Q.Find the maximum profit that a company can make, if the profit function is given by $p(x)=41-72x-18x^2$

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.The maximum value of the function $f(x) = 3\sin^{12}x + 4\cos^{16}x$ is (A) 4 (B) 5 (C) 6 (D) 7

   ›Reveal solutionSolution

   The key idea is to bound each trigonometric term by its maximum possible value (1 for sine or cosine squared) and then check if both maxima can occur simultaneously. The maximum value is 4, corresponding to option (A).

   We want the maximum of $f(x) = 3\sin^{12}x + 4\cos^{16}x$. Since $\sin^2 x$ and $\cos^2 x$ are always between 0 and 1, raising them to higher powers only makes them smaller or keeps them the same. So the largest each term can be is when the base is 1.

   Intuition:

   If $\sin^2 x = 1$, then $\sin^{12}x = 1$ and $\cos^2 x = 0$, so $\cos^{16}x = 0$. That gives $f = 3$.

   If $\cos^2 x = 1$, then $\cos^{16}x = 1$ and $\sin^2 x = 0$, giving $f = 4$.

   So 4 is already larger than 3. Could we get more than 4? For that, both $\sin^{12}x$ and $\cos^{16}x$ would need to be positive simultaneously, but then each is less than 1, so the weighted sum might exceed 4? Let’s check carefully.

   Step-by-step reasoning:

   1. Bound each term individually For any real $x$, $0 \le \sin^2 x \le 1$ and $0 \le \cos^2 x \le 1$. Since $12$ and $16$ are even positive integers,

   $0 \le \sin^{12}x \le 1, \quad 0 \le \cos^{16}x \le 1.$

   Hence

   $f(x) = 3\sin^{12}x + 4\cos^{16}x \le 3\cdot 1 + 4\cdot 1 = 7.$

   But this bound is not attainable because $\sin^{12}x$ and $\cos^{16}x$ cannot both be 1 at the same time (since $\sin^2 x + \cos^2 x = 1$).

   2. Find when each term individually reaches its maximum

      • $\sin^{12}x = 1$ when $\sin^2 x = 1$, i.e., $x = \frac{\pi}{2} + k\pi$. Then $\cos^2 x = 0$, so $\cos^{16}x = 0$. At such $x$, $f = 3\cdot 1 + 4\cdot 0 = 3$.
      • $\cos^{16}x = 1$ when $\cos^2 x = 1$, i.e., $x = k\pi$. Then $\sin^2 x = 0$, so $\sin^{12}x = 0$. At such $x$, $f = 3\cdot 0 + 4\cdot 1 = 4$.

      So far, the largest value we have is 4.

   3. Could a mix give more than 4?

      Suppose both $\sin^2 x$ and $\cos^2 x$ are positive. Let $a = \sin^2 x$, $b = \cos^2 x$, with $a + b = 1$, $a, b \ge 0$.

      Then

      $f = 3a^6 + 4b^8.$ …

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2. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.The maximum volume (in cu. units) of the cylinder which can be inscribed in a sphere of radius 12 units is (A) $384\sqrt{3}\pi$ (B) $768\sqrt{3}\pi$ (C) $\dfrac{768\pi}{\sqrt{3}}$ (D) $\dfrac{1152\pi}{\sqrt{3}}$

   ›Reveal solutionSolution

   The maximum volume of a cylinder inscribed in a sphere of radius 12 is found by expressing the cylinder’s volume in terms of its height, differentiating, and solving. The result is $768\sqrt{3}\pi$, which corresponds to option (B).

   ---

   The key idea is that an inscribed cylinder’s top and bottom circles lie on the sphere’s surface. If we draw a cross-section through the sphere’s center, we see a rectangle (the cylinder’s side view) inside a circle. The cylinder’s height and radius are linked by the sphere’s radius via the Pythagorean theorem. This turns the volume into a function of one variable, which we maximize using calculus.

   ---

   1. Set up the geometry Let the sphere have radius $R = 12$. Let the cylinder have height $h$ and base radius $r$. In a cross-section through the center, the cylinder appears as a rectangle of width $2r$ and height $h$, inscribed in a circle of radius $12$. The center of the sphere is also the midpoint of the cylinder’s axis. Half the height is $h/2$, so by the Pythagorean theorem:

   $r^2 + \left(\frac{h}{2}\right)^2 = 12^2 = 144.$

   Hence,

   $r^2 = 144 - \frac{h^2}{4}.$

   2. Write the volume Volume of a cylinder: $V = \pi r^2 h$. Substitute $r^2$:

   $V(h) = \pi \left(144 - \frac{h^2}{4}\right) h = \pi \left(144h - \frac{h^3}{4}\right).$

   3. Differentiate and find critical points

   $V'(h) = \pi \left(144 - \frac{3h^2}{4}\right).$

   Set $V'(h) = 0$:

   $144 - \frac{3h^2}{4} = 0 \quad\Rightarrow\quad \frac{3h^2}{4} = 144 \quad\Rightarrow\quad h^2 = 192 \quad\Rightarrow\quad h = \sqrt{192} = 8\sqrt{3}.$

   (Only positive height makes sense.)

   4. Find the corresponding radius $$$$

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3. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.The set of all real values of the expression $\dfrac{x^2 - x + 2}{x^2 + x - 2}$ for all $x \in \mathbb{R} - \{-2, 1\}$ is (A) $(-2, 3)$ (B) $\left[\dfrac{7}{9}, \infty\right)$ (C) $(-\infty, -1] \cup \left[\dfrac{7}{9}, \infty\right)$ (D) $(-\infty, -1]$

   ›Reveal solutionSolution

   The expression is a rational function that can be rewritten to reveal its range. By analyzing the quadratic in the denominator and using the discriminant method, the set of all real values is $(-\infty, -1] \cup \left[\frac{7}{9}, \infty\right)$.

   The key idea here is that when you have a rational expression of the form $\frac{ax^2+bx+c}{dx^2+ex+f}$, the range can often be found by setting the expression equal to $y$, cross-multiplying, and then demanding that the resulting quadratic in $x$ has real solutions (since $x$ is real). This is the discriminant method, and it works beautifully here.

   Let’s walk through it.

   1. Set the expression equal to $y$. Let

   $y = \frac{x^2 - x + 2}{x^2 + x - 2}, \quad x \in \mathbb{R} \setminus \{-2, 1\}.$

   The denominator is zero at $x = -2$ and $x = 1$, so those are excluded from the domain.

   2. Cross-multiply and rearrange into a quadratic in $x$.

   $y(x^2 + x - 2) = x^2 - x + 2$

   $y x^2 + y x - 2y = x^2 - x + 2$

   Bring all terms to one side:

   $(y - 1)x^2 + (y + 1)x + (-2y - 2) = 0$

   So we have:

   $(y - 1)x^2 + (y + 1)x - 2(y + 1) = 0$

   3. Consider the case $y = 1$ separately. If $y = 1$, the coefficient of $x^2$ becomes $0$, and the equation reduces to:

   $(1 + 1)x - 2(1 + 1) = 0 \quad \Rightarrow \quad 2x - 4 = 0 \quad \Rightarrow \quad x = 2$

   Since $x = 2$ is in the domain, $y = 1$ is indeed attained. So $1$ is in the range.

   4. For $y \neq 1$, the equation is quadratic in $x$. For $x$ to be real, the discriminant must be non-negative. The discriminant $D$ is:

   $D = (y + 1)^2 - 4(y - 1)[-2(y + 1)]$

   Simplify carefully:

   $D = (y + 1)^2 + 8(y - 1)(y + 1)$

   Factor $(y+1)$:

   $D = (y + 1)[(y + 1) + 8(y - 1)] = (y + 1)(y + 1 + 8y - 8) = (y + 1)(9y - 7)$

   5. Set $D \geq 0$ for real $x$.

   $(y + 1)(9y - 7) \geq 0$

   Solve this inequality. The critical points are $y = -1$ and $y = \frac{7}{9}$.

   Testing intervals:

   • For $y < -1$: both factors negative → product positive.
   • For $-1 < y < \frac{7}{9}$: first factor positive, second negative → product negative.
   • For $y > \frac{7}{9}$: both positive → product positive. …

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4. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $f(x) = (2x-1)(3x+2)(4x-3)$ is a real valued function defined on $\left[\frac{1}{2}, \frac{3}{4}\right]$, then the value(s) of ‘c’ as defined in the statement of Rolle’s theorem (A) Does not exist (B) $\dfrac{7 \pm \sqrt{247}}{36}$ (C) $\dfrac{7 - \sqrt{247}}{36}$ (D) $\dfrac{7 + \sqrt{247}}{36}$

   ›Reveal solutionSolution

   Rolle's theorem applies since $f\!\left(\tfrac12\right)=f\!\left(\tfrac34\right)=0$; solving $f'(c)=72c^2-28c-11=0$ gives $c=\dfrac{7\pm\sqrt{247}}{36}$, and only $\dfrac{7+\sqrt{247}}{36}\approx0.63$ lies in $\left(\tfrac12,\tfrac34\right)$. Answer: (D).

   Concept

   Rolle's theorem: if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then some $c\in(a,b)$ has $f'(c)=0$. The valid $c$ must lie strictly inside the interval.

   Solution

   1. Check the hypotheses. $f(x)=(2x-1)(3x+2)(4x-3)$ is a polynomial, hence continuous and differentiable everywhere. The endpoints are zeros of two factors:

   $f\!\left(\tfrac12\right)=0\ \text{(from }2x-1),\qquad f\!\left(\tfrac34\right)=0\ \text{(from }4x-3),$

   so $f\!\left(\tfrac12\right)=f\!\left(\tfrac34\right)$ and Rolle's theorem applies.

   2. Derivative (product rule on three factors, $u'=2,\ v'=3,\ w'=4$):

   $f'(x)=2(3x+2)(4x-3)+3(2x-1)(4x-3)+4(2x-1)(3x+2).$

   Expanding and adding, each term contributes $24x^2$, so

   $f'(x)=72x^2-28x-11.$

   3. Solve $f'(c)=0$. …

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5. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.If $x^2 - 5x - 14 > 0 \Rightarrow x$ lie outside $[\alpha, \beta]$, then $\dfrac{\alpha}{\beta} =$ (A) $\dfrac{-2}{7}$ (B) $\dfrac{-7}{2}$ (C) $\dfrac{2}{7}$ (D) $\dfrac{7}{2}$

   ›Reveal solutionSolution

   The quadratic inequality $x^2 - 5x - 14 > 0$ holds outside the interval between its roots. The roots are $\alpha = -2$ and $\beta = 7$, so $\frac{\alpha}{\beta} = -\frac{2}{7}$.

   The key idea here is that a quadratic inequality like $x^2 - 5x - 14 > 0$ tells us where the parabola lies above the x-axis. For a quadratic with a positive leading coefficient (which this one has, since the coefficient of $x^2$ is 1), the graph opens upward. That means the expression is positive outside the interval between its two real roots, and negative inside that interval.

   The problem says $x$ lies outside $[\alpha, \beta]$. That interval notation means $\alpha$ is the smaller root and $\beta$ is the larger root. So our job is to find the roots of $x^2 - 5x - 14 = 0$, identify which is $\alpha$ and which is $\beta$, and then compute their ratio.

   1. Find the roots of the quadratic equation. Solve $x^2 - 5x - 14 = 0$. Factor it: we need two numbers that multiply to $-14$ and add to $-5$. Those numbers are $-7$ and $2$. So

   $x^2 - 5x - 14 = (x - 7)(x + 2) = 0$

   Hence the roots are $x = 7$ and $x = -2$.

   2. Identify $\alpha$ and $\beta$. …

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6. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.The quadratic equation whose roots are $\sin^2 18^\circ$ and $\cos^2 36^\circ$ is (A) $16x^2 - 12x - 1 = 0$ (B) $16x^2 - 12x + 4 = 0$ (C) $16x^2 - 12x + 1 = 0$ (D) $16x^2 + 12x + 1 = 0$

   ›Reveal solutionSolution

   The key idea is to find the exact values of $\sin^2 18^\circ$ and $\cos^2 36^\circ$ using known trigonometric identities, then form the quadratic with those roots. The result is $16x^2 - 12x + 1 = 0$, so the correct option is (C).

   We start by recalling that $\sin 18^\circ$ and $\cos 36^\circ$ have well-known exact values. The trick is to compute their squares cleanly, then sum and product to build the quadratic.

   1. Find $\sin 18^\circ$ exactly. A classic derivation uses the fact that $\sin 54^\circ = \cos 36^\circ$ and the triple-angle formula. But the simplest known result is:

   $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}.$

   (One can verify by solving $\sin 5\theta = 0$ for $\theta = 18^\circ$.)

   Hence

   $\sin^2 18^\circ = \left(\frac{\sqrt{5} - 1}{4}\right)^2 = \frac{5 - 2\sqrt{5} + 1}{16} = \frac{6 - 2\sqrt{5}}{16} = \frac{3 - \sqrt{5}}{8}.$

   2. Find $\cos 36^\circ$ exactly. Similarly, $\cos 36^\circ = \frac{\sqrt{5} + 1}{4}$. Therefore

   $\cos^2 36^\circ = \left(\frac{\sqrt{5} + 1}{4}\right)^2 = \frac{5 + 2\sqrt{5} + 1}{16} = \frac{6 + 2\sqrt{5}}{16} = \frac{3 + \sqrt{5}}{8}.$

   3. Sum of the roots. Let $r_1 = \sin^2 18^\circ$ and $r_2 = \cos^2 36^\circ$. Then

   $r_1 + r_2 = \frac{3 - \sqrt{5}}{8} + \frac{3 + \sqrt{5}}{8} = \frac{6}{8} = \frac{3}{4}.$

   4. Product of the roots.

   $r_1 r_2 = \frac{3 - \sqrt{5}}{8} \cdot \frac{3 + \sqrt{5}}{8} = \frac{9 - 5}{64} = \frac{4}{64} = \frac{1}{16}.$

   5. Form the quadratic. …

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7. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The equation of the common tangent to the parabola $y^2 = 8x$ and the circle $x^2 + y^2 = 2$ is $ax + by + 2 = 0$. If $-\dfrac{a}{b} > 0$, then $3a^2 + 2b + 1 =$ (A) $5$ (B) $4$ (C) $3$ (D) $2$

   ›Reveal solutionSolution

   The common tangent to the parabola $y^2 = 8x$ and the circle $x^2 + y^2 = 2$ has the form $y = mx + \frac{2}{m}$ (for the parabola) and must satisfy the circle’s tangency condition, leading to $m = 1$ (since $-\frac{a}{b} > 0$ forces the positive slope). This gives $a = 1, b = -1$, so $3a^2 + 2b + 1 = 2$.

   ---

   Concept & Intuition

   When two curves share a common tangent, the line must satisfy the tangency condition for both curves simultaneously. For a parabola $y^2 = 4ax$, the family of tangents in slope form is $y = mx + \frac{a}{m}$. For a circle, the condition that a line $y = mx + c$ is tangent is that the perpendicular distance from the center equals the radius. Matching these gives the slope, and the sign condition picks the correct one.

   ---

   Step-by-step solution

   1. Identify the parabola’s tangent family The parabola is $y^2 = 8x$, so $4a = 8 \Rightarrow a = 2$. Any tangent to this parabola (with slope $m \neq 0$) is

   $y = mx + \frac{2}{m}.$

   This is the standard formula $y = mx + \frac{a}{m}$ for $y^2 = 4ax$.

   2. Apply the circle’s tangency condition The circle is $x^2 + y^2 = 2$, center $(0,0)$, radius $r = \sqrt{2}$. For the line $y = mx + \frac{2}{m}$ to be tangent to the circle, the distance from the center to the line must equal $\sqrt{2}$. Rewrite the line as:

   $mx - y + \frac{2}{m} = 0.$

   Distance from $(0,0)$ is

   $\frac{\left| \frac{2}{m} \right|}{\sqrt{m^2 + 1}} = \sqrt{2}.$

   3. Solve for $m$ Square both sides:

   $\frac{4}{m^2 (m^2 + 1)} = 2 \quad\Rightarrow\quad \frac{4}{m^2(m^2+1)} = 2.$

   Multiply: $4 = 2m^2(m^2+1)$ $\Rightarrow$ $2 = m^2(m^2+1)$.

   Let $t = m^2$:

   $t(t+1) = 2 \quad\Rightarrow\quad t^2 + t - 2 = 0 \quad\Rightarrow\quad (t+2)(t-1)=0.$

   So $t = 1$ or $t = -2$ (reject negative). Hence $m^2 = 1$, so $m = \pm 1$.

   4. Use the sign condition $-\frac{a}{b} > 0$ The given tangent is $ax + by + 2 = 0$. Compare with $y = mx + \frac{2}{m}$. Rewrite $y = mx + \frac{2}{m}$ as $mx - y + \frac{2}{m} = 0$. …

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8. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.The equation whose roots are squares of the roots of $x^4 - 2x^3 + 6x - 21 = 0$ is (A) $x^4 - 4x^3 - 18x^2 - 36x + 441 = 0$ (B) $x^4 + 18x^3 - 4x^2 + 36x + 441 = 0$ (C) $x^4 - 2x^3 + 4x^2 + 6x + 441 = 0$ (D) $x^4 + 3x^3 - 5x^2 + 6x + 441 = 0$

   ›Reveal solutionSolution

   To find an equation whose roots are the squares of the original roots, we substitute $y = x^2$ into the original equation by isolating terms with odd powers of $x$ and then squaring. The resulting equation is $\boxed{x^4 - 4x^3 - 18x^2 - 36x + 441 = 0}$.

   When we need to find a new polynomial equation whose roots are related to the roots of a given polynomial equation by some function, say $y = f(x)$, the most straightforward approach is often to use substitution. If $x$ is a root of the original equation $P(x) = 0$, and $y$ is a root of the new equation $Q(y) = 0$, and $y = f(x)$, then we can express $x$ in terms of $y$ (i.e., $x = f^{-1}(y)$) and substitute this into $P(x) = 0$. This transforms the equation from one in $x$ to one in $y$.

   In this specific problem, the new roots are the squares of the old roots. If $x$ is a root of the given equation, then $y = x^2$ is a root of the desired equation. Our goal is to eliminate $x$ from the original equation by using this relationship.

   Here's how we can do it:

   1. Define the relationship between the roots:

      Let the roots of the given equation $x^4 - 2x^3 + 6x - 21 = 0$ be $\alpha_1, \alpha_2, \alpha_3, \alpha_4$.

      We want to find an equation whose roots are $\alpha_1^2, \alpha_2^2, \alpha_3^2, \alpha_4^2$.

      Let $y$ be a root of the new equation. Then $y = x^2$, where $x$ is a root of the original equation.

   2. Rearrange the original equation:

      The original equation is $x^4 - 2x^3 + 6x - 21 = 0$.

      We want to substitute $x^2 = y$. Notice that the equation has both even powers ($x^4$) and odd powers ($x^3, x$). To handle the odd powers, we can isolate them on one side of the equation.

      $x^4 - 21 = 2x^3 - 6x$

      Factor out $x$ from the right side:

      $x^4 - 21 = x(2x^2 - 6)$

   3. Substitute $x^2 = y$ and square both sides:

      Now, substitute $x^2 = y$ into the rearranged equation. This means $x^4 = (x^2)^2 = y^2$.

      $y^2 - 21 = x(2y - 6)$

      To eliminate the remaining $x$ term, we square both sides of the equation:

      $(y^2 - 21)^2 = [x(2y - 6)]^2$

      $(y^2 - 21)^2 = x^2 (2y - 6)^2$

      Now, substitute $x^2 = y$ again on the right side:

      $(y^2 - 21)^2 = y (2y - 6)^2$ …

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9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If $m_1$ and $m_2$ are the slopes of the tangents drawn from the point $(1,4)$ to the parabola $y^2 = 11x$ then $2(m_1^2 + m_2^2) =$ (A) $18$ (B) $21$ (C) $24$ (D) $22$

   ›Reveal solutionSolution

   The key idea is to use the equation of a tangent to the parabola $y^2 = 11x$ in slope form, impose that it passes through $(1,4)$, and then use the quadratic in $m$ to find $m_1^2 + m_2^2$ without solving individually. The final result is $2(m_1^2 + m_2^2) = 22$.

   ---

   We are given the parabola $y^2 = 11x$. For a parabola of the form $y^2 = 4ax$, the slope $m$ of a tangent that touches it satisfies the equation $y = mx + \frac{a}{m}$. Here $4a = 11$, so $a = \frac{11}{4}$. Thus any tangent (with slope $m \neq 0$) to this parabola has equation:

   $y = mx + \frac{11}{4m}.$

   We want the tangents that pass through the external point $(1,4)$. Substituting $x=1$, $y=4$ gives:

   $4 = m(1) + \frac{11}{4m}.$

   Multiply through by $4m$ (valid since $m \neq 0$):

   $16m = 4m^2 + 11.$

   Rearrange into a standard quadratic:

   $4m^2 - 16m + 11 = 0.$

   This quadratic has two roots $m_1$ and $m_2$, which are the slopes of the two tangents from $(1,4)$ to the parabola.

   Now we need $2(m_1^2 + m_2^2)$. Instead of solving for $m_1$ and $m_2$ individually, we use the relations from the quadratic:

   $m_1 + m_2 = \frac{16}{4} = 4, \quad m_1 m_2 = \frac{11}{4}.$

   Recall the identity:

   $m_1^2 + m_2^2 = (m_1 + m_2)^2 - 2m_1 m_2.$

   Substitute:

   $$$$

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10. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
    Q.If $f(x) \equiv x^2 + a x + 2 = 0$ and $g(x) \equiv x^2 + 2x + a = 0$ have only one real common root, then sum of the roots of $f(x) + g(x) = 0$ is (A) $\dfrac{-1}{2}$ (B) $0$ (C) $\dfrac{1}{2}$ (D) $1$

    ›Reveal solutionSolution

    Subtracting the equations forces the common root $x=1$, giving $a=-3$; then $f+g = 2x^2-x-1$, whose roots sum to $\tfrac12$.

    Let the two quadratics share a common root. Subtracting them eliminates $x^2$:

    $f(x) - g(x) = (ax + 2) - (2x + a) = (a-2)x - (a-2) = (a-2)(x-1) = 0.$

    So either $a = 2$ or $x = 1$. If $a = 2$ both equations become identical ($x^2+2x+2=0$), sharing both roots — but we are told there is only one common root, so $a \ne 2$ and the common root must be

    $x = 1.$

    Substitute $x=1$ into $f$: $1 + a + 2 = 0 \Rightarrow a = -3$ (and it also satisfies $g$: $1+2+a=0$). …

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11. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.$(1,1)$ is the vertex and $x+y+1=0$ is the directrix of a parabola. If $(a,b)$ is its focus and $(c,d)$ is the point of intersection of the directrix and the axis of the parabola, then $a+b+c+d=$ (A) $6$ (B) $5$ (C) $4$ (D) $3$

    ›Reveal solutionSolution

    The vertex and directrix fix the parabola’s axis and focus; using the vertex as the midpoint of the focus and the directrix’s foot on the axis gives the focus and foot coordinates, whose sum is 4.

    Concept & Intuition

    For any parabola, the vertex lies exactly halfway between the focus and the directrix along the axis. That means the vertex is the midpoint of the segment joining the focus to the point where the axis meets the directrix. If we know the vertex and the directrix line, we can find the axis (perpendicular to the directrix through the vertex), then find the foot of the vertex on the directrix, and finally use the midpoint property to locate the focus. Adding the coordinates of the focus and that foot gives the required sum.

    ---

    1. Find the axis of the parabola. The axis is the line through the vertex $(1,1)$ perpendicular to the directrix $x + y + 1 = 0$. The directrix has slope $-1$, so the axis has slope $1$. Equation of the axis:

    $y - 1 = 1(x - 1) \quad\Rightarrow\quad y = x.$

    2. Find the foot of the vertex on the directrix (point $(c,d)$). This is the intersection of the axis $y = x$ with the directrix $x + y + 1 = 0$. Substitute $y = x$:

    $x + x + 1 = 0 \quad\Rightarrow\quad 2x = -1 \quad\Rightarrow\quad x = -\frac12.$

    Then $y = -\frac12$. So

    $(c,d) = \left(-\frac12, -\frac12\right).$

    3. Use the vertex as the midpoint of the focus and this foot. Let the focus be $(a,b)$. The vertex $(1,1)$ is the midpoint of $(a,b)$ and $\left(-\frac12, -\frac12\right)$:

    $\frac{a + (-\frac12)}{2} = 1, \qquad \frac{b + (-\frac12)}{2} = 1.$

    Solve:

    $a - \frac12 = 2 \quad\Rightarrow\quad a = \frac52,$ …

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.The numerically greatest term in the expansion of $(3x-16y)^{15}$ when $x = \dfrac{2}{3}$ and $y = \dfrac{3}{2}$ is (A) 13th term (B) 14th term (C) 15th term (D) 16th term

    ›Reveal solutionSolution

    The numerically greatest term in the binomial expansion is found by comparing successive term ratios; for the given substitution, the 14th term is the largest, so the answer is option (B).

    We are asked for the numerically greatest term in the expansion of $(3x - 16y)^{15}$ when $x = \frac{2}{3}$ and $y = \frac{3}{2}$.

    The key idea: In a binomial expansion $(a + b)^n$, the terms increase in magnitude up to a point and then decrease. By examining the ratio of consecutive terms, we can locate where the maximum occurs — without computing all 16 terms.

    ---

    1. Substitute the given values and simplify the expression

    First, plug in $x = \frac{2}{3}$ and $y = \frac{3}{2}$:

    $3x = 3 \cdot \frac{2}{3} = 2, \quad 16y = 16 \cdot \frac{3}{2} = 24$

    So the expression becomes:

    $(3x - 16y)^{15} = (2 - 24)^{15} = (-22)^{15}$

    That’s just a single number — but wait: the expansion’s terms are not all equal; they come from the binomial expansion of $(3x - 16y)^{15}$ before substitution. We must substitute into the general term.

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    2. Write the general term

    The general term in $(A + B)^{15}$ is:

    $T_{r+1} = \binom{15}{r} A^{15-r} B^r$

    Here $A = 3x$ and $B = -16y$. So:

    $T_{r+1} = \binom{15}{r} (3x)^{15-r} (-16y)^r$

    Substitute $x = \frac{2}{3}$, $y = \frac{3}{2}$:

    $T_{r+1} = \binom{15}{r} (2)^{15-r} (-24)^r$

    So the magnitude (absolute value) is:

    $|T_{r+1}| = \binom{15}{r} \cdot 2^{15-r} \cdot 24^r$

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    3. Find the ratio of successive terms

    Let $t_r = |T_{r+1}|$. Then:

    $\frac{t_{r+1}}{t_r} = \frac{\binom{15}{r+1} 2^{14-r} 24^{r+1}}{\binom{15}{r} 2^{15-r} 24^r}$

    Simplify:

    $\frac{\binom{15}{r+1}}{\binom{15}{r}} = \frac{15 - r}{r+1}, \quad \frac{2^{14-r}}{2^{15-r}} = \frac{1}{2}, \quad \frac{24^{r+1}}{24^r} = 24$

    Thus:

    $\frac{t_{r+1}}{t_r} = \frac{15 - r}{r+1} \cdot \frac{24}{2} = \frac{15 - r}{r+1} \cdot 12$

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    4. Determine when terms increase or decrease

    Terms increase as long as $\frac{t_{r+1}}{t_r} > 1$:

    $\frac{15 - r}{r+1} \cdot 12 > 1 \quad \Rightarrow \quad 12(15 - r) > r + 1$

    $180 - 12r > r + 1 \quad \Rightarrow \quad 179 > 13r \quad \Rightarrow \quad r < \frac{179}{13} \approx 13.769$

    So for $r = 0, 1, \dots, 13$, the ratio is > 1, meaning terms increase up to $r = 13$.

    At $r = 13$: $\frac{t_{14}}{t_{13}} = \frac{15-13}{13+1} \cdot 12 = \frac{2}{14} \cdot 12 = \frac{24}{14} \approx 1.714 > 1$, so $t_{14} > t_{13}$.

    At $r = 14$: $\frac{t_{15}}{t_{14}} = \frac{15-14}{14+1} \cdot 12 = \frac{1}{15} \cdot 12 = 0.8 < 1$, so $t_{15} < t_{14}$. …

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