a^2+7a+3) Since f is given as a strictly decreasing positive function on R , the inequality sign reverses for the arguments: 3a+15 a^2 + 4a - 12 > 0 (a+6)(a-2) > 0 The solution to this inequality is a (- , -6) (2, ). This can be rewritten in terms of the set difference as a R - [-6, 2]. Comparing this with the given set R - [ , ], we get = -6 and = 2. ^2 + ^2 = (-6)^2 + (2)^2 = 36 + 4 = 40 Answer: 40

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