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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

Let f(x) = (9x)/(25) + c, c > 0 . If the curve y = f ^ - 1 (x)...

Let f(x)=9x25+c,c>0.f\left(x\right) = \dfrac{9x}{25} + c, c > 0 . If the curve y=f−1(x)y = {f}^{- 1}\left(x\right) passes through \left(\dfrac{1}{4}, - \dfrac{5}{4}\right)andg(x)istheantiderivativeofand g\left(x\right) is the antiderivative of{f}^{- 1}\left(x\right)suchthatsuch thatg\left(0\right) = \dfrac{5}{2},thenthevalueof[g(1)]then the value of \left[g\left(1\right)\right] is, (where [.][ . ] represents the greatest integer function)

Correct answer

2

Step-by-step solution

y = f\left(x\right) = \dfrac{9x}{25} + {c}_{1}<br/><br />⇒ x = \dfrac{25}{9}\left(y - {c}_{1}\right) = {f}^{- 1}\left(y\right)<br/><br />⇒ {f}^{- 1}\left(x\right) = \dfrac{25}{9}\left(x - {c}_{1}\right)<br/>Thecurve<br />The curvey = {f}^{- 1}\left(x\right)passesthrough(14,−59)passes through \left(\dfrac{1}{4}, - \dfrac{5}{9}\right)
⇒−59=259(14−c1)⇒ - \dfrac{5}{9} = \dfrac{25}{9}\left(\dfrac{1}{4} - {c}_{1}\right)
⇒c1=14+15=920⇒ {c}_{1} = \dfrac{1}{4} + \dfrac{1}{5} = \dfrac{9}{20}
⇒f−1(x)=259(x−920)⇒ {f}^{- 1}\left(x\right) = \dfrac{25}{9}\left(x - \dfrac{9}{20}\right)= \dfrac{25x}{9} - \dfrac{5}{4}<br/><br />g\left(x\right) = ∫ {f}^{- 1}\left(x\right)dx = ∫ \left(\dfrac{25x}{9} - \dfrac{5}{4}\right)dx<br/><br />= \dfrac{25}{9}\left(\dfrac{{x}^{2}}{2}\right) - \dfrac{5x}{4} + {c}_{2}<br/><br />g\left(0\right) = \dfrac{5}{2} ⇒ {c}_{2} = \dfrac{5}{2}<br/><br />⇒ g\left(x\right) = \dfrac{25}{9}\left(\dfrac{{x}^{2}}{2}\right) - \dfrac{5x}{4} + \dfrac{5}{2}<br/><br />⇒ g\left(1\right) = \dfrac{25}{18} - \dfrac{5}{4} + \dfrac{5}{2} = \dfrac{50 - 45 + 90}{36} = \dfrac{95}{36}<br/><br />⇒ \left[g\left(1\right)\right] = 2
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