# The function \[f\left( x \right) = {{\text{x}}^2} + 2x - 5\] is strictly increasing in the interval

A. $\left( { - \infty , - 1} \right)$

B. $\left( { - \infty , - 1} \right]$

C. $\left[ { - 1,\infty } \right)$

D. $\left( { - 1,\infty } \right)$

Last updated date: 18th Mar 2023

•

Total views: 308.7k

•

Views today: 3.87k

Answer

Verified

308.7k+ views

Hint: Differentiate the equation and equate it with 0. Condition of minima and maxima is when the sign of double derivative is positive it’s minima point and when it's negative its maxima point.

We have been given

\[f\left( x \right) = {{\text{x}}^2} + 2x - 5\]

As coefficient of ${{\text{x}}^2}$ is positive so we will get a local minima in this equation

Now when we differentiate the equation we get

$f'\left( x \right) = 2x + 2$

To find the local minima we need to equate it with 0

So, by equating $f'\left( x \right) = 0$ we get,

$2x + 2 = 0$

$ \Rightarrow x = - 1$

Now that we have the minima we can observe the points in right and left to it.

So for

$x < - 1;$ $f'\left( x \right) < 0$

And

$x > - 1;$ $f'\left( x \right) > 0$

So, as $f'\left( x \right) > 0$ for $x > - 1$

Therefore, $f\left( x \right)$ is strictly increasing in $\left( { - 1,\infty } \right)$.

Hence correct Option is D.

Note: In this question firstly we differentiate the given equation and equate it with 0. After equating them we get the extreme points which in this case is local minima. Now, we observe the neighbourhood of the point and get our answer.

We have been given

\[f\left( x \right) = {{\text{x}}^2} + 2x - 5\]

As coefficient of ${{\text{x}}^2}$ is positive so we will get a local minima in this equation

Now when we differentiate the equation we get

$f'\left( x \right) = 2x + 2$

To find the local minima we need to equate it with 0

So, by equating $f'\left( x \right) = 0$ we get,

$2x + 2 = 0$

$ \Rightarrow x = - 1$

Now that we have the minima we can observe the points in right and left to it.

So for

$x < - 1;$ $f'\left( x \right) < 0$

And

$x > - 1;$ $f'\left( x \right) > 0$

So, as $f'\left( x \right) > 0$ for $x > - 1$

Therefore, $f\left( x \right)$ is strictly increasing in $\left( { - 1,\infty } \right)$.

Hence correct Option is D.

Note: In this question firstly we differentiate the given equation and equate it with 0. After equating them we get the extreme points which in this case is local minima. Now, we observe the neighbourhood of the point and get our answer.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Name the Largest and the Smallest Cell in the Human Body ?

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main

A sample of an ideal gas is expanded from 1dm3 to 3dm3 class 11 chemistry CBSE