← 2020 Paper 1

UPSC 2020 Maths Optional Paper 1 Q3a — Step-by-Step Solution

20 marks · Section A

Maxima and minima of single-variable functions · Calculus · asked 7× in 13 yrs · Read the full method →

Question

Consider the function $f(x)=\int_0^x (t^2-5t+4)(t^2-5t+6)\,dt$. (i) Find the critical points of the function $f(x)$. (ii) Find the points at which local minimum occurs. (iii) Find the points at which local maximum occurs. (iv) Find the number of zeros of the function $f(x)$ in $[0,5]$.

Technique

FTC for $f'$; factor into linear terms; first-derivative sign test for extrema; evaluate $f$ at critical/endpoints to count zeros.

Solution

By the Fundamental Theorem of Calculus,

Factor each quadratic: $x^2-5x+4=(x-1)(x-4)$ and $x^2-5x+6=(x-2)(x-3)$. Hence

(i) Critical points

$f'(x)=0\iff$

(ii)–(iii) Classify via the sign of $f'$

$f'$ is a product of four simple linear factors, so it changes sign at each root. Sign chart:

Reading the transitions of $f'$:

• $x=1$: $+\to-$ ⟹ local maximum.
• $x=2$: $-\to+$ ⟹ local minimum.
• $x=3$: $+\to-$ ⟹ local maximum.
• $x=4$: $-\to+$ ⟹ local minimum.

(iv) Number of zeros of $f$ in $[0,5]$

Integrate $f'(x)=x^4-10x^3+35x^2-50x+24$:

(There is no constant of integration since $f(0)=0$.) Evaluate at the relevant points:

$f(0)=0$, and from $x=0$ the function rises ($f'(0)>0$). The two interior local minima (at $x=2,4$) have values $7.73$ and $7.47$, both well above $0$, so $f(x)$ never returns to $0$ on $(0,5]$. Therefore $f>0$ on $(0,5]$ and the only zero in $[0,5]$ is at $x=0$.

Answer