UPSC NDA I 2016 Mathematics Solved Paper

Concept:A point $c$ is a local minimum if $f'(c)=0$ and $f'(x)$ changes from negative to positive as $x$ increases through $c$.Explanation:Given $f(x) = (x-1)^2 (x+1) (x-2)^3$.Differentiate using product rule:$f'(x) = 2(x-1)(x+1)(x-2)^3 + (x-1)^2 (x-2)^3 + 3(x-1)^2 (x+1)(x-2)^2$.Factor out common terms: $(x-1)(x-2)^2$.$f'(x) = (x-1)(x-2)^2 \left[ 2(x+1)(x-2) + (x-1)(x-2) + 3(x-1)(x+1) \right]$.Simplify the bracket:$2(x^2 - x -2) + (x^2 -3x+2) + 3(x^2 -1) = 2x^2 -2x -4 + x^2 -3x+2 + 3x^2 -3 = 6x^2 -5x -5$.Thus $f'(x) = (x-1)(x-2)^2 (6x^2 -5x -5)$.Set $f'(x)=0$ for critical points:$x=1,\; x=2\;(\text{twice}),\; x=\frac{5\pm\sqrt{145}}{12}$.Since $(x-2)^2 \geq 0$, sign of $f'$ depends on $(x-1)$ and $(6x^2-5x-5)$.The quadratic $6x^2-5x-5=0$ has two real roots, say $\alpha = \frac{5-\sqrt{145}}{12}$ and $\beta = \frac{5+\sqrt{145}}{12}$ with $\alpha < 0 < \beta$.Order the critical points: $\alpha$, $1$, $\beta$, $2$.Test sign of $f'(x)$ in each interval:• Left of $\alpha$: $(x-1)<0$, $\text{quadratic}>0$? Actually for $x < \alpha$, quadratic positive, so product negative? Sign: $\text{(+/-?)}$ Let's do systematically: Use test points.For $x < \alpha$, e.g., $x=-1$: $(x-1)=-2$, $(x-2)^2$ positive, $6(-1)^2-5(-1)-5=6+5-5=6>0$. So $f' = (-)(+)(+)$ = negative.Between $\alpha$ and 1, e.g., $x=0$: $(x-1)=-1$, quadratic: $6(0)-0-5=-5<0$. So $f' = (-)(+)(-)$ = positive. So change from negative to positive at $\alpha$ → local min.Between 1 and $\beta$, e.g., $x=1.5$: $(x-1)=0.5>0$, quadratic: $6(2.25)-7.5-5=13.5-12.5=1>0$. So $f' = (+)(+)(+)$ = positive. No sign change at $x=1$ (both sides positive? Actually check left of 1: between $\alpha$ and 1, $f'>0$; right of 1 up to $\beta$, also $f'>0$ → no extremum at $x=1$).At $x=1$, $f'$ does not change sign → point of inflection.Between $\beta$ and 2, e.g., $x=1.8$ (but $\beta \approx 1.59$, so $x=1.7$): $(x-1)=0.7>0$, quadratic: $6(2.89)-8.5-5=17.34-13.5=3.84>0$. So $f'$ positive → sign from left of $\beta$? Actually between 1 and $\beta$, $f'>0$; after $\beta$, for $x$ just greater than $\beta$, quadratic becomes negative (since leading coefficient positive, quadratic changes from positive to negative at $\beta$). Check $x=2$: but at $x=2$, factor $(x-2)^2=0$. For $x$ just above $\beta$ but less than 2, e.g., $x=1.9$: $(x-1)=0.9>0$, quadratic: $6(3.61)-9.5-5=21.66-14.5=7.16>0$? Wait, need to recalc: quadratic at $x=1.9$: $6(3.61)=21.66$, minus $5(1.9)=9.5$, minus $5$ gives $21.66-14.5=7.16 >0$. Actually the quadratic crosses zero at $\beta \approx 1.59$, so for $x>\beta$, the quadratic becomes negative? Let's check at $x=2$: $6(4)-10-5=24-15=9>0$. That suggests quadratic is actually positive again? That can't be right. Solve: $6x^2-5x-5=0$ has discriminant $25+120=145$, roots: $\frac{5\pm\sqrt{145}}{12}$. $\sqrt{145} \approx 12.04$, so roots: $\frac{5-12.04}{12} \approx -0.587$, and $\frac{5+12.04}{12} \approx 1.42$. So correct values: $\alpha \approx -0.587$, $\beta \approx 1.42$. So order: $\alpha \approx -0.587$, then 1, then $\beta \approx 1.42$, then 2. Now test signs:For $x$ between $\beta$ and 2, say $x=1.5$: $(x-1)=0.5>0$, quadratic: $6(2.25)-7.5-5=13.5-12.5=1>0$ → positive. But quadratic should be negative for $x$ between the roots? Since leading coefficient positive, quadratic is negative between the roots $(-0.587, 1.42)$. At $x=1.5$, quadratic is positive because $1.5 > 1.42$. So correct: between $\beta$ and 2, quadratic positive. So sign of $f'$ is positive in $(\beta,2)$. Now check sign just left of $\beta$: between 1 and $\beta$, $x=1.3$: $(x-1)=0.3>0$, quadratic: $6(1.69)-6.5-5=10.14-11.5=-1.36<0$ → $f'$ negative. So at $\beta$, $f'$ changes from negative (left) to positive (right) → local minimum at $\beta$.At $x=2$, $(x-2)^2$ factor gives no sign change; $f'$ positive on both sides (since $(x-1)>0$, $\text{quadratic}>0$ for $x>\beta$). So no extremum at $x=2$.Thus local minima at $x=\alpha$ and $x=\beta$ → two points.Answer:There are two points of local minima. Hence option C is correct.