If α, β are the distinct roots of x² + bx + c = 0, then $\underset{x \to β}{l i m} \frac{e^{2 (x^{2} + b x + c)} - 1 - 2 (x^{2} + b x + c)}{{(x - β)}^{2}}$ is equal to

Option 1 -

b² – 4c

Option 2 -

b² + 4c

Option 3 -

2(b² + 4c)

Option 4 -

2(b² – 4c)

2 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 months ago

Correct Option - 4

Detailed Solution:

Given let α, β be the roots of the equation x² + bx + c = 0

So, $α^{2} + b α + c = 0 & β^{2} + b β + c = 0$

Also x² + bx + c = (x - α) (x - β)

$N o w L = \underset{x \to β}{l i m} \frac{e^{2 (x^{2} + b x + c)} - 1 - 2 (x^{2} + b x + c)}{{(x - β)}^{2}}$

$\underset{x \to β}{l i m} \frac{2 {(x - α)}^{2} {(x - β)}^{2} + \frac{8}{6} {(x - α)}^{3} {(x - β)}^{3} + . . . . . . . .}{{(x - β)}^{2}}$

= $2 {(β - α)}^{2} = 2 [{(β + α)}^{2} - 4 α β] = 2 [b^{2} - 4 c]$

Given let α, β be the roots of the equation x2 + bx + c = 0So, <math> <mrow> <msup> <mrow> <mi>α</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mi>b</mi> <mi>α</mi> <mo>+</mo> <mi>c</mi> <mo>=</mo> <mn>0</mn> <mo>&</mo> <mtext> </mtext> <mtext> </mtext> <msup> <mrow> <mi>β</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mi>b</mi> <mi>β</mi> <mo>+</mo> <mi>c</mi> <mo>=</mo> <mn>0</mn> </mrow> </math> Also x2 + bx + c = (x - α) (x - β) <math> <mrow> <mi>N</mi> <mi>o</mi> <mi>w</mi> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mi>L</mi> <mo>=</mo> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>x</mi> <mo>→</mo> <mi>β</mi> </mrow> </munder> <mfrac> <mrow> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mi>b</mi> <mi>x</mi> <mo>+</mo> <mi>c</mi> </mrow> <mo>)</mo> </mrow> </mrow> </msup> <mo>−</mo> <mn>1</mn> <mo>−</mo> <mn>2</mn> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mi>b</mi> <mi>x</mi> <mo>+</mo> <mi>c</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mi>β</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math>            <math> <mrow> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>x</mi> <mo>→</mo> <mi>β</mi> </mrow> </munder> <mfrac> <mrow> <mn>2</mn> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mi>α</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mi>β</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mfrac> <mrow> <mn>8</mn> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mi>α</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mi>β</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>+</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> </mrow> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mi>β</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> =  <math> <mrow> <mn>2</mn> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>β</mi> <mo>−</mo> <mi>α</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mn>2</mn> <mrow> <mo>[</mo> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>β</mi> <mo>+</mo> <mi>α</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>4</mn> <mi>α</mi> <mi>β</mi> </mrow> <mo>]</mo> </mrow> <mo>=</mo> <mn>2</mn> <mrow> <mo>[</mo> <mrow> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>4</mn> <mi>c</mi> </mrow> <mo>]</mo> </mrow> </mrow> </math>

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If α, β are the distinct roots of x² + bx + c = 0, then $\underset{x \to β}{l i m} \frac{e^{2 (x^{2} + b x + c)} - 1 - 2 (x^{2} + b x + c)}{{(x - β)}^{2}}$ is equal to

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If α, β are the distinct roots of x2 + bx + c = 0, then l i m x → β e 2 ( x 2 + b x + c ) − 1 − 2 ( x 2 + b x + c ) ( x − β ) 2 is equal to

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If α, β are the distinct roots of x² + bx + c = 0, then $\underset{x \to β}{l i m} \frac{e^{2 (x^{2} + b x + c)} - 1 - 2 (x^{2} + b x + c)}{{(x - β)}^{2}}$ is equal to