Question

find the interval in which the function f(x)=5x^3-15x^2-120x+3 is strictly decreasing
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Answer 1

Given function

$\bold{f(x)=5x^3-15x^2-120x+3}$

The derivative of f(x)

$\bold{f'(x)=15x^2-30x-120}$

The interval of which derivative is negative

$\bold{f'(x)<0}$

$\implies\bold{15x^2-30x-120<0}$

$\implies \bold{x^2-2x-8<0}$

$\implies \bold{(x-4)(x+2)<0}$

$\implies \bold{-2<x<4}$

Therefore, the interval is (-2, 4) in interval notation.

Or, the interval is $\bold{\{x|-2<x<4\}}$ in set notation.

Answer 2

Step-by-step explanation:

Given :

Efficiency of the carnot engine changes from 1/6 to 1/3 when source temperature is raised by 100K.

To Find :

Temperature of sink.

Solution :

❒ Efficiency of carnot's engine (an ideal heat engine), is given by

η = 1 - T'/T

where,

T' denotes temp. of the sink

T denotes temp. of the source

Case - I :

➛ 1/6 = 1 - T'/T

➛ T'/T = 1 - 1/6

➛ T'/T = 5/6

➛ T = 6T'/5 .......... (I)

Case - II :

In this case temperature of the sink is raised by 100K..

➛ 1/3 = 1 - T'/(T + 100)

➛ T'/(T + 100) = 2/3

➛ 3T' = 2T + 200

From the equation (I),

➛ 3T' = 12T'/5 + 200

➛ 15T' - 12T' = 1000

➛ T' = 1000/3

➛ T' = 333.33k