If f : R → [– 1, 2] ,f(x) = x2+9bx+17ax3+x2+bx+33 is onto function and f '(d) = f '(e) = 0. Then

If f : R → [– 1, 2] ,f(x) = \frac{x^{2} + 9 bx + 17}{\left(ax\right)^{3} + x^{2} + bx + 33} is onto function and f '(d) = f '(e) = 0. Then

Engineering

Mathematics

Classification of Functions

Standard Ellipse

Question

If f : R → [– 1, 2] ,f(x) = $\frac{x^{2} + 9 bx + 17}{{ax}^{3} + x^{2} + bx + 33}$ is onto function and f '(d) = f '(e) = 0. Then

eccentricity of curve $\frac{x^{2}}{{(d + e)}^{2}} + \frac{y^{2}}{a + 1} = 1 ca n be \frac{\sqrt{7}}{2} .$

eccentricity of curve $\frac{x^{2}}{{(d + e)}^{2}} + \frac{y^{2}}{b + 1} = 1 can be \frac{1}{2}$

eccentricity of curve $\frac{x^{2}}{{(d + e)}^{2}} + \frac{y^{2}}{a + 1} = 1 can be \frac{\sqrt{3}}{2}$

eccentricity of curve $\frac{x^{2}}{{(d + e)}^{2}} + \frac{y^{2}}{b + 1} = 1 can be \frac{\sqrt{5}}{2}$

$∵$ Domain of function is real number.

$∴$ a = 0

Now, $y = \frac{x^{2} + 9 bx + 17}{x^{2} + bx + 33}$

⇒ (y – 1) x² + (by – 9b) x + (33y – 17) = 0

$∵$ x is real

$∴$ (by – 9b)² – 4 (y – 1) (33y – 17) $\geq$ 0

$∵$ Range is [– 1, 2]

$∴$ – 1 and 2 should be root

$∴$ b = + 2 and – 2

$∴$ for b = 2 ⇒ d = – 5, e = 7

and b = – 2 ⇒ d = – 7, e = + 5.

Understanding the Problem

We are given a function $f : ℝ \to [- 1, 2]$ defined by $f (x) = \frac{x^{2} + 9 b x + 17}{a x^{3} + x^{2} + b x + 33}$ . This function is onto (surjective), meaning its range is exactly the interval $[- 1, 2]$ . We are also told that its derivative is zero at two points, $f' (d) = f' (e) = 0$ . Our goal is to use these conditions to find the values of the parameters $a$ and $b$ , and then evaluate the eccentricity of a given ellipse.

Step-by-Step Solution

Step 1: Analyze the Onto Condition

For $f (x)$ to be onto the interval $[- 1, 2]$ , the function must achieve the values -1 and 2, and all values in between. This implies that the equations $f (x) = - 1$ and $f (x) = 2$ must have real roots. Let's set up these equations.

For $f (x) = - 1$ :

\frac{x^{2} + 9 b x + 17}{a x^{3} + x^{2} + b x + 33} = - 1

Cross-multiplying:

x^{2} + 9 b x + 17 = - 1 \cdot (a x^{3} + x^{2} + b x + 33)

x^{2} + 9 b x + 17 = - a x^{3} - x^{2} - b x - 33

Bringing all terms to one side:

a x^{3} + 2 x^{2} + 10 b x + 50 = 0 (1)

For $f (x) = 2$ :

\frac{x^{2} + 9 b x + 17}{a x^{3} + x^{2} + b x + 33} = 2

Cross-multiplying:

x^{2} + 9 b x + 17 = 2 \cdot (a x^{3} + x^{2} + b x + 33)

x^{2} + 9 b x + 17 = 2 a x^{3} + 2 x^{2} + 2 b x + 66

Bringing all terms to one side:

2 a x^{3} + x^{2} + 7 b x + 49 = 0 (2)

Step 2: Relate the Equations to the Derivative Condition

We are told that the derivative $f' (x)$ is zero at two points, $d$ and $e$ . For a rational function to have its derivative be zero, the numerator of the derivative (after applying the quotient rule) must be zero. The quotient rule states:

f' (x) = \frac{u' (x) v (x) - u (x) v' (x)}{{[v (x)]}^{2}}

where $u (x) = x^{2} + 9 b x + 17$ and $v (x) = a x^{3} + x^{2} + b x + 33$ .

For $f' (<mi$

Detailed Solution

Understanding the Problem

Step-by-Step Solution

Step 1: Analyze the Onto Condition

Step 2: Relate the Equations to the Derivative Condition

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