41 Quadratic Equations
The value of $k$ for which the equation $kx^{2}-6x-4=0$ has real and equal roots, is:
Concept:
1
Discriminant for Equal Roots: For any quadratic equation $ax^2 + bx + c = 0$ to have real and equal roots, its Discriminant ($D$) must be exactly zero. The formula for the discriminant is $D = b^2 - 4ac$.
Hint:
1
Identify $a = k$, $b = -6$, and $c = -4$. Set up the equation: $(-6)^2 - 4(k)(-4) = 0$. Solve this linear equation for $k$.
Mistakes to Avoid:
!
Sign Errors in Multiplication: When calculating $-4ac$, students often forget the negative sign on $c$. Here, $-4 \times k \times -4$ becomes positive $+16k$, not $-16k$. This mistake will give you the positive root (Option A) instead of the negative one!
👉 Scroll horizontally to read the full handwritten solution.
For a quadratic equation to have real and equal roots,
its discriminant $D$ must be zero ($D = 0$).
Comparing $kx^{2}-6x-4=0$ with $ax^2 + bx + c = 0$:
$a = k$, $b = -6$, $c = -4$
$D = b^2 - 4ac = 0$
$(-6)^2 - 4(k)(-4) = 0$
$36 + 16k = 0$ [½ M]
Now, solve for $k$:
$16k = -36$
$k = -\frac{36}{16}$
Simplify the fraction by dividing numerator and denominator by 4:
$k = -\frac{9}{4}$ [½ M]
Correct Option: (C) $-\frac{9}{4}$
42 Coordinate Geometry
The line segment joining the points $P(-4,-2)$ and $Q(10,4)$ is divided by y-axis in the ratio:
Concept:
1
Section Formula & Axis Intersections: If a line segment is divided by the y-axis, the point of intersection lies perfectly on the y-axis, which means its $x$-coordinate is exactly $0$. Use the section formula for $x$: $x = \frac{mx_2 + nx_1}{m+n} = 0$.
Hint:
1
Since $x=0$, the denominator ($m+n$) disappears when you cross-multiply. Just set the numerator to zero: $m(10) + n(-4) = 0$. Now solve for the ratio $m/n$!
Mistakes to Avoid:
!
Using the Y-coordinate: A classic trap! Students hear "y-axis" and immediately try to set up the section formula using the y-coordinates ($-2$ and $4$). This will not work because the $y$-value at the intersection is an unknown variable. You MUST use the $x$-coordinates because you know $x=0$!
👉 Scroll horizontally to read the full handwritten solution.
Let the required ratio be $m:n$.
Since the line segment is divided by the y-axis,
the x-coordinate of the point of intersection must be $0$.
Points given: $P(x_1, y_1) = P(-4, -2)$ and $Q(x_2, y_2) = Q(10, 4)$
Using the section formula for the x-coordinate:
$x = \frac{mx_2 + nx_1}{m+n}$
$0 = \frac{m(10) + n(-4)}{m+n}$
Cross-multiply to eliminate the denominator:
$10m - 4n = 0$
$10m = 4n$ [½ M]
Rearrange to find the ratio $m/n$:
$\frac{m}{n} = \frac{4}{10} = \frac{2}{5}$ [½ M]
Correct Option: (A) $2:5$
43 Polynomials
If the zeroes of a polynomial $p(x)$ are $-3$ and $8$, then $p(x)$ equals:
Concept:
1
Constructing Polynomials from Zeroes: If a polynomial has zeroes $\alpha$ and $\beta$, the polynomial can be written in factored form as $k(x - \alpha)(x - \beta)$, where $k$ is any non-zero real number. The $k$ constant allows for infinite variations of polynomials that share the exact same zeroes.
Hint:
1
Plug in the zeroes to get the base factors: $(x - (-3))(x - 8) = (x+3)(x-8)$. Now, look at the options. Which option is mathematically equivalent to $(x+3)(x-8)$ if you multiply it by some constant $k$ (like $-1$)?
Mistakes to Avoid:
!
Option C Trap: Option C expands to $a(x^2 + 5x - 24)$. If you find the roots of $x^2 + 5x - 24$, you get $-8$ and $+3$. This is the exact opposite of what we want! Always verify the sum and product of zeroes before picking an expanded form.
👉 Scroll horizontally to read the full handwritten solution.
Let the given zeroes be $\alpha = -3$ and $\beta = 8$.
The standard factored form of a polynomial is:
$p(x) = k(x - \alpha)(x - \beta)$
Substitute the zeroes:
$p(x) = k(x - (-3))(x - 8)$
$p(x) = k(x + 3)(x - 8)$
Now, we evaluate the given options to find a match.
If we set the constant $k = -1$, the polynomial becomes:
$p(x) = -1(x + 3)(x - 8)$
Distribute the negative sign into the second bracket:
$p(x) = (x + 3)(-x + 8)$ [1 M]
This matches exactly with Option (B).
Correct Option: (B) $(x+3)(-x+8)$
44 Circles
In the given figure, PQ is tangent to the circle with centre O. S is a point on the circle such that $\angle SQT = 55^\circ$. The measure of $\angle QPS$ is:
Concept:
1
Tangent-Chord Theorem (Alternate Segment Theorem): The angle between a tangent and a chord through the point of contact ($\angle SQT = 55^\circ$) is equal to the angle subtended by that chord in the alternate segment. Alternatively, using central angles, the angle subtended at the center ($\angle QOS$) by the chord is exactly twice the angle between the tangent and the chord: $\angle QOS = 2 \times \angle SQT$.
Hint:
1
First, use the theorem to find $\angle QOS = 2 \times 55^\circ = 110^\circ$. Since $POS$ is a straight line, find $\angle QOP$ by subtracting $110^\circ$ from $180^\circ$. Finally, look at the big right-angled triangle $\Delta PQO$ to find your target angle $\angle QPS$!
Mistakes to Avoid:
!
Assuming Isosceles Triangles: Students sometimes assume $\Delta PQS$ is isosceles and try to split angles evenly. Rely strictly on the linear pair property on the straight line $POS$ and the $90^\circ$ tangent rule at point $Q$!
👉 Scroll horizontally to read the full handwritten solution.
Since $PQ$ is tangent at $Q$, the radius $OQ$ is perpendicular to it.
Therefore, $\angle OQP = 90^\circ$.
Given $\angle SQT = 55^\circ$. By the tangent-chord theorem, the central
angle subtended by chord $QS$ is twice this angle:
$\angle QOS = 2 \times 55^\circ = 110^\circ$. [½ M]
$P$, $O$, and $S$ are collinear (form a straight line).
Hence, linear pair: $\angle QOP = 180^\circ - 110^\circ = 70^\circ$.
Now, consider the right-angled $\Delta PQO$.
The sum of angles in a triangle is $180^\circ$.
$\angle QPS + \angle OQP + \angle QOP = 180^\circ$
$\angle QPS + 90^\circ + 70^\circ = 180^\circ$
$\angle QPS = 180^\circ - 160^\circ = 20^\circ$. [½ M]
Correct Option: (B) $20^\circ$
45 Triangles
Devansh proved that $\Delta ABC \sim \Delta PQR$ using SAS similarity criteria. If he found $\angle C = \angle R$, then which of the following was proved true?
Concept:
1
SAS Similarity Criterion: Two triangles are similar by SAS if one angle of a triangle is equal to one angle of the other triangle AND the sides including (containing) these angles are proportional.
Hint:
1
The equal angles are $\angle C$ and $\angle R$. Which two sides of $\Delta ABC$ meet to form $\angle C$? (It's AC and BC). Which two sides of $\Delta PQR$ meet to form $\angle R$? (It's PR and QR). Set up the ratio using those exact sides!
Mistakes to Avoid:
!
Including the Opposite Side: A common mistake is selecting a ratio that includes the sides $AB$ or $PQ$. Those sides are opposite to the angles $\angle C$ and $\angle R$. SAS strictly requires the sides that "hug" the angle!
👉 Scroll horizontally to read the full handwritten solution.
For the SAS (Side-Angle-Side) similarity criterion,
the ratio of the sides containing the equal included angle
must be strictly proportional.
We are given that $\angle C = \angle R$.
The sides containing $\angle C$ in $\Delta ABC$ are $AC$ and $BC$.
The sides containing $\angle R$ in $\Delta PQR$ are $PR$ and $QR$.
Thus, for SAS similarity to hold true, the proportion must be:
$\frac{AC}{PR} = \frac{BC}{QR}$
We can rewrite this fraction by cross-swapping the terms
(moving $BC$ down and $PR$ up) to match the options:
$\frac{AC}{BC} = \frac{PR}{QR}$ [1 M]
Correct Option: (D)
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