Introduction to Linear PolynomialsClass 9 Mathematics NCERT Solutions

To Do: Write a polynomial with specific properties.

Solution: A polynomial of degree 3 must have its highest power of $x$ as 3. The coefficient of the $x^2$ term must be -7. The other coefficients can be any real numbers (as long as the coefficient of $x^3$ is not zero).

An example of such a polynomial is: $p(x) = 4x^3 - 7x^2 + 5x - 1$

Here, the degree is 3, the variable is $x$, and the coefficient of $x^2$ is -7.

Final Answer: An example is $4x^3 - 7x^2 + 5x - 1$. (Other answers are possible).

To Find: The value of each polynomial for the given value of the variable.

(i) $p(x) = 5 x^{2}-3 x+7$ if $x=1$ Solution: Substitute $x=1$ into the polynomial: $p(1) = 5(1)^2 - 3(1) + 7$ $p(1) = 5(1) - 3 + 7$ $p(1) = 5 - 3 + 7 = 9$ Final Answer: The value is 9.

(ii) $p(t) = 4 t^{3}-t^{2}+6$ if $t=\mathrm{a}$ Solution: Substitute $t=a$ into the polynomial: $p(a) = 4(a)^3 - (a)^2 + 6$ $p(a) = 4a^3 - a^2 + 6$ Final Answer: The value is $4a^3 - a^2 + 6$.

Given: A sequence of operations on an unknown number results in $\frac{-7}{12}$.

To Find: The unknown number.

Let: The number be $x$.

Equation: According to the problem: $(\text{multiply by } \frac{5}{2}) \rightarrow \frac{5}{2}x$ $(\text{add } \frac{2}{3}) \rightarrow \frac{5}{2}x + \frac{2}{3}$ This result is equal to $\frac{-7}{12}$. $\frac{5}{2}x + \frac{2}{3} = -\frac{7}{12}$

Solution: To solve for $x$, first isolate the term with $x$: $\frac{5}{2}x = -\frac{7}{12} - \frac{2}{3}$ To subtract the fractions, find a common denominator, which is 12. $\frac{5}{2}x = -\frac{7}{12} - \frac{2 \times 4}{3 \times 4}$ $\frac{5}{2}x = -\frac{7}{12} - \frac{8}{12}$ $\frac{5}{2}x = -\frac{15}{12}$ Now, multiply both sides by $\frac{2}{5}$ to solve for $x$: $x = -\frac{15}{12} \times \frac{2}{5}$ $x = -\frac{15 \times 2}{12 \times 5} = -\frac{30}{60}$ $x = -\frac{1}{2}$

Final Answer: The number is $-\frac{1}{2}$.

To Find: The original numbers.

Let: The smaller number be $x$. Since it is a positive number, $x>0$. The larger number is $5x$.

After adding 21 to both numbers: The new smaller number is $x+21$. The new larger number is $5x+21$.

Equation: The new larger number is twice the new smaller number. $5x + 21 = 2(x + 21)$

Solution: $5x + 21 = 2x + 42$ $5x - 2x = 42 - 21$ $3x = 21$ $x = \frac{21}{3}$ $x = 7$

The original numbers are: Smaller number: $x = 7$. Larger number: $5x = 5(7) = 35$.

Verification: The numbers are 7 and 35. After adding 21, they become $7+21=28$ and $35+21=56$. We can see that $56 = 2 \times 28$, which is correct.

Final Answer: The numbers are 7 and 35.

Linear Pattern: Let $A(n)$ be the total amount after $n$ months. The total amount is the initial amount plus the total savings. Total savings after $n$ months = $250n$. So, the linear pattern is: $A(n) = 800 + 250n$

(i) Amount after 6 months Solution: Here, $n=6$. $A(6) = 800 + 250(6)$ $A(6) = 800 + 1500$ $A(6) = 2300$ Final Answer (i): After 6 months, you will have ₹2300.

(ii) Amount after 2 years Solution: First, convert 2 years to months: $2 \text{ years} = 2 \times 12 = 24$ months. So, $n=24$. $A(24) = 800 + 250(24)$ $A(24) = 800 + 6000$ $A(24) = 6800$ Final Answer (ii): After 2 years, you will have ₹6800.

Final Answer (Linear Pattern): The linear pattern is $A(n) = 800 + 250n$, where $n$ is the number of months.

To Find: The original number(s).

Let: The digit in the tens place be $x$ and the digit in the units place be $y$. The original number can be written as $10x + y$. The number with interchanged digits is $10y + x$.

Equations: From the given information, we can form two equations:

Solution: First, simplify the second equation: $11x + 11y = 143$ Divide by 11: $x + y = 13 \quad (*)$

Now we have two cases based on the first equation.

Case 1: $x - y = 3$ We have a system of two equations: (a) $x + y = 13$ (b) $x - y = 3$ Adding (a) and (b): $2x = 16 \implies x = 8$ Substituting $x=8$ into (a): $8 + y = 13 \implies y = 5$ The digits are 8 and 5. The original number is $10(8) + 5 = 85$. The interchanged number is 58. Their sum is $85+58=143$. The difference of digits is $8-5=3$. This is a valid solution.

Case 2: $y - x = 3$ We have a system of two equations: (a) $x + y = 13$ (c) $y - x = 3$ or $-x+y=3$ Adding (a) and (c): $2y = 16 \implies y = 8$ Substituting $y=8$ into (a): $x + 8 = 13 \implies x = 5$ The digits are 5 and 8. The original number is $10(5) + 8 = 58$. The interchanged number is 85. Their sum is $58+85=143$. The difference of digits is $8-5=3$. This is also a valid solution.

Final Answer: The two possible numbers are 85 and 58.

To Do: For each equation, find the slope, y-intercept, and the point of intersection with the y-axis. Determine if any lines are parallel.

The standard form of a linear equation is $y = ax + b$, where 'a' is the slope and 'b' is the y-intercept. The line cuts the y-axis at the point $(0, b)$.

Parallel Lines: Lines are parallel if they have the same slope.

Yes, the lines from equations (ii) and (iv) are parallel.

Given: The linear equation relating Kelvin ($x$) and Fahrenheit ($y$) is $y=\frac{9}{5}(x-273)+32$.

(i) Find temperature in Fahrenheit for 313 K. Given: $x = 313$ K. To Find: The value of $y$.

Solution: Substitute $x=313$ into the equation: $y = \frac{9}{5}(313 - 273) + 32$ $y = \frac{9}{5}(40) + 32$ $y = 9 \times 8 + 32$ $y = 72 + 32$ $y = 104$

Final Answer (i): The temperature is 104 °F.

(ii) Find temperature in Kelvin for 158 °F. Given: $y = 158$ °F. To Find: The value of $x$.

Solution: Substitute $y=158$ into the equation: $158 = \frac{9}{5}(x - 273) + 32$ First, subtract 32 from both sides: $158 - 32 = \frac{9}{5}(x - 273)$ $126 = \frac{9}{5}(x - 273)$ Now, multiply both sides by $\frac{5}{9}$: $126 \times \frac{5}{9} = x - 273$ $14 \times 5 = x - 273$ $70 = x - 273$ Now, add 273 to both sides: $x = 70 + 273$ $x = 343$

Final Answer (ii): The temperature is 343 K.

Linear Equation: Substituting $F=3$ into the formula $w = F \times d$, we get the linear equation: $w = 3d$ This is a linear equation in two variables, $w$ and $d$.

Work Done for distance = 2 units: Substitute $d=2$ into the equation: $w = 3(2)$ $w = 6$ Final Answer (Calculation): The work done is 6 units.

Graph and Verification: To describe the graph of $w=3d$, we can find some points. Let the d-axis be the horizontal axis and the w-axis be the vertical axis.

The graph is a straight line passing through the origin (0, 0) with a slope of 3.

To verify the work done when the distance is 2 units, we would plot the line $w=3d$. Then we would find the point on the line where the horizontal coordinate (distance) is 2. The vertical coordinate (work) of that point would be 6. This confirms that the point (2, 6) lies on the graph, verifying that for a distance of 2 units, the work done is 6 units.

Given: A linear polynomial $p(x)$ passes through points (1, 5) and (3, 11).

(i) Find the polynomial $p(x)$. Let: The linear polynomial be $p(x) = ax + b$.

Equations: Since the graph passes through the given points, they must satisfy the equation.

Solution: Subtract equation (1) from equation (2): $(3a + b) - (a + b) = 11 - 5$ $2a = 6$ $a = 3$ Substitute $a=3$ into equation (1): $3 + b = 5$ $b = 2$ So, the polynomial is $p(x) = 3x + 2$.

Final Answer (i): The polynomial is $p(x) = 3x + 2$.

(ii) Find the coordinates where the graph of $p(x)$ cuts the axes. Solution:

Final Answer (ii): The graph cuts the y-axis at (0, 2) and the x-axis at $(-\frac{2}{3}, 0)$.

(iii) Draw the graph of $p(x)$ and verify your answers. Description of Graph and Verification: The graph of $p(x) = 3x + 2$ is a straight line. To draw it, we can use the points we already have:

Plotting these points on a coordinate plane and drawing a line through them would create the graph. We can verify that all these points lie on the same straight line. For example, substituting the coordinates of the given points into our found equation:

To Find: The polynomials $p(x)$ and $q(x)$.

Step 1: Use condition (i) $p(0) = a(0) + b = 5 \implies b=5$. So, $p(x) = ax + 5$.

Step 2: Use condition (iii) $p(x) + q(x) = (ax + 5) + (cx + d) = 6x + 4$ Combine like terms: $(a+c)x + (5+d) = 6x + 4$ By comparing coefficients:

Step 3: Use condition (ii) Let $r(x) = p(x) - q(x)$. $r(x) = (ax+5) - (cx-1) = ax + 5 - cx + 1 = (a-c)x + 6$. The graph of $r(x)$ cuts the x-axis at (3, 0), which means $r(3)=0$. $r(3) = (a-c)(3) + 6 = 0$ $3(a-c) = -6$ $a-c = -2 \quad (B)$

Step 4: Solve for $a$ and $c$ We have a system of two linear equations: (A) $a+c = 6$ (B) $a-c = -2$ Adding (A) and (B): $2a = 4 \implies a=2$. Substitute $a=2$ into (A): $2+c = 6 \implies c=4$.

Step 5: Write the final polynomials We found $a=2, b=5, c=4, d=-1$. $p(x) = ax + b = 2x + 5$ $q(x) = cx + d = 4x - 1$

Final Answer: The polynomials are $p(x) = 2x + 5$ and $q(x) = 4x - 1$.

(ii) Complete the table Based on the pattern identified:

(iii) Find a rule for the $n^{\text{th}}$ stage Let $M(n)$ be the number of matchsticks at stage $n$. The number of matchsticks increases by a constant difference of 5. This is a linear pattern. The first term is 6. The number of matchsticks for stage $n$ can be expressed as: $M(n) = \text{first term} + (\text{common difference}) \times (n-1)$ $M(n) = 6 + 5(n-1)$ $M(n) = 6 + 5n - 5$ $M(n) = 5n + 1$ Final Answer (iii): The rule is $M(n) = 5n + 1$.

(iv) Matchsticks for the 15th stage Using the rule $M(n) = 5n+1$ with $n=15$: $M(15) = 5(15) + 1 = 75 + 1 = 76$. Final Answer (iv): 76 matchsticks will be required for the 15th stage.

(v) Can 200 matchsticks form a stage? We need to check if there is an integer $n$ for which $M(n)=200$. $5n + 1 = 200$ $5n = 199$ $n = \frac{199}{5} = 39.8$ Since $n$ must be a positive integer (representing the stage number), and 39.8 is not an integer, 200 matchsticks cannot form a complete stage in this pattern. Final Answer (v): No, because solving $5n+1=200$ for $n$ gives $n=39.8$, which is not an integer.

To Find: The polynomials $p(x)$ and $q(x)$, and their x-intercepts.

Step 1: Find the polynomial $p(x) = ax+b$ Given it passes through (2, 3) and (6, 11).

Step 2: Find the polynomial $q(x) = cx+d$ Given that the graph of $q(x)$ is parallel to the graph of $p(x)$. Parallel lines have the same slope. The slope of $p(x)$ is $a=2$. Therefore, the slope of $q(x)$ is $c=2$. So, $q(x) = 2x + d$.

Given that $q(x)$ passes through the point (4, -1). $q(4) = -1 \implies 2(4) + d = -1$ $8 + d = -1 \implies d = -9$. So, the polynomial is $q(x) = 2x - 9$.

Step 3: Find the x-intercepts The x-intercept is the point where the graph meets the x-axis, i.e., where the value of the polynomial is 0.

Given: Linear functions of the form $f(x) = ax + a$, with the condition $a > 0$.

To Find: The common property of all such functions.

Analysis: Let's analyze the properties of the function $f(x) = ax + a$.

This means that for any value of $a > 0$, the function will be zero when $x = -1$. In other words, the graph of every function of the form $f(x) = ax + a$ passes through the point $(-1, 0)$.

Final Answer: All linear functions of the form $f(x) = ax + a$ (where $a>0$) pass through the same point on the x-axis, which is $(-1, 0)$. This point is their common x-intercept.

To Find: The degree of each polynomial.

The degree of a polynomial is the highest power of the variable in the polynomial.

(i) $2 x^{2}-5 x+3$ Solution: The terms are $2x^2$, $-5x^1$, and $3x^0$. The powers of the variable $x$ are 2, 1, and 0. The highest power is 2. Final Answer: The degree of the polynomial $2 x^{2}-5 x+3$ is 2.

(ii) $y^{3}+2 y-1$ Solution: The terms are $y^3$, $2y^1$, and $-1y^0$. The powers of the variable $y$ are 3, 1, and 0. The highest power is 3. Final Answer: The degree of the polynomial $y^{3}+2 y-1$ is 3.

(iii) -9 Solution: The polynomial -9 is a constant polynomial. It can be written as $-9x^0$. The power of the variable $x$ is 0. Final Answer: The degree of the polynomial -9 is 0.

(iv) $4 \mathrm{z}-3$ Solution: The terms are $4z^1$ and $-3z^0$. The powers of the variable $z$ are 1 and 0. The highest power is 1. Final Answer: The degree of the polynomial $4 \mathrm{z}-3$ is 1.

To Do: Write one example each for polynomials of degree 1, 2, and 3.

Given: The polynomial $p(x) = x^{4}-3 x^{3}+6 x^{2}-2 x+7$.

To Find: The coefficients of the terms $x^2$ and $x^3$.

Solution: The coefficient is the numerical factor of a term.

Final Answer: The coefficient of $x^3$ is -3. The coefficient of $x^2$ is 6.

Given: The polynomial $p(z) = 4 z^{3}+5 z^{2}-11$.

To Find: The coefficient of the term $z$.

Solution: The given polynomial can be written by including the term with variable $z$ (which is $z^1$) as: $p(z) = 4z^3 + 5z^2 + 0z - 11$ The term containing $z$ is $0z$. The numerical factor of this term is 0.

Final Answer: The coefficient of $z$ is 0.

Given: The polynomial $p(x) = 9 x^{3}+5 x^{2}-8 x-10$.

To Find: The constant term of the polynomial.

Solution: The constant term is the term that does not contain any variable. In the given polynomial, the term -10 is the constant term.

Final Answer: The constant term is -10.

Given: The linear polynomial $p(x) = 5x - 3$.

To Find: The value of the polynomial at different values of $x$.

(i) $x=0$ Solution: Substitute $x=0$ into the polynomial: $p(0) = 5(0) - 3$ $p(0) = 0 - 3 = -3$ Final Answer: The value is -3.

(ii) $x=-1$ Solution: Substitute $x=-1$ into the polynomial: $p(-1) = 5(-1) - 3$ $p(-1) = -5 - 3 = -8$ Final Answer: The value is -8.

(iii) $x=2$ Solution: Substitute $x=2$ into the polynomial: $p(2) = 5(2) - 3$ $p(2) = 10 - 3 = 7$ Final Answer: The value is 7.

Given: The quadratic polynomial $p(s) = 7s^2 - 4s + 6$.

To Find: The value of the polynomial at different values of $s$.

(i) $s=0$ Solution: Substitute $s=0$ into the polynomial: $p(0) = 7(0)^2 - 4(0) + 6$ $p(0) = 0 - 0 + 6 = 6$ Final Answer: The value is 6.

(ii) $s=-3$ Solution: Substitute $s=-3$ into the polynomial: $p(-3) = 7(-3)^2 - 4(-3) + 6$ $p(-3) = 7(9) + 12 + 6$ $p(-3) = 63 + 12 + 6 = 81$ Final Answer: The value is 81.

(iii) $s=4$ Solution: Substitute $s=4$ into the polynomial: $p(4) = 7(4)^2 - 4(4) + 6$ $p(4) = 7(16) - 16 + 6$ $p(4) = 112 - 16 + 6 = 102$ Final Answer: The value is 102.

To Find: Their present ages.

Let: Salil's present age be $x$ years. Then, Salil's mother's present age is $3x$ years.

After 5 years: Salil's age will be $(x+5)$ years. Salil's mother's age will be $(3x+5)$ years.

Equation: According to the problem, the sum of their ages after 5 years is 70. $(x+5) + (3x+5) = 70$

Solution: $4x + 10 = 70$ $4x = 70 - 10$ $4x = 60$ $x = \frac{60}{4}$ $x = 15$

So, Salil's present age is $x=15$ years. Salil's mother's present age is $3x = 3(15) = 45$ years.

Final Answer: Salil's present age is 15 years, and his mother's present age is 45 years.

To Find: The two integers.

Let: The two integers be $2x$ and $5x$, where $x$ is a positive constant. Since the integers are positive, $5x > 2x$.

Equation: The difference between the two integers is 63. $5x - 2x = 63$

Solution: $3x = 63$ $x = \frac{63}{3}$ $x = 21$

Now, we find the integers: The first integer is $2x = 2(21) = 42$. The second integer is $5x = 5(21) = 105$.

Verification: The difference is $105 - 42 = 63$. The ratio is $42:105 = \frac{42}{105} = \frac{2 \times 21}{5 \times 21} = \frac{2}{5}$, which is 2:5.

Final Answer: The two integers are 42 and 105.

To Find: The number of coins of each type.

Let: The number of five-rupee coins be $x$. Then, the number of two-rupee coins is $3x$.

Equation: The total value of the coins is ₹88. Value of five-rupee coins + Value of two-rupee coins = Total value $(5 \times x) + (2 \times 3x) = 88$

Solution: $5x + 6x = 88$ $11x = 88$ $x = \frac{88}{11}$ $x = 8$

So, the number of five-rupee coins is $x = 8$. The number of two-rupee coins is $3x = 3(8) = 24$.

Final Answer: Ruby has 8 five-rupee coins and 24 two-rupee coins.

To Find: The lengths of the two pieces.

Let: The length of the shorter piece be $x$ feet. Then, the length of the longer piece is $4x$ feet.

Equation: The sum of the lengths of the two pieces is the total length of the fence. $x + 4x = 300$

Solution: $5x = 300$ $x = \frac{300}{5}$ $x = 60$

So, the length of the shorter piece is $x = 60$ feet. The length of the longer piece is $4x = 4(60) = 240$ feet.

Final Answer: The two pieces are 60 feet and 240 feet long.

To Find: The dimensions (length and width) of the rectangle.

Let: The width of the rectangle be $w$ cm. Then, the length of the rectangle is $l = (2w + 3)$ cm.

Formula: The perimeter of a rectangle is given by $P = 2(l + w)$.

Equation: $24 = 2((2w+3) + w)$

Solution: $24 = 2(3w + 3)$ Divide both sides by 2: $12 = 3w + 3$ $12 - 3 = 3w$ $9 = 3w$ $w = \frac{9}{3}$ $w = 3$

So, the width of the rectangle is $w = 3$ cm. The length of the rectangle is $l = 2w + 3 = 2(3) + 3 = 6 + 3 = 9$ cm.

Final Answer: The dimensions of the rectangle are length = 9 cm and width = 3 cm.

Solution: Let $A(n)$ be the amount of money at the end of the $n^{\text{th}}$ month. We assume $n=0$ is the initial state.

From the second month onwards, the amounts will be:

Linear Expression: The amount of money $A$ after $n$ months can be represented by the linear expression: $A(n) = 500 + 150n$ where $n$ is the number of months passed.

Final Answer: The linear expression to represent the amount she will have in the $n^{\text{th}}$ month is $500 + 150n$. The amounts from the second month onwards are ₹800 (2nd month), ₹950 (3rd month), etc.

Solution: Let $M(n)$ be the number of members remaining at the end of the $n^{\text{th}}$ hour.