Home
Class 12
MATHS
Consider, f (x) is a function such that ...

Consider, f (x) is a function such that `f(1)=1, f(2)=4 and f(3)=9`
Statement 1 : `f''(x)=2" for some "x in (1,3)`
Statement 2 : `g(x)=x^(2) rArr g''(x)=2, AA x in R`

A

Statement-1 is True, Statement-2 is True and Statement-2 is a correct explanation for Statement-1

B

Statement-1 is True, Statement-2 is True and Statement-2 is NOT a correct explanation for Statement-1

C

Statement-1 is True, Statement-2 is False

D

Statement-1 is False, Statement-2 is True

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will analyze both statements step by step. ### Step 1: Analyze the function \( f(x) \) We are given that: - \( f(1) = 1 \) - \( f(2) = 4 \) - \( f(3) = 9 \) From these values, we can observe that: - \( f(1) = 1^2 \) - \( f(2) = 2^2 \) - \( f(3) = 3^2 \) This suggests that \( f(x) = x^2 \) could be a candidate function. ### Step 2: Determine the first and second derivatives of \( f(x) \) Now, let's find the first and second derivatives of \( f(x) = x^2 \): - The first derivative \( f'(x) = 2x \) - The second derivative \( f''(x) = 2 \) ### Step 3: Verify Statement 1 Statement 1 claims that \( f''(x) = 2 \) for some \( x \) in the interval \( (1, 3) \). Since we found that \( f''(x) = 2 \) for all \( x \) in \( \mathbb{R} \), it is certainly true for \( x \) in the interval \( (1, 3) \). Thus, Statement 1 is **true**. ### Step 4: Analyze Statement 2 Statement 2 states that \( g(x) = x^2 \) implies \( g''(x) = 2 \) for all \( x \in \mathbb{R} \). We already calculated: - \( g'(x) = 2x \) - \( g''(x) = 2 \) This is also true for all \( x \in \mathbb{R} \). Thus, Statement 2 is also **true**. ### Step 5: Conclusion Both statements are true: - Statement 1 is true because \( f''(x) = 2 \) for some \( x \) in \( (1, 3) \). - Statement 2 is true because \( g''(x) = 2 \) for all \( x \in \mathbb{R} \). However, the problem states that Statement 2 does not explain Statement 1, which is also correct. ### Final Answer: Both statements are true, but Statement 2 does not explain Statement 1. ---
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • DIFFERENTIAL CALCULUS 2

    VMC MODULES ENGLISH|Exercise Numerical ValueType for JEE Main|14 Videos
  • DIFFERENTIAL CALCULUS 2

    VMC MODULES ENGLISH|Exercise JEE Main (Archive)|36 Videos
  • DIFFERENTIAL CALCULUS 2

    VMC MODULES ENGLISH|Exercise Level -1|102 Videos
  • DIFFERENTIAL CALCULUS

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive)|75 Videos
  • DIFFERENTIAL EQUATIONS

    VMC MODULES ENGLISH|Exercise JEE ADVANCE (ARCHIVE )|32 Videos

Similar Questions

Explore conceptually related problems

Let f(x) be a function such that f(x).f(y)=f(x+y) , f(0)=1 , f(1)=4 . If 2g(x)=f(x).(1-g(x))

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

STATEMENT - 1 : Let f be a twice differentiable function such that f'(x) = g(x) and f''(x) = - f (x) . If h'(x) = [f(x)]^(2) + [g (x)]^(2) , h(1) = 8 and h (0) =2 Rightarrow h(2) =14 and STATEMENT - 2 : h''(x)=0

Consider the function f(x)=(.^(x+1)C_(2x-8))(.^(2x-8)C_(x+1)) Statement-1: Domain of f(x) is singleton. Statement 2: Range of f(x) is singleton.

Let f be a differential function such that f(x)=f(2-x) and g(x)=f(1 +x) then (1) g(x) is an odd function (2) g(x) is an even function (3) graph of f(x) is symmetrical about the line x= 1 (4) f'(1)=0

Let f be a differential function such that f(x)=f(2-x) and g(x)=f(1 +x) then (1) g(x) is an odd function (2) g(x) is an even function (3) graph of f(x) is symmetrical about the line x= 1 (4) f'(1)=0

The function f (x) satisfy the equation f (1-x)+ 2f (x) =3x AA x in R, then f (0)=

consider the function f:R rarr R,f(x)=(x^(2)-6x+4)/(x^(2)+2x+4) f(x) is

Let f be a function defined by f(x)=(x-1)^(2)+1,(xge1) . Statement 1: The set (x:f(x)=f^(-1)(x)}={1,2} Statement 2: f is a bijection and f^(-1)(x)=1+sqrt(x-1),xge1 .

Statement-1: Function f(x)=sin(x+3 sin x) is periodic . Statement-2: If g(x) is periodic then f(g(x)) periodic