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Question: The minimum value of \( f(a) = (2{a^2} - 3) + 3(3 - a) + 4 \) is \( A)\dfrac{{15}}{2} \) \( B...

The minimum value of f(a)=(2a2−3)+3(3−a)+4f(a) = (2{a^2} - 3) + 3(3 - a) + 4 is
A)152A)\dfrac{{15}}{2}
B)112B)\dfrac{{11}}{2}
C)−132C)\dfrac{{ - 13}}{2}
D)718D)\dfrac{{71}}{8}

Explanation

Solution

First, given that we have the quadratic function f(a)=(2a2−3)+3(3−a)+4f(a) = (2{a^2} - 3) + 3(3 - a) + 4 and we try to simplify the given equation as in the proper form.
After simplified into the proper form of quadratic function, we try to find the unknown variable aa which is in the polynomial f(a)f(a)
Finally, after getting the values of unknown variables aa we just substitute in the given function to get the minimum value of the given function.
Complete step by step answer:
Since from the given that we have to degree at most two-equation which is the quadratic function and it is given as f(a)=(2a2−3)+3(3−a)+4f(a) = (2{a^2} - 3) + 3(3 - a) + 4
Now we are going to simplify the given equation so that it is easy to find the unknown values.
Let us solve the equation f(a)=(2a2−3)+3(3−a)+4f(a) = (2{a^2} - 3) + 3(3 - a) + 4 , now multiplied the terms in the center we get f(a)=(2a2−3)+3(3−a)+4⇒f(a)=(2a2−3)+9−3a+4f(a) = (2{a^2} - 3) + 3(3 - a) + 4 \Rightarrow f(a) = (2{a^2} - 3) + 9 - 3a + 4
Separate the constants terms, power one terms and power two terms, thus we get f(a)=2a2−3a−3+9+4f(a) = 2{a^2} - 3a - 3 + 9 + 4
Then by the addition and subtraction operation, we have f(a)=2a2−3a+10f(a) = 2{a^2} - 3a + 10 which is the proper form of the given quadratic function.
Now apply the first derivative of the given function thus we get f(a)=2a2−3a+10⇒f1(a)=4a−3f(a) = 2{a^2} - 3a + 10 \Rightarrow {f^1}(a) = 4a - 3
The first derivative value is positive because 4>04 > 0 (at the degree one) and hence we get f1(a)=4a−3>0{f^1}(a) = 4a - 3 > 0is also the positive term. Also, the second derivative of the function is f11(a)=4>0{f^{11}}(a) = 4 > 0
If the function needs to be minimum then we assume the first derivative is equal to zero; then we have f1(a)=0{f^1}(a) = 0
Substitute this value in the above first derivative, then we get f1(a)=4a−3⇒0=4a−3⇒a=34{f^1}(a) = 4a - 3 \Rightarrow 0 = 4a - 3 \Rightarrow a = \dfrac{3}{4}
(this process done just to get the minimum value of the unknown variables)
Thus, apply the value of a=34a = \dfrac{3}{4}in the proper form of a quadratic given function, then we get f(a)=2a2−3a+10⇒f(a)min⁡=2(34)2−3(34)+10f(a) = 2{a^2} - 3a + 10 \Rightarrow f{(a)_{\min }} = 2{(\dfrac{3}{4})^2} - 3(\dfrac{3}{4}) + 10
Further solving we have, f(a)min⁡=2(34)2−3(34)+10⇒f(a)min⁡=(98)−(94)+10⇒718f{(a)_{\min }} = 2{(\dfrac{3}{4})^2} - 3(\dfrac{3}{4}) + 10 \Rightarrow f{(a)_{\min }} = (\dfrac{9}{8}) - (\dfrac{9}{4}) + 10 \Rightarrow \dfrac{{71}}{8} (by multiplication, division operation, and LCM method)

So, the correct answer is “Option D”.

Note: LCM method is to find the common multiples between two or more numbers and then multiply with the least possible ways.
Quadratic function means at most degree two and the general equations ax2+bx+c=0a{x^2} + bx + c = 0
Where the viable a=0a = 0 is not possible because if a=0a = 0 then we have bx+c=0bx + c = 0 which is a linear function and hence a=0a = 0 is never zero.