Class X CBSE one mark questions

This lesson works through ten one mark questions taken from older Class X CBSE board papers, spanning polynomials, mensuration, number theory, probability, trigonometry, and coordinate geometry.

Sum of the zeros of a quadratic

Find the sum of the zeros of the polynomial $2x^2 + 9x - 10$. Here $a = 2$, $b = 9$, $c = -10$. The sum of the zeros is

$-\frac{b}{a} = -\frac{9}{2}.$

Forming a polynomial from its zeros

Form a quadratic polynomial whose zeros are $5 + \sqrt{2}$ and $5 - \sqrt{2}$.

Sum of zeros $= (5 + \sqrt{2}) + (5 - \sqrt{2}) = 10.$

Product of zeros $= (5 + \sqrt{2})(5 - \sqrt{2}) = 25 - 2 = 23.$

A quadratic polynomial is $x^2 - (\text{sum})x + (\text{product})$, so

$x^2 - 10x + 23.$

Area of a sector from its arc length

Find the area of a sector of radius $6\ \text{cm}$ whose arc length is $5\ \text{cm}$. With $r = 6$ and arc length $\ell = 5$,

$\text{Area} = \tfrac{1}{2}\,\ell r = \tfrac{1}{2}\times 5 \times 6 = 15\ \text{cm}^2.$

LCM and HCF of three primes

Find the LCM and HCF of $17$, $23$, $29$ by the factorization method. Each number is prime, so they share no common factor and

$\text{HCF} = 1.$

The LCM is their product,

$\text{LCM} = 17 \times 23 \times 29 = 11339.$

Probability of losing, and a trig simplification

If the probability of winning a game is $0.3$, the probability of losing is

$1 - 0.3 = 0.7.$

For the trigonometric part, using $1 - \sin^2\theta = \cos^2\theta$,

$\frac{1 - \sin^2\theta}{\cos\theta} = \frac{\cos^2\theta}{\cos\theta} = \cos\theta.$

Rational numbers and the discriminant

Is the real number $5.235$ rational? Yes. It is a terminating decimal, so it is rational.

Find the discriminant of $3x^2 - 2x + \tfrac{1}{3} = 0$. Here $a = 3$, $b = -2$, $c = \tfrac{1}{3}$, so

$b^2 - 4ac = (-2)^2 - 4(3)\left(\tfrac{1}{3}\right) = 4 - 4 = 0.$

A point a fraction along a segment

Find the point that is two thirds of the way from $P(0, 1)$ to $Q(1, 0)$. Take the dividing ratio $m_1 : m_2 = 2 : 3$. By the section formula,

$\left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2},\ \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right).$

x-coordinate $= \dfrac{2(1) + 3(0)}{2 + 3} = \dfrac{2}{5}.$

y-coordinate $= \dfrac{2(0) + 3(1)}{2 + 3} = \dfrac{3}{5}.$

So the point is $\left(\tfrac{2}{5}, \tfrac{3}{5}\right).$

Value of k for equal roots

Find $k$ so that $2x^2 + 3x + k = 0$ has equal roots. Equal roots means the discriminant is zero, with $a = 2$, $b = 3$, $c = k$:

$b^2 - 4ac = 0 \implies 3^2 - 4(2)(k) = 0 \implies 9 - 8k = 0.$

Therefore

$k = \tfrac{9}{8}.$

Key takeaways

• For $ax^2 + bx + c$, the sum of the zeros is $-\tfrac{b}{a}$, and a polynomial from given zeros is $x^2 - (\text{sum})x + (\text{product})$.
• A sector's area is $\tfrac{1}{2}\ell r$, and for distinct primes the HCF is $1$ while the LCM is their product.
• A quadratic has equal roots exactly when its discriminant $b^2 - 4ac$ is zero.