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JEE Main 20209 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Differentiation

JEE Main 2020 · 9 January · Morning Shift · Mathematics · Differentiation

Let f be any function continuous on [a, b] and twice differentiable...

Let ff be any function continuous on \left[a, b\right]andtwicedifferentiableon(a,b)and twice differentiable on \left(a, b\right) . If all x∈(a,b),f′(x)>0x \in \left(a, b\right), {f}^{'}\left(x\right) > 0 and f′′(x)<0{f}^{''}\left(x\right) < 0 , then for any c∈(a,b),f(c)−f(a)f(b)−f(c)c \in \left(a, b\right), \dfrac{f\left(c\right) - f\left(a\right)}{f\left(b\right) - f\left(c\right)}

Options

  1. Ab+ab−a\dfrac{b + a}{b - a}
  2. B11
  3. Cb−cc−a\dfrac{b - c}{c - a}
  4. Dc−ab−c\dfrac{c - a}{b - c}

Correct answer

D. c−ab−c\dfrac{c - a}{b - c}

Step-by-step solution

Let’s use LMVT for x∈[a,c]x \in \left[a, c\right]

f(c)−f(a)c−a=f′(α),α∈(a,c)\dfrac{f\left(c\right) - f ( a )}{c - a} = {f}^{'}\left(\alpha\right), \alpha \in \left(a, c\right)

Also use LMVT for x∈[c,b]x \in \left[c, b\right]

f(b)−f(c)b−c=f′(β),β∈(c,b)\dfrac{f\left(b\right) - f ( c )}{b - c} = {f}^{'}\left(\beta\right), \beta \in \left(c, b\right)

∵f′′(x)<0⇒f′(x)∵ {f}^{''}\left(x\right) < 0 ⇒ {f}^{'}\left(x\right) is decreasing

f′(α)>f′(β){f}^{'}\left(\alpha\right) > {f}^{'}\left(\beta\right)

f(c)−f(a)c−a>f(b)−f(c)b−c\dfrac{f\left(c\right) - f ( a )}{c - a} > \dfrac{f\left(b\right) - f ( c )}{b - c}

f(c)−f(a)f(b)−f(c)>c−ab−c\dfrac{f\left(c\right) - f ( a )}{f\left(b\right) - f ( c )} > \dfrac{c - a}{b - c} ( ∵f(x)∵ f ( x ) is increasing)

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