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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2019 · Mathematics · Application of Derivatives

If f and g are differentiable function in [0, 1] satisfying f(0) =...

If fandgf and g are differentiable function in [0,1]\left[0, 1\right] satisfying f(0)=f\left(0\right) =2 = g \left(1\right),g(0)=, g \left(0\right) =0 and f(1)=6f\left(1\right) = 6, then for some c∈(0,1)c \in \left(0, 1\right)

Options

  1. A2f′(c)=g(c)2f^{'} \left(c\right) = g \left(c\right)
  2. B2f′(c)=3g′(c)2f^{'} \left(c\right) = 3g^{'} \left(c\right)
  3. Cf′(c)=g′(c)f^{'} \left(c\right) = g^{'} \left(c\right)
  4. Df′(c)=2g′(c)f^{'} \left(c\right) = 2g^{'} \left(c\right)

Correct answer

D. f′(c)=2g′(c)f^{'} \left(c\right) = 2g^{'} \left(c\right)

Step-by-step solution

Let

hh\left(x\right) = f \left(x\right) - 2g \left(x\right)

hh\left(0\right)=f= f\left(0\right)−2g- 2g\left(0\right)</p><p></p><p>h \left(0\right) = 2 - 2 \left(0\right) & \left[∵ f \left(0\right) =2, g \left(0\right)=0\right]

hh\left(0\right) =22

h(1)=fh\left(1\right)=f\left(1\right)- 2g \left(1\right)

h(1)=6−2(2)h ( 1 ) = 6 - 2 ( 2 )\left[∵ f (1)=6,g(1)\left(1\right) = 6, g \left(1\right) = 2\right]</p><p></p><p>h (1)\left(1\right) = 2</p><p></p><p>h(0)\left(0\right) = h (1)</p><p>\left(1\right)</p><p>h^{'} (x)=f′\left(x\right) = f^{'}\left(x\right) - 2g^{'}\left(x\right)

By Rolle's theorem h′(c)h^{'}\left(c\right) = 0</p><p></p><p>∴ f^{'} (c)−2g′\left(c\right) - 2g^{'}\left(c\right)=0,f′= 0, f^{'}(c)=\left(c\right) =2g^{'}\left(c\right)</p>

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