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$5x^3), 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 $x^2+1$x^2+1 is quadratic, $5$5 is a constant (degree $0$0), and $2x^3-x$2x^3-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=-54, 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.

Multiply both sides by the LCM $6$6:

$2(2x+1) - 3(x-1) = 12.$2(2x+1) - 3(x-1) = 12.

Expand: $4x + 2 - 3x + 3 = 12 \Rightarrow x + 5 = 12 \Rightarrow x = 7$4x + 2 - 3x + 3 = 12 x + 5 = 12 x = 7.

Verification: $\dfrac{2(7)+1}{3} - \dfrac{7-1}{2} = \dfrac{15}{3} - \dfrac{6}{2} = 5 - 3 = 2.$2(7)+1/3 - 7-1/2 = 15/3 - 6/2 = 5 - 3 = 2. ✓

(a) $\sqrt{x} + 3 = x^{1/2} + 3$√x + 3 = x^1/2 + 3: the power $\tfrac12$12 is not a whole number, so this is not a polynomial.

(b) $2x^2 - 5x + 1$2x^2 - 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.

Let the smaller number be $x$x. Then the larger number is $x + 15$x + 15.

Their sum is $95$95:

$x + (x + 15) = 95.$x + (x + 15) = 95.

This is a linear equation in one variable, of degree $1$1.

Solve: $2x + 15 = 95 \Rightarrow 2x = 80 \Rightarrow x = 40$2x + 15 = 95 2x = 80 x = 40.

So the smaller number is $40$40 and the larger number is $40 + 15 = 55$40 + 15 = 55.

Check: $40 + 55 = 95$40 + 55 = 95 and $55 - 40 = 15$55 - 40 = 15. ✓

Let the son's present age be $x$x years. Then the father's present age is $3x$3x years.

After $12$12 years: son's age $= x + 12$= x + 12, father's age $= 3x + 12$= 3x + 12.

The father will then be twice as old as the son:

$3x + 12 = 2(x + 12).$3x + 12 = 2(x + 12).

Expand: $3x + 12 = 2x + 24$3x + 12 = 2x + 24.

Solve: $3x - 2x = 24 - 12 \Rightarrow x = 12$3x - 2x = 24 - 12 x = 12.

So the son is $12$12 years old and the father is $3\times 12 = 36$3× 12 = 36 years old.

Check: After $12$12 years, son $=24$=24, father $=48$=48, and $48 = 2\times 24$48 = 2× 24. ✓

(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: $C = 2(15) + 100 = 30 + 100 = \overline{\rm Rs}\,130$C = 2(15) + 100 = 30 + 100 = Rs\,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) The zero of $2x - 40$2x - 40 is $x = 20$x = 20. It is simply the value of $x$x that makes that expression zero; it represents the data usage ($20$20 GB) at which the quantity $2x-40$2x-40 vanishes.