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Ellipse Condition with Decreasing f, Major Axis y-axis | JEE

JEE Maths question with a full step-by-step solution.

Question
If x2f(4a)+y2f(a2−5)=1\dfrac{x^2}{f(4a)}+\dfrac{y^2}{f(a^2-5)}=1 represents an ellipse with major axis as yy-axis and ff is a decreasing function, then
Aa∈(−∞,1)a\in(-\infty,1)
Ba∈(5,∞)a\in(5,\infty)
Ca∈(1,4)a\in(1,4)
Da∈(−1,5)a\in(-1,5)correct
Solution
Step 1: The major axis is the yy-axis, so the denominator under y2y^2 exceeds the one under x2x^2.
f(a2−5)>f(4a).f(a^2-5)>f(4a).
 Step 2: Since ff is decreasing, the inequality on outputs reverses on inputs.
a2−5<4a  ⇒  a2−4a−5<0.a^2-5<4a \;\Rightarrow\; a^2-4a-5<0.
 Step 3: Factor.
(a−5)(a+1)<0  ⇒  −1<a<5.(a-5)(a+1)<0 \;\Rightarrow\; -1<a<5.
 ∴a∈(−1,5)\therefore a\in(-1,5). Correct answer: (4)
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