GATE 2018 IN Q.4: solution and answer

GATE 2018 · INQ.4 · 1 mark · NAT

Consider two functions $f (x) = (x - 2)^{2}$ and $g (x) = 2 x - 1$ , where $x$ is real. The smallest value of $x$ for which $f (x) = g (x)$ is ______.

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

The smallest real $x$ with $f (x) = g (x)$ is $x = 1$ . The two curves meet at $x = 1$ and $x = 5$ .

Official answer: 1Source: Official answer key, GATE 2018 IN (IIT Guwahati)

Set the two functions equal: $(x - 2)^{2} = 2 x - 1$ .
Expand the left side: $x^{2} - 4 x + 4 = 2 x - 1$ .
Move every term to one side to get a quadratic: $x^{2} - 6 x + 5 = 0$ .
Factorise: $(x - 1) (x - 5) = 0$ , so $x = 1$ or $x = 5$ . Both roots are real, which the question requires.
Check both roots: at $x = 1$ , $f (1) = (1 - 2)^{2} = 1$ and $g (1) = 2 (1) - 1 = 1$ . At $x = 5$ , $f (5) = (5 - 2)^{2} = 9$ and $g (5) = 2 (5) - 1 = 9$ . Both are genuine intersection points.
The smaller of the two roots is $x = 1$ .

KEY IDEA

Where two curves meet, their values are equal, so setting $f (x) = g (x)$ gives the quadratic $x^{2} - 6 x + 5 = 0$ . It has two real roots, $1$ and $5$ , and the question asks for the smaller one.

WATCH OUT FOR

Three slips are common. Some students answer $5$ , the larger root. Others answer $3$ , which is where $f (x) - g (x)$ is smallest (the vertex of $x^{2} - 6 x + 5$ ), not where it equals zero. Others expand $(x - 2)^{2}$ wrongly, as $x^{2} + 4$ or $x^{2} - 4 x - 4$ , and end up with the wrong quadratic.

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 39 Calculus questions, with notes

GATE 2018 IN Q.4: solution and answer

Answer and solution

More Calculus questions