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Let $f$ be the differentiable for $\forall \: x.$ If $f\left ( 1 \right )= -2$ and $f{}'\left ( x \right )\geq 2$ for $\left [ 1,6 \right ],$ then

Option 1)
$f\left ( 6 \right )< 8$

Option 2)
$f\left ( 6 \right )\geq 8$

Option 3)
$f\left ( 6 \right )= 5$

Option 4)
$f\left ( 6 \right )< 5$

Answers (1)

best_answer

As we have learned

Lagrange's mean value theorem -

If a function f(x)

1. is continuous in the closed interval [a,b] and

2. is differentiable in the open interval (a, b) then

$f'(c)=\frac{f(b)-f(a)}{b-a}\:\:\:\:where\:\:c\epsilon (a,b)$

-

Geometrical interpretation of Lagrange's theorem -

Let A, B be the points on the curve y = f(x) at x = a and x = b so that A [a, f(a)], and B [b, f(b)]

$\therefore \:\:slope\:of\:chord\:AB$

$=\frac{f(b)-f(a)}{b-a}$

So slope of the chord AB = f'(c) the slope of the tangent to the curve at x = c.

- wherein

We have

$f'(x)\geq 2$

f(x) is an increasing function ,

$So, \frac{f(6)-f(1)}{6-1} = f'(c )$

For some $c \: \: \epsilon \: \: [1,6]\: \: where \: \:$

$f' (c)\geq 2 \\$

$so, \frac{f(6)-f(2)}{5}\geq 2$

$\\f(6 )\geq 8$

Option 1)

$f\left ( 6 \right )< 8$

Option 2)

$f\left ( 6 \right )\geq 8$

Option 3)

$f\left ( 6 \right )= 5$

Option 4)

$f\left ( 6 \right )< 5$

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Himanshu

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