Let $\mathbb{R}$ denote the set of all real numbers. For a real number $x$, let $[x]$ denote the greatest integer less than or equal to $x$. Let $n$ denote a natural number.
Match each entry in List-I to the correct entry in List-II and choose the correct option.
**List-I**
(P) The minimum value of $n$ for which the function
$f(x) = \left[\dfrac{10x^3-45x^2+60x+35}{n}\right]$
is continuous on the interval $[1,2]$, is
(Q) The minimum value of $n$ for which
$g(x)=(2n^2-13n-15)(x^3+3x)$, $x\in\mathbb{R}$, is an increasing function on $\mathbb{R}$, is
(R) The smallest natural number $n$ which is greater than 5, such that $x=3$ is a point of local minima of
$h(x)=(x^2-9)^n(x^2+2x+3)$, is
(S) Number of $x_0\in\mathbb{R}$ such that
$l(x)=\displaystyle\sum_{k=0}^{4}\left(\sin|x-k|+\cos\left|x-k+\dfrac{1}{2}\right|\right)$, $x\in\mathbb{R}$, is NOT differentiable at $x_0$, is
**List-II**
(1) 8
(2) 9
(3) 5
(4) 6
(5) 10
(P)→(1) (Q)→(3) (R)→(2) (S)→(5)
(P)→(2) (Q)→(1) (R)→(4) (S)→(3)
(P)→(5) (Q)→(1) (R)→(4) (S)→(3)
(P)→(2) (Q)→(3) (R)→(1) (S)→(5)
Step-by-Step Solution
Key Concept: Floor function continuity requires avoiding integers in range; increasing function needs positive leading coefficient; local min at zero of even-power factor; differentiability of absolute value compositions
(P) $p(x)=10x^3-45x^2+60x+35$ on $[1,2]$. $p'(x)=30x^2-90x+60=30(x-1)(x-2)$, so $p$ increases then decreases: $p(1)=60$, $p(2)=57$ (max $p(1.5)=...)$. Actually $p(1)=10-45+60+35=60$, $p(2)=80-180+120+35=55$. For $[p(x)/n]$ to be continuous, $p(x)/n$ must not cross any integer on $[1,2]$. The range of $p$ on $[1,2]$ is $[55,60]$ (since $p(1)=60$, $p(2)=55$... wait need to check max). $p'=0$ at $x=1,2$ (endpoints), so $p$ is monotone on $[1,2]$? $p'(x)=30(x-1)(x-2)\leq0$ on $[1,2]$. So $p$ decreases from 60 to 55. Range = $[55,60]$. For $[p/n]$ continuous, $p/n$ must avoid integers: need $\lfloor55/n\rfloor=\lfloor60/n\rfloor$ OR the function $p/n$ is continuous except at floor-jump points. The minimum $n$ such that no integer lies in $(55/n, 60/n)$... alternatively, we need $n>60-55=5$, i.e., $n\geq6$? Check $n=9$: $55/9=6.1$, $60/9=6.6$, no integer in between. Check $n=8$: $55/8=6.875$, $60/8=7.5$ — integer 7 in between. $n=9$: works. → (P)→9→(2). ✓
(Q) $g(x)=(2n^2-13n-15)(x^3+3x)$ increasing: need coefficient $>0$ since $x^3+3x$ is increasing. $2n^2-13n-15>0$. Roots: $n=\dfrac{13\pm\sqrt{169+120}}{4}=\dfrac{13\pm17}{4}$. So $n=\dfrac{30}{4}=7.5$ or $n=-1$. For $n>7.5$, i.e., minimum natural number $n=8$. → (Q)→8→(1). ✓
(R) $h(x)=(x^2-9)^n(x^2+2x+3)$. At $x=3$: $(x^2-9)=0$. For $x=3$ to be local minimum with $n>5$. $h(x)=(x-3)^n(x+3)^n(x^2+2x+3)$. Near $x=3$: $h(x)\approx(x-3)^n\cdot6^n\cdot18$. For local min at $x=3$: need $n$ even (so $(x-3)^n\geq0$ on both sides, with minimum 0 at $x=3$). Smallest even $n>5$ is $n=6$. → (R)→6→(4). ✓
(S) $l(x)=\sum_{k=0}^4(\sin|x-k|+\cos|x-k+1/2|)$. Non-differentiable points from $|x-k|$: at $x=0,1,2,3,4$ (5 points). From $|x-k+1/2|$: at $x=k-1/2$, i.e., $x=-0.5,0.5,1.5,2.5,3.5$ (5 points). But $\cos|u|$ is differentiable at $u=0$ since $\cos|u|=\cos u$ is even and smooth. So only $\sin|x-k|$ contributes non-differentiable points at $x=0,1,2,3,4$. That's 5 points. → (S)→5→(3). ✓
Answer: B → (P)→(2), (Q)→(1), (R)→(4), (S)→(3).
Correct Answer: B