Question

Given that,

x	10	15	20
f(x)	14	18	28

The estimated value of f(12) is
(a) 15.520
(b) 14.88
(c) 15.37
(d) 14.78

Answer 1

To solve the given question, we will first find out the function f(x). We will assume that the function f(x) is of the form ${{a}_{1}}{{x}^{n}}+{{a}_{2}}{{x}^{n-1}}+{{a}_{3}}{{x}^{n-2}}+.....+{{a}_{n}}x+{{a}_{n+1}}.$ Now, as there are only three pairs of x and f(x) are given, we will assume the value of n such that we are able to find all the variables $\left( {{a}_{1}},{{a}_{2}}......{{a}_{n}} \right).$ Now, after finding the values of all the variables, we will put them in f(x). After doing this, we will put x = 12 in f(x) and we will find its value.

Complete step-by-step answer:
To start with, we will assume that the function f(x) is a polynomial of n order i.e. $f\left( x \right)={{a}_{1}}{{x}^{n}}+{{a}_{2}}{{x}^{n-1}}+{{a}_{3}}{{x}^{n-2}}+.....+{{a}_{n}}x+{{a}_{n+1}}.$ Now, we have to find this function f(x) and three pairs of x and f(x) are given in the question. So, we will have to take the value of n such that we are able to find the values of all the variables $\left( {{a}_{1}},{{a}_{2}}......{{a}_{n}} \right).$ Thus, we will choose n = 2 as we have three unknown variables ${{a}_{1}},{{a}_{2}},{{a}_{3}}.$ So, after putting n = 2 in f(x), we get the following equation.
$f\left( x \right)={{a}_{1}}{{x}^{2}}+{{a}_{2}}x+{{a}_{3}}......\left( i \right)$
Now, we have to find the values of ${{a}_{1}},{{a}_{2}},{{a}_{3}}.$ For this, we are given three pairs of x and f(x). So, when we put x = 10, we get f(x) = 14. Thus, we will get the following equation
$f\left( 10 \right)={{a}_{1}}{{\left( 10 \right)}^{2}}+{{a}_{2}}\left( 10 \right)+{{a}_{3}}$
$\Rightarrow 14=100{{a}_{1}}+10{{a}_{2}}+{{a}_{3}}$
$\Rightarrow 100{{a}_{1}}+10{{a}_{2}}+{{a}_{3}}=14.......\left( ii \right)$
Now, it is given that when we put x = 15, we get f(x) = 18. Thus, we will get the following equation
$f\left( 15 \right)={{a}_{1}}{{\left( 15 \right)}^{2}}+{{a}_{2}}\left( 15 \right)+{{a}_{3}}$
$\Rightarrow 18=225{{a}_{1}}+15{{a}_{2}}+{{a}_{3}}$
$\Rightarrow 225{{a}_{1}}+15{{a}_{2}}+{{a}_{3}}=18......\left( iii \right)$
Now, it is given that when we put x = 20, we get f(x) = 28. Thus, we will get the following equation
$f\left( 20 \right)={{a}_{1}}{{\left( 20 \right)}^{2}}+{{a}_{2}}\left( 20 \right)+{{a}_{3}}$
$\Rightarrow 28=400{{a}_{1}}+20{{a}_{2}}+{{a}_{3}}$
$\Rightarrow 400{{a}_{1}}+20{{a}_{2}}+{{a}_{3}}=28.......\left( iv \right)$
Now, we will subtract (ii) from (iii). Thus, we will get the following equation.
$\Rightarrow \left( 225{{a}_{1}}+15{{a}_{2}}+{{a}_{3}} \right)-\left( 100{{a}_{1}}+10{{a}_{2}}+{{a}_{3}} \right)=18-14$
$\Rightarrow 225{{a}_{1}}-100{{a}_{1}}+15{{a}_{2}}-10{{a}_{2}}+{{a}_{3}}-{{a}_{3}}=4$
$\Rightarrow 125{{a}_{1}}+5{{a}_{2}}=4......\left( v \right)$
Now, we will subtract (iii) from (iv). Thus, we will get the following equation
$\Rightarrow \left( 400{{a}_{1}}+20{{a}_{2}}+{{a}_{3}} \right)-\left( 225{{a}_{1}}+15{{a}_{2}}+{{a}_{3}} \right)=28-18$
$\Rightarrow 400{{a}_{1}}-225{{a}_{1}}+20{{a}_{2}}-15{{a}_{2}}+{{a}_{3}}-{{a}_{3}}=10$
$\Rightarrow 175{{a}_{1}}+5{{a}_{2}}=10......\left( vi \right)$
Now, we will subtract (v) from (vi). Thus, we will get the following equation.
$\Rightarrow \left( 175{{a}_{1}}+5{{a}_{2}} \right)-\left( 125{{a}_{1}}+5{{a}_{2}} \right)=10-4$
$\Rightarrow 175{{a}_{1}}-125{{a}_{1}}+5{{a}_{2}}-5{{a}_{2}}=6$
$\Rightarrow 50{{a}_{1}}=6$
$\Rightarrow {{a}_{1}}=\dfrac{6}{50}$
$\Rightarrow {{a}_{1}}=\dfrac{3}{25}......\left( vii \right)$
Now, we will put the value ${{a}_{1}}$ from (vii) to (v). Thus, we will get the following equation.
$\Rightarrow 125\left( \dfrac{3}{25} \right)+5{{a}_{2}}=4$
$\Rightarrow 15+5{{a}_{2}}=4$
$\Rightarrow 5{{a}_{2}}=4-15$
$\Rightarrow 5{{a}_{2}}=-11$
$\Rightarrow {{a}_{2}}=\dfrac{-11}{5}....\left( viii \right)$
Now, we will put the values of ${{a}_{1}}$ and ${{a}_{2}}$ from (vii) and (viii) to (ii). Thus, we will get the following equation.
$\Rightarrow 100\left( \dfrac{3}{25} \right)+10\left( \dfrac{-11}{5} \right)+{{a}_{3}}=14$
$\Rightarrow 12-22+{{a}_{3}}=14$
$\Rightarrow -10+{{a}_{3}}=14$
$\Rightarrow {{a}_{3}}=24......\left( ix \right)$
Now, we will put the values of ${{a}_{1}},{{a}_{2}},{{a}_{3}}$ from (vii), (viii) and (ix) into (i). Thus, we will get,
$f\left( x \right)=\dfrac{3}{25}{{x}^{2}}-\dfrac{11}{5}x+24$
Now, we have to find the value of f(12). For this, we will simply put x = 12. Thus, we will get,
$f\left( 12 \right)=\dfrac{3}{25}{{\left( 12 \right)}^{2}}-\dfrac{11}{5}\left( 12 \right)+24$
$\Rightarrow f\left( 12 \right)=\dfrac{3}{25}\left( 144 \right)-\dfrac{11\times 12}{5}+24$
$\Rightarrow f\left( 12 \right)=\dfrac{432}{25}-\dfrac{132}{5}+24$
$\Rightarrow f\left( 12 \right)=17.28-26.5+24$
$\Rightarrow f\left( 12 \right)=14.88$
Hence, option (b) is the right answer.

So, the correct answer is “Option (b)”.

Note: While solving the question, we have assumed that f(x) is a quadratic polynomial. There may be chances that f(x) is a polynomial of a higher order. Also, it may be possible that f(x) is not even a polynomial, it is some other function. That’s why the value of f(12) obtained is not exact, it is estimated on the basis that f(x) is a quadratic polynomial.