NCERT Solutions · Class 9 Mathematics Introduction to Linear Polynomials

Turn each of the three conditions into an equation in \(\displaystyle a, b, c, d\), then solve.We are given \(\displaystyle p(x) = ax + b\) and \(\displaystyle q(x) = cx + d\).Condition (i): \(\displaystyle p(0) = 5\). \[p(0) = a(0) + b = b, \qquad \text{so } b = 5 \]Condition (iii): \(\displaystyle p(x) + q(x) = 6x + 4\) for all real \(\displaystyle x\).Add the two polynomials by collecting like terms: \[p(x) + q(x) = (ax + b) + (cx + d) = (a + c)x + (b + d) \] Two polynomials that are equal for all \(\displaystyle x\) must have matching coefficients — the \(\displaystyle x\)-terms must agree and the constant terms must agree. Comparing with \(\displaystyle 6x + 4\): \[a + c = 6 \qquad \ldots (1) \] \[b + d = 4 \] Since \(\displaystyle b = 5\), the second of these gives \[d = 4 - 5 = -1 \]Condition (ii): \(\displaystyle p(x) - q(x)\) cuts the x-axis at \(\displaystyle (3, 0)\).Subtract: \[p(x) - q(x) = (a - c)x + (b - d) \] and \(\displaystyle b - d = 5 - (-1) = 6\), so \(\displaystyle p(x) - q(x) = (a - c)x + 6\).Cutting the x-axis at \(\displaystyle (3, 0)\) means the value of this polynomial is $\displaystyle 0$ when \(\displaystyle x = 3\): \[(a - c)(3) + 6 = 0 \quad \Longrightarrow \quad 3(a - c) = -6 \quad \Longrightarrow \quad a - c = -2 \qquad \ldots (2) \]Solve ($\displaystyle 1$) and ($\displaystyle 2$) together.Add them: \(\displaystyle (a + c) + (a - c) = 6 + (-2)\), so \(\displaystyle 2a = 4\) and \(\displaystyle a = 2\). Then from ($\displaystyle 1$), \(\displaystyle c = 6 - 2 = 4\).So \[p(x) = 2x + 5, \qquad q(x) = 4x - 1 \]Check all three conditions. Answer: \(\displaystyle p(x) = 2x + 5\) and \(\displaystyle q(x) = 4x - 1\).

Growing by the same amount each time.Read the building rule carefully first, because the whole question follows from it. Stage $\displaystyle 1$ is a single hexagon. A hexagon has $\displaystyle 6$ sides, so stage $\displaystyle 1$ uses \(\displaystyle 6\) matchsticks.At every later stage one more hexagon is joined on at the end, and the new hexagon shares a side with the hexagon already there. That shared side is a matchstick that has already been counted once. So the new hexagon does not cost $\displaystyle 6$ fresh matchsticks — it costs only its other \[6-1=5 \] sides. This is the key idea: every stage after the first needs exactly $\displaystyle 5$ more matchsticks than the stage before it.So \[\text{Stage }1=6,\qquad \text{Stage }2=6+5=11,\qquad \text{Stage }3=11+5=16 . \](i) The next two stages.To draw stage $\displaystyle 4$, copy the stage-$\displaystyle 3$ strip of three hexagons and attach a fourth hexagon at the free end so that it touches the third hexagon along exactly one matchstick. Stage $\displaystyle 5$ is the same strip with a fifth hexagon attached in the same way. (Keep the hexagons in a straight row, each one sharing exactly one side with its neighbour.)Adding $\displaystyle 5$ each time, \[\text{Stage }4=16+5=21,\qquad \text{Stage }5=21+5=26 . \]So stage $\displaystyle 4$ needs \(\displaystyle 21\) matchsticks and stage $\displaystyle 5$ needs \(\displaystyle 26\) matchsticks.(ii) The completed table.

Stage Number	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	...	\(\displaystyle n\)
Number of matchsticks	$\displaystyle 6$	$\displaystyle 11$	$\displaystyle 16$	$\displaystyle 21$	$\displaystyle 26$	...	\(\displaystyle 5n+1\)

(iii) The rule.Start at $\displaystyle 6$ matchsticks and add $\displaystyle 5$ once for every stage after the first. Going from stage $\displaystyle 1$ up to stage \(\displaystyle n\) we add $\displaystyle 5$ exactly \(\displaystyle n-1\) times, so the number of matchsticks is \[M(n)=6+5(n-1)=6+5n-5=5n+1 . \]Here is a second way to see the same rule, which explains where the \(\displaystyle +1\) comes from. Stand one single matchstick upright at the left end of the strip. Now build the hexagons one after another from left to right: each hexagon closes itself off using $\displaystyle 5$ new matchsticks, because its left-hand side is already in place. With \(\displaystyle n\) hexagons that is \(\displaystyle 5n\) matchsticks, plus the one we started with: \[M(n)=5n+1 . \]Checking against the counts we already know: \(\displaystyle M(1)=6\), \(\displaystyle M(2)=11\), \(\displaystyle M(3)=16\), \(\displaystyle M(4)=21\), \(\displaystyle M(5)=26\). All five agree, so the rule is right.Notice that \(\displaystyle M(n)=5n+1\) is a linear polynomial in \(\displaystyle n\) — that is exactly what a pattern with a constant increase of $\displaystyle 5$ per step looks like, and the $\displaystyle 5$ is the coefficient of \(\displaystyle n\).(iv) The 15th stage. \[M(15)=5\times 15+1=75+1=76 . \](v) Can $\displaystyle 200$ matchsticks form a stage?A stage uses $\displaystyle 200$ matchsticks only if \(\displaystyle M(n)=200\) for some whole number \(\displaystyle n\) (there is no such thing as stage \(\displaystyle 39.8\)). So solve \[5n+1=200 \;\Longrightarrow\; 5n=199 \;\Longrightarrow\; n=\frac{199}{5}=39.8 , \] which is not a whole number. Hence no stage of this pattern uses $\displaystyle 200$ matchsticks.A neater way to say the same thing: \(\displaystyle 5n+1\) always leaves remainder \(\displaystyle 1\) on division by \(\displaystyle 5\) (the counts run \(\displaystyle 6,11,16,21,26,\dots\), always ending in $\displaystyle 1$ or $\displaystyle 6$), whereas $\displaystyle 200$ is exactly divisible by 5. The pattern jumps straight past it: stage $\displaystyle 39$ uses \(\displaystyle 5\times 39+1=196\) matchsticks and stage $\displaystyle 40$ uses \(\displaystyle 5\times 40+1=201\).Stage $\displaystyle 4$ needs $\displaystyle 21$ matchsticks and stage $\displaystyle 5$ needs $\displaystyle 26$; the rule is \(\displaystyle M(n)=5n+1\); the 15th stage needs $\displaystyle 76$ matchsticks; and no, $\displaystyle 200$ matchsticks cannot form a stage, because \(\displaystyle 5n+1=200\) gives \(\displaystyle n=39.8\), which is not a whole number.

Two points fix a line; parallel lines share the same coefficient of \(\displaystyle x\).Step $\displaystyle 1$: turn "the graph passes through" into equations.Saying the graph of \(\displaystyle p\) passes through \(\displaystyle (2,3)\) means that when \(\displaystyle x=2\) the value of the polynomial is \(\displaystyle 3\), i.e. \(\displaystyle p(2)=3\). Similarly \(\displaystyle p(6)=11\). Substituting into \(\displaystyle p(x)=ax+b\): \[2a+b=3 \qquad \text{and} \qquad 6a+b=11 . \]Step $\displaystyle 2$: solve for \(\displaystyle a\) and \(\displaystyle b\).Both equations contain \(\displaystyle +b\), so subtracting the first from the second removes \(\displaystyle b\): \[(6a+b)-(2a+b)=11-3 \;\Longrightarrow\; 4a=8 \;\Longrightarrow\; a=2 . \] Putting \(\displaystyle a=2\) back into \(\displaystyle 2a+b=3\): \[4+b=3 \;\Longrightarrow\; b=-1 . \] So \[p(x)=2x-1 . \] Check both given points: \(\displaystyle p(2)=4-1=3\) and \(\displaystyle p(6)=12-1=11\). Both are correct.Step $\displaystyle 3$: use "parallel" to get \(\displaystyle c\).For a linear polynomial \(\displaystyle cx+d\), the coefficient \(\displaystyle c\) is what controls the steepness and direction of the line: increasing \(\displaystyle x\) by $\displaystyle 1$ changes the value by \(\displaystyle c\). Two lines are parallel exactly when they rise at the same rate, so they must have the same coefficient of \(\displaystyle x\). (If the coefficients were different, one line would gain on the other and they would have to cross.) Hence \[c=a=2 . \]Step $\displaystyle 4$: use the point \(\displaystyle (4,-1)\) to get \(\displaystyle d\).\(\displaystyle q(4)=-1\), and \(\displaystyle q(x)=2x+d\), so \[2(4)+d=-1 \;\Longrightarrow\; 8+d=-1 \;\Longrightarrow\; d=-9 . \] So \[q(x)=2x-9 . \] Check: \(\displaystyle q(4)=8-9=-1\), which is correct. Also \(\displaystyle d=-9\) is different from \(\displaystyle b=-1\), so \(\displaystyle p\) and \(\displaystyle q\) really are two different parallel lines and not the same line drawn twice.Step $\displaystyle 5$: where each line meets the \(\displaystyle x\)-axis.Every point of the \(\displaystyle x\)-axis has \(\displaystyle y\)-coordinate \(\displaystyle 0\). So the line meets the \(\displaystyle x\)-axis where the polynomial takes the value \(\displaystyle 0\) — that is, at its zero.For \(\displaystyle p\): \[2x-1=0 \;\Longrightarrow\; 2x=1 \;\Longrightarrow\; x=\tfrac12 , \] so the point is \(\displaystyle \left(\tfrac12,\,0\right)\).For \(\displaystyle q\): \[2x-9=0 \;\Longrightarrow\; 2x=9 \;\Longrightarrow\; x=\tfrac92 , \] so the point is \(\displaystyle \left(\tfrac92,\,0\right)\).These are two different points, as they must be: parallel lines can never meet the \(\displaystyle x\)-axis at the same place, otherwise they would meet each other there.\(\displaystyle p(x)=2x-1\) and \(\displaystyle q(x)=2x-9\). The graph of \(\displaystyle p(x)\) meets the \(\displaystyle x\)-axis at \(\displaystyle \left(\tfrac12,\,0\right)\) and the graph of \(\displaystyle q(x)\) meets it at \(\displaystyle \left(\tfrac92,\,0\right)\).

Take out the common factor \(\displaystyle a\).The two coefficients in \(\displaystyle f(x)=ax+a\) are the same number, so \(\displaystyle a\) can be factored out: \[f(x)=ax+a=a(x+1). \] Every function in this family is therefore just the single expression \(\displaystyle x+1\) multiplied by a positive number. That one observation explains everything they share.1. They all have the same zero, so all their graphs pass through the point \(\displaystyle (-1,0)\).Put \(\displaystyle x=-1\): \[f(-1)=a(-1+1)=a\times 0=0 . \] The value of \(\displaystyle a\) never enters, because it is multiplied by \(\displaystyle 0\). So \(\displaystyle x=-1\) is a zero of every function of this form, and every one of these lines cuts the \(\displaystyle x\)-axis at exactly the same point \(\displaystyle (-1,0)\). Picture them as a fan of lines all pivoting about that one point.For example, \(\displaystyle f(x)=x+1\) (with \(\displaystyle a=1\)), \(\displaystyle f(x)=3x+3\) (with \(\displaystyle a=3\)) and \(\displaystyle f(x)=\tfrac12 x+\tfrac12\) (with \(\displaystyle a=\tfrac12\)) are three different lines, but each one gives \(\displaystyle 0\) at \(\displaystyle x=-1\).2. They are all increasing (rising) lines.The coefficient of \(\displaystyle x\) is \(\displaystyle a\), and we are told \(\displaystyle a>0\). A positive coefficient of \(\displaystyle x\) means the value goes up by \(\displaystyle a\) each time \(\displaystyle x\) goes up by $\displaystyle 1$, so every graph in this family slopes upward from left to right.3. The value at \(\displaystyle x=0\) equals the coefficient of \(\displaystyle x\).\(\displaystyle f(0)=a\), and the coefficient of \(\displaystyle x\) is also \(\displaystyle a\). So each of these graphs crosses the \(\displaystyle y\)-axis at \(\displaystyle (0,a)\), at exactly the height given by its own steepness — a steeper line in this family is also the one that meets the \(\displaystyle y\)-axis higher up. This crossing is always above the origin, since \(\displaystyle a>0\).4. They all take positive values to the right of \(\displaystyle -1\) and negative values to the left.Since \(\displaystyle a>0\), the sign of \(\displaystyle a(x+1)\) is the sign of \(\displaystyle x+1\): so \(\displaystyle f(x)>0\) when \(\displaystyle x>-1\) and \(\displaystyle f(x)<0\) when \(\displaystyle x<-1\), for every member of the family. In particular \(\displaystyle f(x)>0\) for all \(\displaystyle x>0\), so none of these graphs ever enters the fourth quadrant.The single most important shared feature is the first one: the common zero.Every function \(\displaystyle f(x)=ax+a=a(x+1)\) with \(\displaystyle a>0\) has \(\displaystyle -1\) as its zero, so all of their graphs are straight lines passing through the one common point \(\displaystyle (-1,0)\); and because \(\displaystyle a>0\), all of them are increasing lines that cross the \(\displaystyle y\)-axis at \(\displaystyle (0,a)\) — the height there being equal to the line's own coefficient of \(\displaystyle x\).