Question

JEE ADVANCED MATHS QUESTION SEPTEMBER 2020

Answer 1

AnswEr:-

$\longrightarrow \: \sf \: f(x) + f(1 - x) = \frac{ {4}^{x} }{4 {}^{x} + 2 } + \frac{4 {}^{1 - \times } }{ {4}^{1 - \times } + 2}$

$\implies \sf \: \frac{ {4}^{x} }{ {4}^{x} + 2} + \frac{4}{4 + {2.4}^{x} }$

$\implies \: \frac{4 {}^{x} }{ {4}^{x} + 2} + \frac{2}{2 + {4}^{x} } \\ \\ \\ = 1$

Hence,

$\implies \sf \: f( \frac{1}{40} ) + f( \frac{2}{40} ) + ...... + f( \frac{39}{40} ) - f( \frac{1}{2} )$

$\implies \sf \: 19 + f( \frac{1}{2} ) - f( \frac{1}{2} ) \\ \\ \implies \: 19.$

Therefore,

Answer :- 19.

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Answer 2

Let the function f : [0, 1] → R be defined by

f(x) = $\frac{4^x}{4^x+2}$

Then the value of

f(1/40) + f(2/40) + f(3/40) + .... f(39/40) - f(1/2) is ...

solution : f(x) = 4^x/(4^x + 2)

f(1 - x) = 4^(1 - x)/{4^(1 - x) + 2}

= 4^1 × 4^-x/{4^1 × 4^-x + 2}

= 4/{4^x ×( 4^1 × 1/4^x + 2)}

= 4/{4 + 2 4^x}

now f(x) + f(1 - x) = 4^x/(4^x + 2) + 4/(4 + 2 4^x)

= (4 4^x + 2 4^2x + 4 4^x + 4 2)/(4^x + 2)(4 + 2 4^x)

= {2(4^2x + 4 4^x + 4)}/{2(4^2x + 4 4^x + 4)}

= 1

so we get, f(x) + f(1 - x) = 1

putting x = 1/40

f(1/40) + f(39/40) = 1

putting, x = 2/40

f(2/40) + f(38/40) = 1

........

.............

putting x = 19/40

f(19/40) + f(21/40) = 1

putting x = 20/40 = 1/2

f(1/2) + f(1/2) = 1

⇒2f(1/2) = 1

⇒f(1/2) = 1/2

now f(1/40) + f(2/40) + f(3/40) + .... + f(39/40) - f(1/2)

= 19 + f(1/2) - f(1/2)

= 19 + 1/2 - 1/2

= 19

Therefore the value of f(1/40) + f(2/40) + f(3/40) + .... + f(39/40) - f(1/2) is 19.