6 Arithmetic Progressions
The common difference of the AP: $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, \dots$ is:
Concept:
1
Common Difference (d): In an Arithmetic Progression, the common difference is calculated by subtracting any term from its consecutive succeeding term ($d = a_2 - a_1 = a_3 - a_2$).
Hint:
1
Subtract the first term ($\sqrt{2}$) from the second term ($2\sqrt{2}$). Treat $\sqrt{2}$ like a variable, such as $x$. What is $2x - x$?
Mistakes to Avoid:
!
Ignoring Surd Rules: A common mistake is subtracting the numbers inside or outside the square roots incorrectly, like thinking $2\sqrt{2} - \sqrt{2} = 2$. Remember, you are just subtracting the coefficients!
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Let the terms of the A.P. be $a_1, a_2, a_3, \dots$
Given:
$a_1 = \sqrt{2}$
$a_2 = 2\sqrt{2}$
$a_3 = 3\sqrt{2}$
The common difference $d$ is:
$d = a_2 - a_1$
$d = 2\sqrt{2} - 1\sqrt{2}$
$d = (2 - 1)\sqrt{2} = \sqrt{2}$ [1 M]
Correct Option: (a) $\sqrt{2}$
7 Triangles
If $\Delta ABC$ and $\Delta DEF$ are similar such that $2AB = DE$ and $BC = 8 \text{ cm}$, then $EF$ is equal to:
Concept:
1
Similar Triangles: When two triangles are similar, the ratio of their corresponding sides is strictly equal. If $\Delta ABC \sim \Delta DEF$, then $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$.
Hint:
1
Rearrange the given relation $2AB = DE$ into a ratio: $\frac{AB}{DE} = \frac{1}{2}$. Set this equal to $\frac{BC}{EF}$ and solve for $EF$!
Mistakes to Avoid:
!
Inverting the Ratio: A common error is looking at $2AB = DE$ and writing the ratio as $\frac{AB}{DE} = 2$ instead of $\frac{1}{2}$. This leads to the incorrect answer of 4 cm. Always cross-multiply to verify your ratio!
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Given that $\Delta ABC \sim \Delta DEF$, the ratio of corresponding sides is equal:
$\frac{AB}{DE} = \frac{BC}{EF}$
We are given the relationship $2AB = DE$.
This can be rewritten as a ratio:
$\frac{AB}{DE} = \frac{1}{2}$
Substitute this ratio and the given length $BC = 8 \text{ cm}$ into the proportion:
$\frac{1}{2} = \frac{8}{EF}$
Cross-multiplying gives:
$EF = 8 \times 2 = 16 \text{ cm}$ [1 M]
Correct Option: (d) 16 cm
8 Coordinate Geometry
The mid-point of the line segment joining the points $(5, -4)$ and $(6, 4)$ lies on:
Concept:
1
Midpoint Formula: The coordinates of the midpoint $M(x, y)$ of a segment connecting $(x_1, y_1)$ and $(x_2, y_2)$ are $M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$. If the resulting y-coordinate is 0, the point lies on the x-axis.
Hint:
1
Apply the midpoint formula to the given points $(5, -4)$ and $(6, 4)$. Focus particularly on the y-coordinate calculation: $\frac{-4 + 4}{2}$.
Mistakes to Avoid:
!
Axis Confusion: A point with coordinates $(x, 0)$ lies on the x-axis. A point with coordinates $(0, y)$ lies on the y-axis. Do not mix these up!
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The coordinates of the mid-point $M(x, y)$ are given by:
$M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
Given the points are $(5, -4)$ and $(6, 4)$:
Substitute the values into the formula:
$x = \frac{5 + 6}{2} = \frac{11}{2} = 5.5$
$y = \frac{-4 + 4}{2} = \frac{0}{2} = 0$
The coordinates of the mid-point are $(5.5, 0)$.
Since the y-coordinate of the mid-point is $0$, the point lies exactly on the x-axis. [1 M]
Correct Option: (a) x-axis
9 Trigonometry
Given that $\sin \theta = \frac{a}{b}$, then $\cos \theta$ is equal to:
Concept:
1
Trigonometric Identities & Ratios: You can solve this using either the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, or by defining a right triangle where Perpendicular = $a$ and Hypotenuse = $b$, and finding the Base using Pythagoras theorem.
Hint:
1
Use $\cos \theta = \sqrt{1 - \sin^2 \theta}$. Substitute $\sin \theta = \frac{a}{b}$ into this formula and simplify the fraction under the square root.
Mistakes to Avoid:
!
Inverting the Ratio: A common mistake is finding the Base correctly as $\sqrt{b^2 - a^2}$ but then writing the ratio upside down as Hypotenuse / Base. Remember, $\cos \theta = \text{Base} / \text{Hypotenuse}$.
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Method 1: Using Trigonometric Identities
We know the fundamental trigonometric identity:
$\sin^2 \theta + \cos^2 \theta = 1$
Rearranging to solve for $\cos \theta$:
$\cos^2 \theta = 1 - \sin^2 \theta$
$\cos \theta = \sqrt{1 - \sin^2 \theta}$
Substitute $\sin \theta = \frac{a}{b}$ into the equation:
$\cos \theta = \sqrt{1 - \left(\frac{a}{b}\right)^2}$
$\cos \theta = \sqrt{1 - \frac{a^2}{b^2}}$
$\cos \theta = \sqrt{\frac{b^2 - a^2}{b^2}}$
$\cos \theta = \frac{\sqrt{b^2 - a^2}}{b}$ [1 M]
Correct Option: (c) $\frac{\sqrt{b^2 - a^2}}{b}$
10 Trigonometry
If $\cos A = \frac{1}{2}$, then the value of $\sin^2 A + 2\cos^2 A$ is:
Concept:
1
Simplifying Trig Expressions: Before substituting numerical values, look for algebraic simplifications. The expression $\sin^2 A + 2\cos^2 A$ can be split into $(\sin^2 A + \cos^2 A) + \cos^2 A$, which instantly reduces to $1 + \cos^2 A$ due to the fundamental identity.
Hint:
1
Use the shortcut mentioned in the concept: $1 + \cos^2 A$. Now simply square the given value ($\frac{1}{2}$) and add it to 1!
Mistakes to Avoid:
!
Long Calculations: While finding angle $A = 60^\circ$, then finding $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, squaring it, and plugging everything in will yield the right answer, it takes much longer and increases the chance of arithmetic errors. Work smart by simplifying expressions first!
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Method 1: Simplifying the Expression First (Faster)
The given expression is $\sin^2 A + 2\cos^2 A$.
Rewrite $2\cos^2 A$ as $\cos^2 A + \cos^2 A$:
$= (\sin^2 A + \cos^2 A) + \cos^2 A$
Since $\sin^2 A + \cos^2 A = 1$, the expression simplifies to:
$= 1 + \cos^2 A$
Substitute the given value $\cos A = \frac{1}{2}$:
$= 1 + \left(\frac{1}{2}\right)^2 = 1 + \frac{1}{4} = \frac{5}{4}$ [1 M]
Correct Option: (b) $\frac{5}{4}$
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