If log?(x+?) x² < log?(x+?) (2x+3) then value of x

l o g ( 2 x + 3 ) ? x 2 < l o g ( 2 x + 3 ) ? ( 2 x + 3 ) <!-- [if gte msEquation 12]>log(2x+3)?x2<log(2x+3)?(2x+3)

If log?(x+?) x² &lt; log?(x+?) (2x+3) then value of x &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

x ∈ (-3/2, -1) U (-1, 3) - {0}

x ∈ [-3/2, -1) U (1, 3)

) x ∈ [-3/2, -1] U [-1, 3)

x ∈ [-3/2, -1] U (-1, 3)

Ask & Answer Question

Maths NCERT Exemplar Solutions Class 11th Chapter Two Maths Class 11th Ncert Solutions Maths class 11th

If log?(x+?) x² < log?(x+?) (2x+3) then value of x

Option 1 -

x ∈ [-3/2, -1) U (1, 3)

Option 2 -

x ∈ (-3/2, -1) U (-1, 3) - {0}

Option 3 -

) x ∈ [-3/2, -1] U [-1, 3)

Option 4 -

x ∈ [-3/2, -1] U (-1, 3)

2 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 2

Detailed Solution:

${l o g}_{(2 x + 3)} ? x^{2} < {l o g}_{(2 x + 3)} ? (2 x + 3)$

Case-I: $0 < 2 x + 3 < 1$

$x^{2} > 2 x + 3$

Case-II: $2 x + 3 > 1$

$0 < x^{2} < 2 x + 3$

By solving (1) & (2) $\begin{array}{r} \frac{- 3}{2} < x < - 1 & (x - 3) (x + 1) > 0 . \\ \frac{- 3}{2} < x < - 1, x > 3 \end{array}$

Equation (4) has no solution $\frac{- 3}{2} < x < - 1, x < - 1$

.(5) By equation (5) . $x \in (\frac{- 3}{2}, - 1)$

We obtain solving equation (2) $x > - 1 x > - 1 \Rightarrow x \in (- 1,3)$

or $(x - 3) (x + 1) < 0 - 1 < x < 3, x^{2} > 0 \Rightarrow x \neq 0$

$x \in (\frac{- 3}{2}, - 1) \cup (- 1,3) - {0}$

<math> <msub> <mrow> <mrow> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">g</mi> </mrow> </mrow> <mrow> <mrow> <mo>(</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>?</mo> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo><</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">g</mi> </mrow> </mrow> <mrow> <mrow> <mo>(</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow> </msub> <mo>?</mo> <mo>(</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mn>3</mn> <mo>)</mo> </math>  Case-I: <math> <mn>0</mn> <mo><</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mn>3</mn> <mo><</mo> <mn>1</mn> </math>    <math> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>></mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mn>3</mn> </math>  Case-II: <math> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mn>3</mn> <mo>></mo> <mn>1</mn> </math> <math> <mn>0</mn> <mo><</mo> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo><</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mn>3</mn> </math>    By solving (1) & (2) <math> <mtable columnalign="right"> <mtr> <mtd> <mrow></mrow> </mtd> <mtd> <mrow> <mfrac> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo><</mo> <mi>x</mi> <mo><</mo> <mo>-</mo> <mn>1</mn> <mo>&</mo> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mn>3</mn> <mo>)</mo> <mo>(</mo> <mi>x</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>></mo> <mn>0</mn> <mo>.</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> <mtd> <mrow> <mfrac> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo><</mo> <mi>x</mi> <mo><</mo> <mo>-</mo> <mn>1</mn> <mo>,</mo> <mi>x</mi> <mo>></mo> <mn>3</mn> </mrow> </mtd> </mtr> </mtable> </math>  Equation (4) has no solution <math> <mfrac> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo><</mo> <mi>x</mi> <mo><</mo> <mo>-</mo> <mn>1</mn> <mo>,</mo> <mi>x</mi> <mo><</mo> <mo>-</mo> <mn>1</mn> </math>  .(5) By equation (5)  . <math> <mi>x</mi> <mo>∈</mo> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>,</mo> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </mfenced> </math> We obtain solving equation (2) <math> <mi>x</mi> <mo>></mo> <mo>-</mo> <mn>1</mn> <mrow> <mi mathvariant="normal"> </mi> </mrow> <mi>x</mi> <mo>></mo> <mo>-</mo> <mn>1</mn> <mrow> <mi mathvariant="normal"> </mi> </mrow> <mo>⇒</mo> <mi>x</mi> <mo>∈</mo> <mo>(</mo> <mo>-</mo> <mn>1,3</mn> <mo>)</mo> </math>  or <math> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mn>3</mn> <mo>)</mo> <mo>(</mo> <mi>x</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo><</mo> <mn>0</mn> <mrow> <mi mathvariant="normal"> </mi> </mrow> <mo>-</mo> <mn>1</mn> <mo><</mo> <mi>x</mi> <mo><</mo> <mn>3</mn> <mo>,</mo> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>></mo> <mn>0</mn> <mo>⇒</mo> <mi>x</mi> <mo>≠</mo> <mn>0</mn> </math> <math> <mi>x</mi> <mo>∈</mo> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>,</mo> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </mfenced> <mo>∪</mo> <mo>(</mo> <mo>-</mo> <mn>1,3</mn> <mo>)</mo> <mo>-</mo> <mo>{</mo> <mn>0</mn> <mo>}</mo> </math>