NIMCET 2016 — Mathematics PYQ

Concept:
The function f(x) has a global maximum at the point 'a' in the interval I if f(a) ≥ f(x), for all x ∈ I. Similarly, f(x) has a global minimum at the point 'a' if f(a) ≤ f(x), for all x ∈ I.
Global maxima or minima in [a, b] will always occur either at the critical points of f(x) within [a, b] or at the endpoints of the interval.

Calculations:
Given, function is $f(x)=x^2-bx+c$.
Consider α, β be the roots of the function $f(x)$.

$\Rightarrow \alpha +\beta =$b

Here, f(x)=0 has to prime numbers as roots and b is an odd positive integer.
$\Rightarrow$One root of the function f$(\mathbf{x})={\mathbf{x}}^2-\mathbf{bx}+$c as 2.
$\Rightarrow$f(2)=0

$$\Rightarrow 2^2-\mathrm{b}(2)+\mathrm{c}=0$$

$$\Rightarrow 2\mathrm{b}-\mathrm{c}=4....(1)$$

andb+c=35...(2)from equation (1)and(2),wehave ST OOO K.com

$\Rightarrow b=13$ and c=22

The global minimum value of f(x) is attained x= {13}

$\mathbf{f}(\frac{13}{2})=($ $\frac{13}{2}{)}^2-13.(\frac{13}{2})+22$
$$\mathrm{f}(\frac{13}{2})=-\frac{81}{4}$$

Hence,  if f( x) = $x^2$- bx+ c,  b is an odd positive integer.  If f( x) = 0 has to prime numbers as roots and b+ c= 35,  then Hence,  if f( x) = $x^2$- bx+ c,  b is an odd positive integer.  If f( x) = 0 has to prime numbers as root and b+ c= 35,  then the global minimum

value of f(x) is$-\frac{81}{4}$