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826

Given that X  : R \(\to\) R is defined by x = \(\frac{y + 1}{5 - y}\) , y \(\in\) R, find the domain of x.

  • A. {y : y \(\in\) R, y \(\neq\) 0}
  • B. {y : y \(\in\) R, y \(\neq\) 1}
  • C. {y : y \(\in\) R, y \(\neq\) 5}
  • D. {y : y \(\in\) R, y \(\neq\) 7}
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827

Simplify; \(\frac{\sqrt{5} + 3}{4 - \sqrt{10}}\) 

 

  • A. \(\frac{2}{3}\)\(\sqrt{5}\) + \(\frac{5}{6}\sqrt{2}\) + 2
  • B. \(\frac{2}{3}\)\(\sqrt{5}\) + \(\frac{5}{6}\sqrt{2}\) + \(\frac{1}{2}\sqrt{10}\)
  • C. \(\frac{2}{3}\)\(\sqrt{5}\) + \(\frac{5}{6}\sqrt{2}\) + \(\frac{1}{2}\sqrt{10}\) + 2
  • D. \(\frac{2}{3}\)\(\sqrt{5}\) - \(\frac{5}{6}\sqrt{2}\) + \(\frac{1}{2}\sqrt{10}\) + 2
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828

If \(\frac{6x + k}{2x^2 + 7x - 15}\)  = \(\frac{4}{x + 5} - \frac{2}{2x - 3}\). Find the value of k. 

  • A. - 21
  • B. - 22
  • C. - 24
  • D. - 25
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WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free
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829

Differentiate \(\frac{x}{x + 1}\) with respect to x. 

 

  • A. \(\frac{x}{x + 1}\)
  • B. \(\frac{-1}{x + 1}\)
  • C. \(\frac{1 - x}{(x + 1)^2}\)
  • D. \(\frac{1}{(x + 1)^2}\)
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830

Given that 2x + 3y - 10 and 3x = 2y - 11, calculate the value of (x - y). 

  • A. 5
  • B. 3
  • C. - 3
  • D. - 5
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Download WAEC May/June App - Get all past questions and answers, 100% offline - 43208
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts