<p><strong>For <math><mrow><mi>k</mi><mo>∈</mo><mi>R</mi><mo>,</mo></mrow></math> let the solutions of the equation <math><mrow><mi>c</mi><mi>o</mi><mi>s</mi><mrow><mo>(</mo><mrow><mi>s</mi><mi>i</mi><msup><mrow><mi>n</mi></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo>(</mo><mrow><mi>x</mi><mi>c</mi><mi>o</mi><mi>t</mi><mrow><mo>(</mo><mrow><mi>t</mi><mi>a</mi><msup><mrow><mi>n</mi></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo>(</mo><mrow><mi>c</mi><mi>o</mi><mi>s</mi><mrow><mo>(</mo><mrow><mi>s</mi><mi>i</mi><msup><mrow><mi>n</mi></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mi>x</mi></mrow><mo>)</mo></mrow></mrow><mo>)</mo></mrow></mrow><mo>)</mo></mrow></mrow><mo>)</mo></mrow></mrow><mo>)</mo></mrow></mrow></math> =, 0 < <math><mrow><mrow><mo>|</mo><mrow><mi>x</mi></mrow><mo>|</mo></mrow><mo><</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mroot><mrow><mn>2</mn></mrow><mrow></mrow></mroot></mrow></mfrac></mrow></math> be and , where the inverse trigonometric functions take only principal values. If the solutions of the equation x2 – bx – 5 = 0 are <math><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><msup><mrow><mi>α</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><msup><mrow><mi>β</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mi>a</mi><mi>n</mi><mi>d</mi><mtext> </mtext><mtext> </mtext><mfrac><mrow><mi>α</mi></mrow><mrow><mi>β</mi></mrow></mfrac><mtext> </mtext><mi>t</mi><mi>h</mi><mi>e</mi><mi>n</mi><mtext> </mtext><mtext> </mtext><mfrac><mrow><mi>b</mi></mrow><mrow><msup><mrow><mi>k</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></math> is equal to</strong></p>

use sin−1x=cos−11−x2tan−11−x2=cot−111−x2sin−1x1−x2=cos−11−2x21−x2Sum of roots b = 1 + 2 (k2−1)k2−2Product of roots 5 = 2 (k2−1k2−2)b = 4, k2 = 13

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For $k \in R,$ let the solutions of the equation $c o s (s i n^{- 1} (x c o t (t a n^{- 1} (c o s (s i n^{- 1} x)))))$ =, 0 < $| x | < \frac{1}{\sqrt[]{2}}$ be and , where the inverse trigonometric functions take only principal values. If the solutions of the equation x2 – bx – 5 = 0 are $\frac{1}{α^{2}} + \frac{1}{β^{2}} a n d \frac{α}{β} t h e n \frac{b}{k^{2}}$ is equal to

9 Views | Posted 3 months ago

Answered by

Payal Gupta | Contributor-Level 10

3 months ago

use $s i n^{- 1} x = c o s^{- 1} \sqrt[]{1 - x^{2}}$

$t a n^{- 1} \sqrt[]{1 - x^{2}} = c o t^{- 1} \frac{1}{\sqrt[]{1 - x^{2}}}$

$s i n^{- 1} \frac{x}{\sqrt[]{1 - x^{2}}} = c o s^{- 1} \frac{\sqrt[]{1 - 2 x^{2}}}{\sqrt[]{1 - x^{2}}}$

Sum of roots b = 1 + 2 $\frac{(k^{2} - 1)}{k^{2} - 2}$

Product of roots 5 = 2 $(\frac{k^{2} - 1}{k^{2} - 2})$

b = 4, k² = $\frac{1}{3}$

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