<p><strong> <math><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mrow><mo>(</mo><mrow><mn>2</mn><mn>0</mn><mo>−</mo><mi>a</mi></mrow><mo>)</mo></mrow><mrow><mo>(</mo><mrow><mn>4</mn><mn>0</mn><mo>−</mo><mi>a</mi></mrow><mo>)</mo></mrow></mrow></mfrac><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mrow><mo>(</mo><mrow><mn>4</mn><mn>0</mn><mo>−</mo><mi>a</mi></mrow><mo>)</mo></mrow><mrow><mo>(</mo><mrow><mn>6</mn><mn>0</mn><mo>−</mo><mi>a</mi></mrow><mo>)</mo></mrow></mrow></mfrac><mo>+</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mrow><mo>(</mo><mrow><mn>1</mn><mn>8</mn><mn>0</mn><mo>−</mo><mi>a</mi></mrow><mo>)</mo></mrow><mrow><mo>(</mo><mrow><mn>2</mn><mn>0</mn><mn>0</mn><mo>−</mo><mi>a</mi></mrow><mo>)</mo></mrow></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn><mn>5</mn><mn>6</mn></mrow></mfrac><mo>,</mo></mrow></math> then the maximum value of a is:</strong></p>

1(20−a)(40−a)+1(40−a)(60−a)+1(60−a)(80−a)+.....+1(180−a)(200−a)=1256LHS = 120(120−a−140−a)+120(140−a−160−a)+...+120(1180−a−1200−a)=120(120−a−1200−a)=120.180(20−a)(200−a)⇒a2−220a+4000−2304=0⇒a2220a+1696=0a=220±2048=2122

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Maths Class 12th Relations and Functions Maths NCERT Exemplar Solutions Class 12th Chapter One

$\frac{1}{(20 - a) (40 - a)} + \frac{1}{(40 - a) (60 - a)} + . . . + \frac{1}{(180 - a) (200 - a)} = \frac{1}{256},$ then the maximum value of a is:

Option 1 -

198

Option 2 -

202

Option 3 -

212

Option 4 -

218

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

$\frac{1}{(20 - a) (40 - a)} + \frac{1}{(40 - a) (60 - a)} + \frac{1}{(60 - a) (80 - a)} + . . . . . + \frac{1}{(180 - a) (200 - a)} = \frac{1}{256}$

LHS = $\frac{1}{20} (\frac{1}{20 - a} - \frac{1}{40 - a}) + \frac{1}{20} (\frac{1}{40 - a} - \frac{1}{60 - a}) + . . . + \frac{1}{20} (\frac{1}{180 - a} - \frac{1}{200 - a})$

$= \frac{1}{20} (\frac{1}{20 - a} - \frac{1}{200 - a}) = \frac{1}{20} . \frac{180}{(20 - a) (200 - a)}$

$\Rightarrow a^{2} - 220 a + 4000 - 2304 = 0 \Rightarrow a^{2} 220 a + 1696 = 0$

$a = \frac{220 \pm 204}{8} = 21 2^{2}$

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$\frac{1}{(20 - a) (40 - a)} + \frac{1}{(40 - a) (60 - a)} + . . . + \frac{1}{(180 - a) (200 - a)} = \frac{1}{256},$ then the maximum value of a is:

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1(20−a)(40−a)+1(40−a)(60−a)+...+1(180−a)(200−a)=1256, then the maximum value of a is:

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$\frac{1}{(20 - a) (40 - a)} + \frac{1}{(40 - a) (60 - a)} + . . . + \frac{1}{(180 - a) (200 - a)} = \frac{1}{256},$ then the maximum value of a is: