Question

minimum value of x^(2)-4x+5 at x=p is equal to q then p^(2)+q^(2) is equal to​

Answer 1

Answer:

Correct option is

D

5

We have to find the absolute maximum value of f(x)=x

2

−4x+5 on [0,3].

Consider f(x)=x

2

−4x+5

Since f(x) is a polynomial it is continuous everywhere.

Differentiate f(x) with respect to x on both sides we get

f

′

(x)=2x−4

Now to find the critical point equate f

′

(x) to 0.

⇒2x−4=0

⇒2x=4

⇒x=2

The only critical point is 2.

Now let us find the function value at the endpoints and the critical value to find the absolute maximum.

At x=0, f(0)=0

2

−4(0)+5=5

At x=2, f(2)=2

2

−4(2)+5=4−8+5=1

At x=3, f(3)=3

2

−4(3)+5=9−12+5=2

From the above values we see that the absolute maximum is 5 and it occur at x=0.

Answer 2

Answer:

The value of p² + q² is 5.

Step-by-step explanation:

Second Derivative Test for Maxima and Minima :

• Let f(x) be a continuous and differentiable function on an interval I .
• Let, a ∈ I is such that f''(x) is continuous at x = a. Then,
• x = a is a point of local maxima, if f'(a) = 0 and f"(a) < 0.
• x = a is a point of local minima, if f'(a) = 0 and f"(a) > 0.
• where, f'(x) is first order derivative with respect to x and f"(x) is second order derivative with respect to x.

Given that :

• f(x) = x² - 4x + 5.
• minima at x = p is equal to q.

To find :

• Value of p² + q²

Solution :

• The given function is f(x) = x² - 4x + 5. Using second derivative test;
• f'(x) = 2x - 4
• Now, f'(x) = 0 gives, x = 2.
• Again differentiating f'(x), we get;
• f"(x) = 2.
• f"(2) = 2.
• Since, f"(x) > 0 at x = 2. This is a point of local minima.
• Point x = 2 is a point of local minima. Hence, p = 2 ----(1)
• At local minima, value of f(x) is equal to q.
• Putting x = 2 in f(x), we get;
• f(2) = (2)² - 4(2) + 5

⇒f(2) = (4 - 8 + 5) = 1 .

⇒ q = 1 ----(2).

• The value of p² + q² = (2)² + (1)²
• = 4 + 1 = 5.
• Hence, value of p² + q² is 5.