Let f:(0,∞)→(0,∞) be a differentiable function such that f(1)=e and limt→x t2f2(x)-x2f2(t)t-x=0 . If f(x)=1 , then x is equal to:

Since (3,3) lies on x2a2-y2b2=1 9a2-9b2=1 Now, normal at (3,3) is y-3=-a2b2 (x-3) , which passes through (9,0)⇒b2=2a2 So, e2=1+b2a2=3 Also, a2=92 (From (i) and (ii) Thus, a2, e2=92, 3

Let $f : (0, \infty) \to (0, \infty)$ be a differentiable function such that $f (1) = e$ and ${l i m}_{t \to x} \frac{t^{2} f^{2} (x) - x^{2} f^{2} (t)}{t - x} = 0$ . If $f (x) = 1$ , then $x$ is equal to:

Option 1 - e

Option 2 - <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">e</mi> </mrow> </mrow> </mfrac> </math>

Option 3 - <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi mathvariant="normal">e</mi> </mrow> </mrow> </mfrac> </math>

Option 4 - <math> <mn>2</mn> <mi mathvariant="normal">e</mi> </math> (Limits)

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Alok Kumar Singh | Contributor-Level 10

8 months ago

Correct Option - 4

Detailed Solution:

Since $(3,3)$ lies on $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$\frac{9}{a^{2}} - \frac{9}{b^{2}} = 1$

Now, normal at $(3,3)$ is $y - 3 = - \frac{a^{2}}{b^{2}} (x - 3)$ ,

which passes through $(9,0) \Rightarrow b^{2} = 2 a^{2}$

So, $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 3$

Also, $a^{2} = \frac{9}{2}$

(From (i) and (ii)

Thus, $(a^{2}, e^{2}) = (\frac{9}{2}, 3)$

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