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) $x^2+x-6$x^2+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)=x^2-(sum)x+product=x^2-(-1)x+(-6)=x^2+x-6.

Answer: (a) $\dfrac{3}{4}$3/4 — Here $\alpha+\beta=6$+=6 and $\alpha\beta=8$=8, so $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{6}{8}=\dfrac{3}{4}$1/+1/=+/=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 $x^2+4$x^2+4, $\ b^2-4ac=0-4(1)(4)=-16<0$b^2-4ac=0-4(1)(4)=-16<0, so its graph never meets the $x$x-axis and it has no real zeroes.

$x^2-2x-8=(x-4)(x+2)$x^2-2x-8=(x-4)(x+2), so the zeroes are $x=4$x=4 and $x=-2$x=-2.

Verification: Sum $=4+(-2)=2=-\dfrac{(-2)}{1}=-\dfrac{b}{a}$=4+(-2)=2=-(-2)/1=-b/a ✓ and product $=4\times(-2)=-8=\dfrac{-8}{1}=\dfrac{c}{a}$=4×(-2)=-8=-8/1=c/a ✓.

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(x^2-(sum)x+product)=k(x^2-2\,x+1/3). Taking $k=3$k=3 to clear the fraction gives $3x^2-3\sqrt2\,x+1$3x^2-32\,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)$6x^2-7x-3=6x^2-9x+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: $x^2-p(x+1)-c=x^2-px-(p+c)$x^2-p(x+1)-c=x^2-px-(p+c). Hence $\alpha+\beta=p$+=p and $\alpha\beta=-(p+c)$=-(p+c).

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

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

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

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

Since $\sqrt2$2 and $-\sqrt2$-2 are zeroes, $(x-\sqrt2)(x+\sqrt2)=x^2-2$(x-2)(x+2)=x^2-2 is a factor.

Dividing $2x^4-3x^3-3x^2+6x-2$2x^4-3x^3-3x^2+6x-2 by $x^2-2$x^2-2:
$2x^4-3x^3-3x^2+6x-2=(x^2-2)(2x^2-3x+1).$2x^4-3x^3-3x^2+6x-2=(x^2-2)(2x^2-3x+1).
Now $2x^2-3x+1=(2x-1)(x-1)$2x^2-3x+1=(2x-1)(x-1), giving zeroes $x=\dfrac{1}{2}$x=1/2 and $x=1$x=1.

All zeroes: $\sqrt2,\ -\sqrt2,\ 1,\ \dfrac{1}{2}$2, -2, 1, 1/2.

Sum of zeroes: $(a-b)+a+(a+b)=3a=-\dfrac{(-3)}{1}=3\Rightarrow a=1$(a-b)+a+(a+b)=3a=-(-3)/1=3 a=1.

Product of zeroes: $(a-b)(a)(a+b)=a(a^2-b^2)=-\dfrac{1}{1}=-1$(a-b)(a)(a+b)=a(a^2-b^2)=-1/1=-1.

Substituting $a=1$a=1: $\ 1(1-b^2)=-1\Rightarrow 1-b^2=-1\Rightarrow b^2=2\Rightarrow b=\pm\sqrt2$1(1-b^2)=-1 1-b^2=-1 b^2=2 b=±2.

Hence $a=1$a=1 and $b=\pm\sqrt2$b=±2.

(i) $y=-x^2+8x-12=-(x-2)(x-6)$y=-x^2+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.