Question:
Absolute minimum value of \( f(x) = (x - 2)^2 + 5 \) in the interval \( [-3, 2] \) is :
Absolute minimum value of \( f(x) = (x - 2)^2 + 5 \) in the interval \( [-3, 2] \) is :
Show Hint
For a parabola of the form \( (x - h)^2 + k \), the vertex \( (h, k) \) is the absolute minimum point if the parabola opens upward.
Since \( (x - 2)^2 \) is always \( \geq 0 \), the expression is minimum when the squared term is zero.
Since \( (x - 2)^2 \) is always \( \geq 0 \), the expression is minimum when the squared term is zero.
Updated On: Sep 10, 2026
- \( -3 \)
- \( 2 \)
- \( 5 \)
- \( 30 \)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
• To find the absolute minimum in a closed interval, we must check the functional values at the critical points and the endpoints.
• Critical points are where \( f'(x) = 0 \).
Step 1: Find the critical points by differentiating \( f(x) \)
\[ f(x) = (x - 2)^2 + 5 \] \[ f'(x) = 2(x - 2) \] Setting \( f'(x) = 0 \): \[ 2(x - 2) = 0 \Rightarrow x = 2 \] The critical point \( x = 2 \) is within the given interval \( [-3, 2] \).
Step 2: Calculate the functional values at critical points and endpoints
The endpoints are \( x = -3 \) and \( x = 2 \). At \( x = -3 \): \[ f(-3) = (-3 - 2)^2 + 5 = (-5)^2 + 5 = 25 + 5 = 30 \] At \( x = 2 \) (which is both a critical point and an endpoint): \[ f(2) = (2 - 2)^2 + 5 = 0^2 + 5 = 5 \]
Step 3: Identify the absolute minimum value
Comparing the values: Values are \( \{30, 5\} \). The smallest value is \( 5 \). So, the absolute minimum value is \( 5 \), matching option (C).
• To find the absolute minimum in a closed interval, we must check the functional values at the critical points and the endpoints.
• Critical points are where \( f'(x) = 0 \).
Step 1: Find the critical points by differentiating \( f(x) \)
\[ f(x) = (x - 2)^2 + 5 \] \[ f'(x) = 2(x - 2) \] Setting \( f'(x) = 0 \): \[ 2(x - 2) = 0 \Rightarrow x = 2 \] The critical point \( x = 2 \) is within the given interval \( [-3, 2] \).
Step 2: Calculate the functional values at critical points and endpoints
The endpoints are \( x = -3 \) and \( x = 2 \). At \( x = -3 \): \[ f(-3) = (-3 - 2)^2 + 5 = (-5)^2 + 5 = 25 + 5 = 30 \] At \( x = 2 \) (which is both a critical point and an endpoint): \[ f(2) = (2 - 2)^2 + 5 = 0^2 + 5 = 5 \]
Step 3: Identify the absolute minimum value
Comparing the values: Values are \( \{30, 5\} \). The smallest value is \( 5 \). So, the absolute minimum value is \( 5 \), matching option (C).
Was this answer helpful?
0
0
Top CBSE CLASS XII Mathematics Questions
- 2, b, c are in A.P. and the range of determinant $\begin{vmatrix}1&1&1\\ 2&b&c\\ 4&b^{2}&c^{2}\end{vmatrix}$ is [2, 16]. Then find range of c is
- A committee of 11 members is to be formed from 8 males and 5 females. Let m denotes the number of ways of selecting committee having atleast 6 males and n denotes the number of ways of selecting atleast 3 females then
Determine whether each of the following relations are reflexive, symmetric, and transitive.
- Relation R in the set A={1,2,3,...13,14} defined as R={(x,y): 3x-y=0}.
- Relation R in the set of N natural numbers defined as R={(x,y): y=x+5 and x<4}.
- Relation R in the set A={1,2,3,4,5,6} defined as R={(x,y): y is divisible by x}.
- Relation R in the set of Z integers defined as R={(x,y): x-y is an integer}
- Relation R in the set of human beings in a town at a particular time given by
- R={(x,y): x and y work at same place}
- R={(x,y): x and y live in the same locality}
- R={(x,y): x is exactly 7 am taller that y}
- R={(x,y): x is wife of y}
- R={(x,y):x is father of y}
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
View More Questions
Top CBSE CLASS XII Absolute maxima and Absolute minima Questions
- Let \( \mathbb{R}_+ \) denote the set of all non-negative real numbers. Then the function \( f : \mathbb{R}_+ \to \mathbb{R}_+ \) defined as \( f(x) = x^2 + 1 \) is:
- A function \( f(x) = |1 - x + |x|| \) is:
- If the sides of a square are decreasing at the rate of \( 1.5 \, \mathrm{cm/s} \), the rate of decrease of its perimeter is:
- \( \int_{-a}^a f(x) \, dx = 0 \), if:
- \( x \log x \frac{dy}{dx} + y = 2 \log x \) is an example of a:
View More Questions
Top CBSE CLASS XII Questions
- 2, b, c are in A.P. and the range of determinant $\begin{vmatrix}1&1&1\\ 2&b&c\\ 4&b^{2}&c^{2}\end{vmatrix}$ is [2, 16]. Then find range of c is
- A boy of mass 50 kg is standing at one end of a, boat of length 9 m and mass 400 kg. He runs to the other, end. The distance through which the centre of mass of the boat boy system moves is
- A capillary tube of radius r is dipped inside a large vessel of water. The mass of water raised above water level is M. If the radius of capillary is doubled, the mass of water inside capillary will be
- A circular disc is rotating about its own axis. An external opposing torque 0.02 Nm is applied on the disc by which it comes rest in 5 seconds. The initial angular momentum of disc is
- A circular disc is rotating about its own axis at uniform angular velocity \(\omega.\) The disc is subjected to uniform angular retardation by which its angular velocity is decreased to \(\frac {\omega}{2}\) during 120 rotations. The number of rotations further made by it before coming to rest is
View More Questions