Sure 2-Mark Questions for Class 10 CBSE

This lesson works through several sure-shot two-mark questions taken from previous CBSE Class 10 board papers, covering trigonometry, probability, circle geometry, quadratics, arithmetic progressions, and coordinate geometry.

Proving a trigonometric identity

Prove that

$\left(1 - \sin\theta + \cos\theta\right)^2 = 2\left(1 + \cos\theta\right)\left(1 - \sin\theta\right).$

Start from the left side and expand it as $\big((1 - \sin\theta) + \cos\theta\big)^2$:

$(1 - \sin\theta)^2 + \cos^2\theta + 2\cos\theta(1 - \sin\theta).$

Expanding $(1 - \sin\theta)^2$ and using $\sin^2\theta + \cos^2\theta = 1$:

$1 - 2\sin\theta + \sin^2\theta + \cos^2\theta + 2\cos\theta - 2\sin\theta\cos\theta$

$= 2 - 2\sin\theta + 2\cos\theta - 2\sin\theta\cos\theta.$

Take out a factor of $2$ and group the terms:

$2\big[(1 - \sin\theta) + \cos\theta(1 - \sin\theta)\big] = 2(1 - \sin\theta)(1 + \cos\theta).$

This equals the right side, so the identity holds.

Evaluating a trigonometric expression

Given $\sin a = \tfrac{1}{\sqrt{2}}$, find

$\frac{2\sin^2 a + 3\cos^2 a}{4\tan^2 a - \cos^2 a}.$

Finding the ratios. Using $\cos^2 a = 1 - \sin^2 a$:

$\cos^2 a = 1 - \tfrac{1}{2} = \tfrac{1}{2}, \qquad \cos a = \tfrac{1}{\sqrt{2}}, \qquad \tan a = \frac{\sin a}{\cos a} = 1.$

Substituting.

$\frac{2\cdot\tfrac{1}{2} + 3\cdot\tfrac{1}{2}}{4\cdot 1 - \tfrac{1}{2}} = \frac{1 + \tfrac{3}{2}}{4 - \tfrac{1}{2}} = \frac{\tfrac{5}{2}}{\tfrac{7}{2}} = \frac{5}{7}.$

Probability with a card problem

All aces, jacks, and queens are removed from a standard deck, and one card is drawn from the rest. This removes $3 \times 4 = 12$ cards, leaving $52 - 12 = 40$ cards.

The only face cards left are the $4$ kings, so the favourable outcomes number $4$, giving

$P(\text{a king}) = \frac{4}{40} = \frac{1}{10},$

$P(\text{not a king}) = 1 - \frac{1}{10} = \frac{9}{10}.$

Tangents at the ends of a diameter

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Let the circle have centre $O$, with tangents $PR$ and $QS$ drawn at the ends $A$ and $B$ of a diameter. A radius meets its tangent at a right angle, so

$\angle 1 = \angle 2 = 90^\circ.$

These are co-interior angles for the transversal $AB$, and

$\angle 1 + \angle 2 = 90^\circ + 90^\circ = 180^\circ.$

Since the co-interior angles are supplementary, $PR \parallel QS$.

Solving a quadratic with literal coefficients

Solve $(a+b)^2 x^2 - (a+b)x - 6 = 0$.

Here the coefficients are $A = (a+b)^2$, $B = -(a+b)$, and $C = -6$. The discriminant is

$B^2 - 4AC = (a+b)^2 + 24(a+b)^2 = 25(a+b)^2,$

so $\sqrt{B^2 - 4AC} = 5(a+b)$. Then

$x = \frac{(a+b) \pm 5(a+b)}{2(a+b)^2}.$

The two roots are

$x = \frac{6(a+b)}{2(a+b)^2} = \frac{3}{a+b}, \qquad x = \frac{-4(a+b)}{2(a+b)^2} = \frac{-2}{a+b}.$

Arithmetic progression

If $k+2$, $4k-6$, and $3k-2$ are in arithmetic progression, find $k$. For three terms in AP, twice the middle term equals the sum of the outer terms:

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

So $8k - 12 = 4k$, giving $4k = 12$ and $k = 3$.

A point on the x-axis equidistant from two points

Find the point on the x-axis equidistant from $(7, 6)$ and $(-3, 4)$. Take the point as $(x, 0)$ and set the squared distances equal:

$(x - 7)^2 + 6^2 = (x + 3)^2 + 4^2.$

$x^2 - 14x + 49 + 36 = x^2 + 6x + 9 + 16.$

The $x^2$ terms cancel, leaving $-20x = -60$, so $x = 3$. The point is $(3, 0)$.

Evaluating a trig ratio from cotangent

If $\cot x = \tfrac{3}{4}$, find $\dfrac{\sin x - \cos x}{\sin x + \cos x}$. Then $\tan x = \tfrac{4}{3}$. Dividing numerator and denominator by $\cos x$:

$\frac{\tan x - 1}{\tan x + 1} = \frac{\tfrac{4}{3} - 1}{\tfrac{4}{3} + 1} = \frac{\tfrac{1}{3}}{\tfrac{7}{3}} = \frac{1}{7}.$

Checking the zeros of a polynomial

Show that $3$ and $-\tfrac{3}{4}$ are zeros of $p(x) = 4x^2 - 9x - 9$.

$p(3) = 4(9) - 9(3) - 9 = 36 - 27 - 9 = 0.$

$p\!\left(-\tfrac{3}{4}\right) = 4\cdot\tfrac{9}{16} + \tfrac{27}{4} - 9 = \tfrac{9}{4} + \tfrac{27}{4} - 9 = 9 - 9 = 0.$

Both give $0$, so both are zeros.

Expressing a fraction as a decimal

Express $\tfrac{22}{8}$ as a decimal. Reduce to $\tfrac{11}{4}$ and divide to get $2.75$.

Key takeaways

• To prove a trig identity, expand one side, apply $\sin^2\theta + \cos^2\theta = 1$, and factor toward the other side.
• Three numbers are in AP exactly when twice the middle term equals the sum of the other two.
• A number is a zero of a polynomial precisely when substituting it gives $0$.