Question

Let f be a differentiable function on [1,5], f(1)=0, |f'(x)| ≥ 2 |f(x)|, then find the value of 3f(3)-2f(2)+f'(4)-f(5)

Answer 1

this is a. tricky question type

Answer 2

Answer:

The value if given expression is

$3f(3) - 2f(2) + f'(4) - f(5) = 0$

Step-by-step explanation:

Given that,

f is a differentiable function on [1,5] and also given that

$f(1) = 0 \\ |f'(x)| \geqslant 2 |f(x)|$

We can rewrite the above relation as follows

$|f'(x)| \geqslant 2 | \frac{f(x) - f(1)}{1 - 0} |$

From rules of differentiability,we can write the relation as follows

$|f'(x)| \geqslant 2|f'(x)| \\ = > |f'(x)| = 0 \\ = > f'(x) = 0$

If the differential of a function is zero,then the function is equal to a constant.We know that f(1)=0.Hence the function is a constant function and is always equal to zero.

$f'(x) = 0 \\ f(x) = 0$

Hence,the value of given expression becomes

$3f(3) - 2f(2) + f'(4) - f(5) = 0$

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