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Maths Question 20 – JEE-MAIN 2026

Let x2f(a2+7a+3)+y2f(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[α,β], then α2+β2 is equal to :

Recall the conditions for an ellipse and how the major axis orientation relates to the denominators in the standard equation.

Step 1: Identify conditions for the ellipse and major axis✦ Active

The given equation is x2f(a2+7a+3)+y2f(3a+15)=1. For this to represent an ellipse, the denominators must be positive. Since f is a positive function, f(a2+7a+3)>0 and f(3a+15)>0 are always satisfied. The major axis is along the y-axis, which implies that the denominator of the y2 term must be greater than the denominator of the x2 term.

f(3a+15)>f(a2+7a+3)
Step 2: Apply the property of a strictly decreasing function○ Expand

Given that f is a strictly decreasing function, if f(x1)>f(x2), then it must be that x1<x2. Applying this property to the inequality from Step 1:

3a+15<a2+7a+3

Rearranging the terms to form a quadratic inequality:

a2+7a+33a15>0 a2+4a12>0
💡 Teacher's Secret Hint

Remember that the inequality sign flips when dealing with a strictly decreasing function.

Step 3: Solve the quadratic inequality and calculate α2+β2○ Expand

Factor the quadratic expression:

(a+6)(a2)>0

This inequality holds when a<6 or a>2. The set of all possible values of a is (,6)(2,). This can be expressed as R[6,2]. Comparing this with the given form R[α,β], we identify α=6 and β=2. Now, calculate α2+β2:

α2+β2=(6)2+(2)2=36+4=40
💡 Teacher's Secret Hint

Ensure correct identification of α and β from the interval notation.

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