- JEE Main 2026
- 08 April Evening
- MCQ
- 4 marks
Ellipse with an Abstract Decreasing Function – JEE Main 2026 PYQ
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Quick Summary
The ellipse x²/f(a²+7a+3) + y²/f(3a+15) = 1 has its major axis along the y-axis exactly when the y-denominator exceeds the x-denominator: f(3a+15) > f(a²+7a+3). Since f is strictly decreasing, this output inequality REVERSES on the inputs: f(3a+15)>f(a²+7a+3) ⟺ 3a+15 < a²+7a+3 ⟺ a²+4a−12>0 ⟺ (a+6)(a−2)>0. This holds for a<−6 or a>2, so the excluded interval is [α,β]=[−6,2], giving α²+β² = 36+4 = 40.
Question Replay
Let x²/f(a²+7a+3) + y²/f(3a+15) = 1 represent an ellipse with major axis along the 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:
Options
Correct Answer
Correct Answer: Option (B) — 40
Why This Answer
The question deliberately hides the denominators behind an abstract function f rather than concrete numbers — this is a signal that the ONLY property of f that matters is its monotonicity, not its formula. Because f is positive everywhere on ℝ, both denominators are automatically valid (positive) for every real a, which removes an entire layer of case-checking that a more concrete version of this question would require.
The one real piece of work is correctly reversing the inequality: a strictly decreasing function flips the direction of an inequality when you move from outputs back to inputs. Forgetting this flip is the single most likely way to get the wrong quadratic inequality (and hence the wrong excluded interval) here.
Core Concept
Major axis condition for an ellipse. For x²/A + y²/B = 1 with A,B>0, the major axis lies along the y-axis exactly when B > A (the larger denominator sits under the axis containing the major axis).
Monotonic functions and inequality reversal. If f is strictly decreasing, then f(u) > f(v) ⟺ u < v — larger inputs give smaller outputs, so an inequality on the f-values corresponds to the OPPOSITE inequality on the raw inputs. This is the single tool needed to strip the abstract function away entirely.
f strictly decreasing ⇒ 3a+15 < a²+7a+3 ⇒ 0 < a²+4a-12 = (a+6)(a-2) ⇒ a < -6 or a > 2
Excluded interval: [α,β] = [-6, 2]
α² + β² = 36 + 4 = 40
Step-by-Step Solution
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State the major-axis condition. For x²/f(a²+7a+3) + y²/f(3a+15) = 1 to have its major axis along the y-axis, need f(3a+15) > f(a²+7a+3) (the y-denominator must be the larger one).
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Reverse the inequality using strict monotonicity. Since f is strictly decreasing, f(u)>f(v) ⟺ u<v. Applying this with u=3a+15, v=a²+7a+3:
3a + 15 < a² + 7a + 3 -
Simplify to a standard quadratic inequality:
0 < a² + 7a + 3 - 3a - 15 = a² + 4a - 12 -
Factor and solve. a²+4a−12 = (a+6)(a−2) > 0 ⇒ a < −6 or a > 2.
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Read off the excluded interval. The valid set is ℝ − [−6,2], so α=−6, β=2 (note f being positive everywhere means no further positivity constraint narrows this set further).
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Compute α² + β² = (−6)² + 2² = 36 + 4 = 40.
Final Answer
Final Answer: Option (B) — 40
α²+β² = 40. In the exam, whenever a problem hides expressions behind an abstract monotonic function, immediately translate the condition into a statement about the RAW inputs using the monotonicity direction — the function itself almost never needs to be found.
Key Facts to Remember
- x²/A+y²/B=1 has major axis along y ⇔ B>A>0; along x ⇔ A>B>0.
- For strictly decreasing f: f(u)>f(v) ⟺ u<v (inequality reverses).
- For strictly increasing f: f(u)>f(v) ⟺ u>v (inequality preserved).
- A function stated to be "positive on ℝ" removes the need to separately verify denominators are positive.
Common Mistakes
Mistake 1: Forgetting that f is DECREASING and keeping the inequality direction the same when moving from f(3a+15)>f(a²+7a+3) to the raw inputs, giving 3a+15 > a²+7a+3 instead.
Fix: Always explicitly state "f decreasing ⇒ reverse the inequality" as a written step before simplifying — this catches the sign flip every time.
Mistake 2: Trying to separately verify a²+7a+3>0 and 3a+15>0 as extra constraints, not realising "f positive on ℝ" already guarantees both denominators are positive for any real input.
Fix: Read the problem statement carefully — "f is a positive function on ℝ" is doing exactly this job; no extra denominator-positivity case-work is needed.
FAQs
Q1. How do you compare f(u) and f(v) when only f's monotonicity is known, not its formula?
For a strictly decreasing function, f(u) > f(v) if and only if u < v — the inequality on the outputs reverses to give the opposite inequality on the inputs. This lets you solve the problem entirely in terms of u and v without ever knowing f itself.
Q2. What is the correct answer to this question?
α²+β² = 40, option (2), from the excluded interval [−6, 2].
Q3. Why don't we need to check the denominators are positive separately?
Because f is given to be a positive function on all of ℝ, f(a²+7a+3) and f(3a+15) are automatically positive for every real a — the only real constraint left is which denominator is larger, which determines the major axis direction.
Q4. How many marks is this question worth, and how long should it take?
It is a 4-mark MCQ with -1 negative marking. Once the decreasing-function trick is recognised, it reduces to a 60-second quadratic inequality.
Prerequisites
Before practising this question type, make sure the following are in place:
- Major/minor axis conditions for a standard ellipse from its denominators.
- Monotonic function inequality reversal for strictly increasing vs. strictly decreasing functions.
- Solving quadratic inequalities by factoring.
Revise these from the Conic Sections chapter, then attempt the related questions below.
Solved by Nishant Kumar Gupta for padholikhojee.in · JEE Main 2026 08 April Evening · Last updated: 2026-09-02 · Verified against the official question paper and answer key.