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Let \(f\) be a strictly decreasing function defined on \(\mathbb{R}\) such that \(f(x) > 0, \forall x \in \mathbb{R}\). Let \(\frac{x^2}{f\left(a^2+5 a+3\right)}+\frac{y^2}{f(a+15)}=1\) be an ellipse with major axis along the \(y\)-axis. The value of ' \(a\) ' can lie in the interval (s)
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2534 Upvotes
Verified Answer
The correct answers are:
\((-\infty,-6)\), \((2, \infty)\)
Hint : \(f \Rightarrow\) strictly decreasing function \(\forall x \in R\)
\(\begin{aligned}
& f(x) > 0 \Rightarrow f^{\prime}(x) < 0 \\
& \frac{x^2}{f\left(a^2+5 a+3\right)}+\frac{y^2}{f(a+15)}=1
\end{aligned}\)
As major axis is \(y\)-axis
\(\begin{aligned}
& f(a+15) > f\left(a^2+5 a+3\right) \\
& \Rightarrow a+15 < a^2+5 a+3 \quad (\because f \text{ is decreasing) }\\
& \Rightarrow a^2+4 a-12 > 0 \\
& \Rightarrow a < -6 \text { or } a > 2
\end{aligned}\)
\(\begin{aligned}
& f(x) > 0 \Rightarrow f^{\prime}(x) < 0 \\
& \frac{x^2}{f\left(a^2+5 a+3\right)}+\frac{y^2}{f(a+15)}=1
\end{aligned}\)
As major axis is \(y\)-axis
\(\begin{aligned}
& f(a+15) > f\left(a^2+5 a+3\right) \\
& \Rightarrow a+15 < a^2+5 a+3 \quad (\because f \text{ is decreasing) }\\
& \Rightarrow a^2+4 a-12 > 0 \\
& \Rightarrow a < -6 \text { or } a > 2
\end{aligned}\)
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