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5f(x) + 4f $(\frac{1}{x})$ = x² – 4 and y = 9f(x) × x². If y is strictly increasing function, find interval of x.

Option 1 -

$(- \infty, \frac{- 1}{\sqrt[]{5}}] \cup (\frac{- 1}{\sqrt[]{5}}, 0)$

Option 2 -

$(\frac{- 1}{\sqrt[]{5}}, 0) \cup (0, \frac{1}{\sqrt[]{5}})$

Option 3 -

$(0, \frac{1}{\sqrt[]{5}}) \cup (\frac{1}{\sqrt[]{5}}, \infty)$

Option 4 -

$(- \sqrt[]{\frac{2}{5}}, 0) \cup (\sqrt[]{\frac{2}{5}}, \infty)$

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

5f(x) + 4f $(\frac{1}{x})$ = x² – 4 .(1)

Replace x by $\frac{1}{x}$

5f $(\frac{1}{x})$ + 4f(x) = $\frac{1}{x^{2}}$ – 4 .(2)

5 × equation (1) – 4 × equation (2)

$9 f (x) = 5 x^{2 -} \frac{4}{x^{2}} - 4$

$y = 9 f (x) \cdot x^{2} = \frac{5 x^{4} - 4 - 4 x^{2}}{x^{2}} x^{2}$

y = 5x⁴ – 4 – 4x²

y = 20x³ – 8x > 0

4x(5x² – 2) > 0

$x \in (- \sqrt[]{\frac{2}{5}}, 0) \cup (\sqrt[]{\frac{2}{5}}, \infty)$

5f(x) + 4f <math> <mrow> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>x</mi> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math>  = x2 – 4           .(1)Replace x by  <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>x</mi> </mrow> </mfrac> </mrow> </math> 5f  <math> <mrow> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>x</mi> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math>  + 4f(x) = <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math>  – 4   .(2)5 × equation (1) – 4 × equation (2) <math> <mrow> <mn>9</mn> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>5</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> <mo>−</mo> </mrow> </msup> <mfrac> <mrow> <mn>4</mn> </mrow> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <mo>−</mo> <mn>4</mn> </mrow> </math>             <math> <mrow> <mi>y</mi> <mo>=</mo> <mn>9</mn> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>⋅</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>5</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> <mo>−</mo> <mn>4</mn> <mo>−</mo> <mn>4</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>            y = 5x4 – 4 – 4x2y = 20x3 – 8x > 04x(5x2 – 2) > 0<picture><img src="https://images.shiksha.com/mediadata/images/1758801580phpa1HNFo.jpeg" alt="" width="316" height="133" loading="lazy"></picture>    <math> <mrow> <mi>x</mi> <mo>∈</mo> <mrow> <mo>(</mo> <mrow> <mo>−</mo> <mroot> <mrow> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> <mo>,</mo> <mtext> </mtext> <mn>0</mn> </mrow> <mo>)</mo> </mrow> <mo>∪</mo> <mrow> <mo>(</mo> <mrow> <mroot> <mrow> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> <mo>,</mo> <mtext> </mtext> <mo>∞</mo> </mrow> <mo>)</mo> </mrow> </mrow> </math>

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5f(x) + 4f $(\frac{1}{x})$ = x² – 4 and y = 9f(x) × x². If y is strictly increasing function, find interval of x.

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5f(x) + 4f ( 1 x ) = x2 – 4 and y = 9f(x) × x2. If y is strictly increasing function, find interval of x.

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5f(x) + 4f $(\frac{1}{x})$ = x² – 4 and y = 9f(x) × x². If y is strictly increasing function, find interval of x.