Polynomials — CBSE Class 10 Maths Important Questions

Answer: (c) 2 — The graph meets the $x$x-axis at exactly two points, so $p(x)$p(x) has $2$2 real zeroes.

Answer: (a) $x^2+x-6$x²+x-6 — Sum $=2+(-3)=-1$=2+(-3)=-1 and product $=2\times(-3)=-6$=2×(-3)=-6. So $p(x)=x^2-(\text{sum})x+\text{product}=x^2-(-1)x+(-6)=x^2+x-6$p(x)=x²-(sum)x+product=x²-(-1)x+(-6)=x²+x-6.

Answer: (a) $\dfrac{3}{4}$3/4 — Here $\alpha+\beta=6$α+β=6 and $\alpha\beta=8$α β=8, so $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{6}{8}=\dfrac{3}{4}$1/α+1/β=(α+β)/(α β)=6/8=3/4.

Answer: (d) $-\dfrac{2}{3}$-2/3 — Sum $=-\dfrac{2}{k}$=-2/k and product $=\dfrac{3k}{k}=3$=3k/k=3. Setting them equal: $-\dfrac{2}{k}=3\Rightarrow k=-\dfrac{2}{3}$-2/k=3⇒ k=-2/3.

Answer: (a) Both A and R are true and R correctly explains A. For $x^2+4$x²+4, $\ b^2-4ac=0-4(1)(4)=-16<0$b²-4ac=0-4(1)(4)=-16<0, so its graph never meets the $x$x-axis and it has no real zeroes.

Let the zeroes be $\alpha$α and $\dfrac{1}{\alpha}$1/α. Their product is $\alpha\times\dfrac{1}{\alpha}=1$α × 1/α=1.

But the product of the zeroes is $\dfrac{c}{a}=\dfrac{k}{3}$c/a=k/3. So $\dfrac{k}{3}=1\Rightarrow k=3$k/3=1⇒ k=3.

Then $3x^2-10x+3=3x^2-9x-x+3=3x(x-3)-1(x-3)=(3x-1)(x-3)$3x²-10x+3=3x²-9x-x+3=3x(x-3)-1(x-3)=(3x-1)(x-3), so the zeroes are $3$3 and $\dfrac{1}{3}$1/3. (Check: $3\times\dfrac{1}{3}=1$3× 1/3=1 ✓.)

The required polynomial is $k\left(x^2-(\text{sum})x+\text{product}\right)=k\left(x^2-\sqrt2\,x+\dfrac{1}{3}\right)$k(x²-(sum)x+product)=k(x²-√2 x+1/3). Taking $k=3$k=3 to clear the fraction gives $3x^2-3\sqrt2\,x+1$3x²-3√2 x+1.

Splitting the middle term: $6x^2-7x-3=6x^2-9x+2x-3=3x(2x-3)+1(2x-3)=(2x-3)(3x+1)$6x²-7x-3=6x²-9x+2x-3=3x(2x-3)+1(2x-3)=(2x-3)(3x+1).

Zeroes: $x=\dfrac{3}{2}$x=3/2 and $x=-\dfrac{1}{3}$x=-1/3.

Verification: Sum $=\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{7}{6}=-\dfrac{(-7)}{6}=-\dfrac{b}{a}$=3/2-1/3=7/6=-(-7)/6=-b/a ✓; product $=\dfrac{3}{2}\times\left(-\dfrac{1}{3}\right)=-\dfrac{1}{2}=\dfrac{-3}{6}=\dfrac{c}{a}$=3/2× (-1/3)=-1/2=-3/6=c/a ✓.

Rewrite the polynomial: $x^2-p(x+1)-c=x^2-px-(p+c)$x²-p(x+1)-c=x²-px-(p+c). Hence $\alpha+\beta=p$α+β=p and $\alpha\beta=-(p+c)$α β=-(p+c).

Now $(\alpha+1)(\beta+1)=\alpha\beta+(\alpha+\beta)+1=-(p+c)+p+1=1-c.$(α+1)(β+1)=α β+(α+β)+1=-(p+c)+p+1=1-c. Hence proved.

From the polynomial, $\alpha+\beta=5$α+β=5; and it is given that $\alpha-\beta=1$α-β=1.

Adding the two: $2\alpha=6\Rightarrow\alpha=3$2α=6⇒ α=3, so $\beta=2$β=2.

Then $k=\alpha\beta=3\times2=6$k=α β=3×2=6.

Table of values:

Plot these points and join them with a smooth curve. The graph cuts the $x$x-axis at $(-1,0)$(-1,0) and $(3,0)$(3,0), so the zeroes of $p(x)$p(x) are $-1$-1 and $3$3.

(i) $x^2-2x-3=x^2-3x+x-3=x(x-3)+1(x-3)=(x-3)(x+1)$x²-2x-3=x²-3x+x-3=x(x-3)+1(x-3)=(x-3)(x+1), which gives the zeroes $3$3 and $-1$-1 ✓.

(ii) Sum of zeroes $=-1+3=2=-\dfrac{(-2)}{1}=-\dfrac{b}{a}$=-1+3=2=-(-2)/1=-b/a ✓; product of zeroes $=(-1)(3)=-3=\dfrac{-3}{1}=\dfrac{c}{a}$=(-1)(3)=-3=-3/1=c/a ✓.

From the coefficients, $\alpha+\beta=-\dfrac{(-5)}{2}=\dfrac{5}{2}$α+β=-(-5)/2=5/2 and $\alpha\beta=\dfrac{-3}{2}$α β=-3/2.

(i) $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\dfrac{25}{4}+3=\dfrac{37}{4}$α²+β²=(α+β)²-2α β=25/4+3=37/4.

(ii) $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=\dfrac{125}{8}-3\left(-\dfrac{3}{2}\right)\left(\dfrac{5}{2}\right)=\dfrac{125}{8}+\dfrac{45}{4}=\dfrac{215}{8}$α³+β³=(α+β)³-3α β(α+β)=125/8-3(-3/2)(5/2)=125/8+45/4=215/8.

(iii) Sum of the new zeroes $=\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}=\dfrac{\alpha^2+\beta^2}{\alpha\beta}=\dfrac{37/4}{-3/2}=-\dfrac{37}{6}$=α/β+β/α=(α²+β²)/(α β)=37/4/-3/2=-37/6; product $=\dfrac{\alpha}{\beta}\cdot\dfrac{\beta}{\alpha}=1$=α/β· β/α=1.

So the polynomial is $k\left(x^2+\dfrac{37}{6}x+1\right)$k(x²+37/6x+1). Taking $k=6$k=6 gives $6x^2+37x+6$6x²+37x+6.

Check: $2x^2-5x-3=2x^2-6x+x-3=(2x+1)(x-3)$2x²-5x-3=2x²-6x+x-3=(2x+1)(x-3), so the zeroes are $3$3 and $-\dfrac{1}{2}$-1/2. Then $\alpha^2+\beta^2=9+\dfrac{1}{4}=\dfrac{37}{4}$α²+β²=9+1/4=37/4 ✓.

(i) $y=-x^2+8x-12=-(x-2)(x-6)$y=-x²+8x-12=-(x-2)(x-6), so the zeroes are $x=2$x=2 and $x=6$x=6 (where the ball is on the ground).

(ii) Sum of zeroes $=2+6=8=-\dfrac{b}{a}=-\dfrac{8}{-1}$=2+6=8=-b/a=-8/-1 ✓; product $=2\times6=12=\dfrac{c}{a}=\dfrac{-12}{-1}$=2×6=12=c/a=-12/-1 ✓.

(iii) The ball returns to the ground at $x=6$x=6 m. The maximum height occurs midway at $x=4$x=4, giving $y=-(4)^2+8(4)-12=4$y=-(4)²+8(4)-12=4 m.