Introduction to Linear Polynomials — CBSE Class 9 Maths Important Questions

Answer: (c) $3$3.

The degree is the highest power of the variable present. Here the largest exponent is $3$3 (in the term $5x^3$5x³), so the degree is $3$3.

Answer: (b) $3x - 7$3x - 7.

A linear polynomial has degree exactly $1$1. Only $3x-7$3x-7 has highest power $1$1. Here $x^2+1$x²+1 is quadratic, $5$5 is a constant (degree $0$0), and $2x^3-x$2x³-x is cubic.

Answer: (b) $-3$-3.

Set $p(x)=0$p(x)=0: $2x+6=0 \Rightarrow 2x=-6 \Rightarrow x=-3$2x+6=0 ⇒ 2x=-6 ⇒ x=-3.

Answer: (a) $0$0.

$p(1) = 1 - 3 + 2 = 0$p(1) = 1 - 3 + 2 = 0 and $p(2) = 4 - 6 + 2 = 0$p(2) = 4 - 6 + 2 = 0. Hence $p(1)+p(2) = 0+0 = 0$p(1)+p(2) = 0+0 = 0.

Answer: (a) Both A and R are true and R is the correct explanation of A. Since $4x+5$4x+5 is linear with $a=4\neq0$a=4≠0, it has a single zero $x=-\tfrac{5}{4}$x=-5/4, exactly as the general rule states.

$p(2) = 3(2) - 4 = 6 - 4 = 2$p(2) = 3(2) - 4 = 6 - 4 = 2.

$p(-1) = 3(-1) - 4 = -3 - 4 = -7$p(-1) = 3(-1) - 4 = -3 - 4 = -7.

Rewriting in standard form: $p(x) = -5x + 7$p(x) = -5x + 7.

Coefficient of $x$x is $-5$-5; constant term is $7$7. The highest power of $x$x is $1$1, so the degree is $1$1 (it is a linear polynomial).

Since $x=2$x=2 is a zero, $p(2)=0$p(2)=0.

$p(2) = k(2) + 3(2) - 10 = 2k + 6 - 10 = 2k - 4$p(2) = k(2) + 3(2) - 10 = 2k + 6 - 10 = 2k - 4.

Setting $p(2)=0$p(2)=0: $2k - 4 = 0 \Rightarrow 2k = 4 \Rightarrow k = 2$2k - 4 = 0 ⇒ 2k = 4 ⇒ k = 2.

So $k = 2$k = 2.

(i) Comparing with $y=ax+b$y=ax+b: slope $a=-3$a=-3 and $y$y-intercept $b=6$b=6. The negative slope means $y$y falls by $3$3 for every increase of $1$1 in $x$x.

(ii) $y=-3(-1)+6=3+6=9$y=-3(-1)+6=3+6=9.

(iii) On the $x$x-axis $y=0$y=0: $-3x+6=0\Rightarrow x=2$-3x+6=0⇒ x=2. The line crosses the $x$x-axis at $(2,0)$(2,0).

(a) $\sqrt{x} + 3 = x^{1/2} + 3$√x + 3 = x^1/2 + 3: the power $\tfrac12$1/2 is not a whole number, so this is not a polynomial.

(b) $2x^2 - 5x + 1$2x² - 5x + 1: all powers are whole numbers; highest power $2$2, so a quadratic polynomial (degree $2$2).

(c) $x + \dfrac{3}{x} = x + 3x^{-1}$x + 3/x = x + 3x^-1: the power $-1$-1 is negative, so this is not a polynomial.

(d) $-7$-7: a non-zero constant, a constant polynomial of degree $0$0.

Plot the points and join them with a straight line:

$y$y-intercept: the line meets the $y$y-axis at $(0,1)$(0,1), so the $y$y-intercept is $1$1.

Slope: for every increase of $1$1 in $x$x, $y$y increases by $2$2, so the slope is $2$2 (the coefficient of $x$x).

Point $(4,9)$(4,9): $2(4)+1=9$2(4)+1=9, so $(4,9)$(4,9) lies on the line.

(i) $h=20-2.5t$h=20-2.5t.

(ii) Linear decay: the height falls by the same amount, $2.5$2.5 cm, every hour (the coefficient of $t$t is negative).

(iii) $h=20-2.5(3)=20-7.5=12.5$h=20-2.5(3)=20-7.5=12.5 cm.

(iv) $20-2.5t=0\Rightarrow t=8$20-2.5t=0⇒ t=8 hours.

(v) The slope $-2.5$-2.5 is the rate of burning, $2.5$2.5 cm lost per hour; the intercept $20$20 is the starting height at $t=0$t=0.

(i) Fixed rental $100$100 plus $2$2 per GB gives $C = 2x + 100$C = 2x + 100. The highest power of $x$x is $1$1, so it is a linear polynomial of degree $1$1.

(ii) At $x = 15$x = 15: $C = 2(15) + 100 = 30 + 100 = 130$C = 2(15) + 100 = 30 + 100 = 130, that is ₹$130$130.

(iii) Set $C = 150$C = 150: $2x + 100 = 150 \Rightarrow 2x = 50 \Rightarrow x = 25$2x + 100 = 150 ⇒ 2x = 50 ⇒ x = 25 GB.

(iv) $2$2 is the slope: the cost goes up by ₹$2$2 for every extra GB. $100$100 is the intercept: the fixed rental, the cost when no data is used ($x=0$x=0).