Home
Class 12
MATHS
The function f(x) given by f(x)=|{:(x+1...

The function f(x) given by `f(x)=|{:(x+1" "1" "1),(1" "x+1" "1),(1" "1" " x+1):}|` is increasing on

A

R

B

(-2,0)

C

R-[-2,0]

D

none of these

Text Solution

AI Generated Solution

The correct Answer is:
To determine the intervals on which the function \( f(x) \) is increasing, we will follow these steps: ### Step 1: Define the function The function is given by the determinant: \[ f(x) = \begin{vmatrix} x + 1 & 1 & 1 \\ 1 & x + 1 & 1 \\ 1 & 1 & x + 1 \end{vmatrix} \] ### Step 2: Differentiate the function To find where the function is increasing, we need to compute the derivative \( f'(x) \). We will use the properties of determinants to differentiate. Using the determinant expansion, we can differentiate row-wise. The derivative of the first row is: \[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & x + 1 & 1 \\ 1 & 1 & x + 1 \end{vmatrix} \] The derivative of the second row is: \[ \begin{vmatrix} x + 1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & x + 1 \end{vmatrix} \] The derivative of the third row is: \[ \begin{vmatrix} x + 1 & 1 & 1 \\ 1 & x + 1 & 1 \\ 0 & 0 & 1 \end{vmatrix} \] ### Step 3: Expand the determinant Expanding the determinant, we can compute \( f(x) \): \[ f(x) = (x + 1)((x + 1)(x + 1) - 1) - 1((x + 1) - 1) + 1((x + 1) - 1) \] After simplification, we find: \[ f(x) = 3(x + 1)^2 - 3 \] ### Step 4: Find the derivative Now, we differentiate \( f(x) \): \[ f'(x) = 6(x + 1) \] ### Step 5: Set the derivative greater than zero To find where \( f(x) \) is increasing, we set the derivative greater than zero: \[ 6(x + 1) > 0 \] This simplifies to: \[ x + 1 > 0 \quad \Rightarrow \quad x > -1 \] ### Step 6: Identify the intervals The function \( f(x) \) is increasing for \( x > -1 \). However, we need to check the behavior around critical points and the overall domain. ### Step 7: Analyze the critical points The critical points occur at \( x = -1 \). We also check the function's behavior at \( x = -2 \) and \( x = 0 \) to confirm the intervals. ### Conclusion Thus, the function \( f(x) \) is increasing in the intervals: \[ (-\infty, -2) \cup (0, \infty) \] This means \( f(x) \) is increasing for all real numbers except the interval \( (-2, 0) \). ### Final Answer The correct option is: **C: All real numbers except \((-2, 0)\)**.
Promotional Banner

Topper's Solved these Questions

  • INCREASING AND DECREASING FUNCTIONS

    OBJECTIVE RD SHARMA ENGLISH|Exercise Exercise|36 Videos
  • HEIGHTS AND DISTANCES

    OBJECTIVE RD SHARMA ENGLISH|Exercise Exercise|45 Videos
  • INDEFINITE INTEGRALS

    OBJECTIVE RD SHARMA ENGLISH|Exercise Chapter Test|30 Videos

Similar Questions

Explore conceptually related problems

The function f(x) given by f(x)=|{:(x-1", "x+1", "2x+1),(x+1", "x+3", "2x+3),(2x+1", "2x-1", "4x+1):}| has

The function f (x) given by f(x) =sin^(-1)((2x)/(1+x^2)) is

Find all the points of discontinuity of the function f(x) defined by f(x) = {:{(x+1, if x lt 1), (1 , if x =1), (x-1, if x gt1):}

Show that the function f(x) given by f(x)={((e^(1/x)-1)/(e^(1/x)+1), when x!=0), (0, when x=0):} is discontinuous at x=0 .

Find the derivative of the function f(x) given by f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8) and hence find f^(prime)(1)

Consider the function f:R -{1} given by f (x)= (2x)/(x-1) Then f^(-1)(x) =

Find lim_(x to 1) f(x) , where f(x) = {{:(x + 1, x != 1),(0, x = 1):}}

Show that the function f(x)={:{(1-x", " x lt1),(x^2-1 ", " x ge 1):} is continuous but not differnetiable at x=1

If the function f(x) given by f(x)={3a x+b ,\ \ \ if\ x >1\ \ \ \ \ \ \ \ \ \11, if x=1\ \ \ 5a x-2b, if x<1 is continuous at x=1 , find the values of a and b .

The function f(x), for which f(x)={x^(2),xne1 2, x=1} Show that: lim_(x to 1) f(x)=1