NCERT Solutions · Class 9 Mathematics Introduction to Linear Polynomials

A fixed starting height plus the same growth every month.The plant is $\displaystyle 1.75$ feet tall to begin with, and each month adds another $\displaystyle 0.5$ feet. Write \(\displaystyle t\) for the number of months and \(\displaystyle h\) for the height in feet.(i) Height after $\displaystyle 7$ months.In $\displaystyle 7$ months the plant gains \(\displaystyle 0.5 \times 7 = 3.5\) feet, on top of the $\displaystyle 1.75$ feet it already had: \[h = 1.75+3.5 = 5.25 \ \text{feet} \] In fractions this is \(\displaystyle \tfrac{7}{4}+\tfrac{7}{2} = \tfrac{7}{4}+\tfrac{14}{4} = \tfrac{21}{4} = 5\tfrac{1}{4}\) feet, which agrees.(ii) Table of heights for \(\displaystyle t = 0\) to \(\displaystyle t = 10\).Each row is obtained from the row above by adding $\displaystyle 0.5$ feet.

Time \(\displaystyle t\) (months)	Height \(\displaystyle h\) (feet)
$\displaystyle 0$	$\displaystyle 1.75$
$\displaystyle 1$	$\displaystyle 2.25$
$\displaystyle 2$	$\displaystyle 2.75$
$\displaystyle 3$	$\displaystyle 3.25$
$\displaystyle 4$	$\displaystyle 3.75$
$\displaystyle 5$	$\displaystyle 4.25$
$\displaystyle 6$	$\displaystyle 4.75$
$\displaystyle 7$	$\displaystyle 5.25$
$\displaystyle 8$	$\displaystyle 5.75$
$\displaystyle 9$	$\displaystyle 6.25$
$\displaystyle 10$	$\displaystyle 6.75$

The value at \(\displaystyle t=7\) is $\displaystyle 5.25$ feet, matching part (i).(iii) The expression, and why the growth is linear.After \(\displaystyle t\) months the plant has gained \(\displaystyle 0.5t\) feet, so \[h = 1.75+0.5t \]Two reasons this is linear growth:First, as an algebraic expression, \(\displaystyle 1.75+0.5t\) is a polynomial of degree $\displaystyle 1$ in \(\displaystyle t\) — the variable \(\displaystyle t\) appears only to the first power, with no \(\displaystyle t^{2}\), no \(\displaystyle t^{3}\) and no \(\displaystyle t\) in a denominator. That is exactly the definition of a linear polynomial.Second, and this is what "linear" means in the growing plant itself, the increase over any one month is always the same: \[h(t+1)-h(t) = \big(1.75+0.5(t+1)\big)-\big(1.75+0.5t\big) = 0.5 \] The \(\displaystyle 1.75\) and the \(\displaystyle 0.5t\) cancel out and only \(\displaystyle 0.5\) is left, whatever \(\displaystyle t\) is. So the plant grows by $\displaystyle 0.5$ feet in the first month, $\displaystyle 0.5$ feet in the fifth month, $\displaystyle 0.5$ feet in the tenth — equal increases in equal intervals of time, which is precisely the constant-rate growth the column of numbers in the table shows.It is worth seeing what would not be linear: if the plant instead grew by $\displaystyle 10$% of its current height each month, the increase would be larger every month (bigger plant, bigger $\displaystyle 10$%), the differences between successive rows would not be constant, and the pattern would not be a linear one.In \(\displaystyle h = 1.75+0.5t\), the constant \(\displaystyle 1.75\) is the height at \(\displaystyle t=0\) (where the plant starts) and the coefficient \(\displaystyle 0.5\) is the fixed monthly rate of growth.(i) $\displaystyle 5.25$ feet. (ii) See the table above: heights $\displaystyle 1.75$, $\displaystyle 2.25$, $\displaystyle 2.75$, …, $\displaystyle 6.75$ feet for \(\displaystyle t = 0\) to \(\displaystyle 10\). (iii) \(\displaystyle h = 1.75+0.5t\) — linear because \(\displaystyle t\) appears only to the first power, so the height rises by the same $\displaystyle 0.5$ feet in every month.

Constant yearly drop — linear decay.The phone loses the same amount, ₹$\displaystyle 800$, in every single year. That constant change per year is exactly what makes the pattern linear.(i) Value after $\displaystyle 3$ years.In $\displaystyle 3$ years the total drop is \(\displaystyle 3 \times 800 = 2400\). \[v = 10000 - 2400 = 7600 \] So the phone is worth ₹$\displaystyle 7,600$ after $\displaystyle 3$ years.(ii) Table of values.After \(\displaystyle t\) years the total drop is \(\displaystyle 800t\), so the value left is \(\displaystyle 10000 - 800t\).

\(\displaystyle t\) (years)	$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 8$
\(\displaystyle v\) (in ₹)	$\displaystyle 10000$	$\displaystyle 9200$	$\displaystyle 8400$	$\displaystyle 7600$	$\displaystyle 6800$	$\displaystyle 6000$	$\displaystyle 5200$	$\displaystyle 4400$	$\displaystyle 3600$

Read the second row across: each entry is exactly $\displaystyle 800$ less than the one before it. The column \(\displaystyle t = 3\) again gives ₹$\displaystyle 7,600$, which checks part (i).(iii) The expression relating \(\displaystyle v\) and \(\displaystyle t\). \[v = 10000 - 800t \]Why this is linear decay: One sensible-model remark: \(\displaystyle v = 0\) when \(\displaystyle 800t = 10000\), that is \(\displaystyle t = 12.5\) years, so the formula only describes the phone up to about $\displaystyle 12$ and a half years — after that the value cannot keep falling below zero.Answer: (i) ₹$\displaystyle 7$,600. (iii) \(\displaystyle v = 10000 - 800t\), a degree-$\displaystyle 1$ polynomial in \(\displaystyle t\) with the negative constant rate \(\displaystyle -800\) per year, hence linear decay.

Constant yearly increase — linear growth.The same number of people, $\displaystyle 50$, is added every year. A fixed change per year makes the pattern linear.(i) Population after $\displaystyle 6$ years.In $\displaystyle 6$ years the total number who move in is \(\displaystyle 6 \times 50 = 300\). \[P = 750 + 300 = 1050 \] So after $\displaystyle 6$ years the village has $\displaystyle 1050$ people.(ii) Table of values.After \(\displaystyle t\) years the increase is \(\displaystyle 50t\), so the population is \(\displaystyle 750 + 50t\).

\(\displaystyle t\) (years)	$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 8$	$\displaystyle 9$	$\displaystyle 10$
\(\displaystyle P\) (people)	$\displaystyle 750$	$\displaystyle 800$	$\displaystyle 850$	$\displaystyle 900$	$\displaystyle 950$	$\displaystyle 1000$	$\displaystyle 1050$	$\displaystyle 1100$	$\displaystyle 1150$	$\displaystyle 1200$	$\displaystyle 1250$

Each entry in the second row is $\displaystyle 50$ more than the previous one, and the column \(\displaystyle t = 6\) reads $\displaystyle 1050$, which checks part (i).(iii) The expression relating \(\displaystyle P\) and \(\displaystyle t\). \[P = 750 + 50t \]Why this is linear growth: (This is a model of migration only. It assumes nobody is born, dies, or leaves — so like all such models it should not be pushed too far into the future.)Answer: (i) $\displaystyle 1050$ people. (iii) \(\displaystyle P = 750 + 50t\), a degree-$\displaystyle 1$ polynomial in \(\displaystyle t\) with the positive constant rate \(\displaystyle +50\) per year, hence linear growth.

Start value minus a constant daily amount.(i) The model.The balance starts at ₹$\displaystyle 600$ and loses ₹$\displaystyle 15$ on each day of use. After \(\displaystyle x\) days the total spent is \(\displaystyle 15x\), so \[b(x) = 600 - 15x \]Why this represents linear decay: (ii) When does the balance run out?The balance runs out when \(\displaystyle b(x) = 0\): \[600 - 15x = 0 \quad \Longrightarrow \quad 15x = 600 \quad \Longrightarrow \quad x = \frac{600}{15} = 40 \] Check: \(\displaystyle 15 \times 40 = 600\), so exactly the whole recharge is used up.So the balance runs out after $\displaystyle 40$ days. (For \(\displaystyle x > 40\) the formula would give a negative balance, which has no meaning here, so the model is valid for \(\displaystyle 0 \le x \le 40\).)(iii) Table of values.

\(\displaystyle x\) (days)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 8$	$\displaystyle 9$	$\displaystyle 10$
\(\displaystyle b(x)\) (in ₹)	$\displaystyle 585$	$\displaystyle 570$	$\displaystyle 555$	$\displaystyle 540$	$\displaystyle 525$	$\displaystyle 510$	$\displaystyle 495$	$\displaystyle 480$	$\displaystyle 465$	$\displaystyle 450$

Each value is $\displaystyle 15$ less than the one before it — the constant step that shows the decay is linear.Answer: (i) \(\displaystyle b(x) = 600 - 15x\); linear decay because it is degree $\displaystyle 1$ in \(\displaystyle x\) with the constant negative rate \(\displaystyle -15\) per day. (ii) After $\displaystyle 40$ days.