Q.Maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ is:
(A) $0$
(B) $12$
(C) $16$
(D) $32$

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 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.

   ---

   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:

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

   ---

   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:

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

   ---

   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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4. 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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5. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 4x^2 - 9x - 36 = 0$ such that $\alpha + \beta = 0$, then $\alpha^2 + 2\beta^2 + 3\gamma^2 =$ (A) 75 (B) 61 (C) 34 (D) 27

   ›Reveal solutionSolution

   Factoring the cubic gives roots $3,-3,-4$; the condition $\alpha+\beta=0$ forces $\{\alpha,\beta\}=\{3,-3\}$ and $\gamma=-4$, so $\alpha^2+2\beta^2+3\gamma^2=75$ — option (A).

   Factor $x^3+4x^2-9x-36$ by grouping:

   $x^3+4x^2-9x-36 = x^2(x+4)-9(x+4)=(x^2-9)(x+4).$

   Setting each factor to zero:

   $x^2-9=0\Rightarrow x=\pm3,\qquad x+4=0\Rightarrow x=-4.$ …

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6. 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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7. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.If the length and breadth of a rectangle of maximum area that can be inscribed in an ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ are $8\sqrt{2}$ and $4\sqrt{2}$ respectively, then the eccentricity of that ellipse is (A) $\dfrac{1}{2}$ (B) $\dfrac{\sqrt{3}}{2}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{1}{\sqrt{3}}$

   ›Reveal solutionSolution

   The maximum-area rectangle inscribed in an ellipse has sides parallel to the axes, with vertices at $(a\cos\theta, b\sin\theta)$. For a given rectangle area $4ab\sin\theta\cos\theta$, the maximum occurs at $\theta = 45^\circ$, giving sides $2a/\sqrt{2}$ and $2b/\sqrt{2}$. Equating these to the given dimensions yields $a$ and $b$, from which eccentricity $e = \sqrt{1 - b^2/a^2} = \frac{\sqrt{3}}{2}$.

   The problem gives the dimensions of the rectangle of maximum area that can be inscribed in an ellipse. The key insight is that for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, any inscribed rectangle with sides parallel to the axes has its vertices at $(\pm a\cos\theta, \pm b\sin\theta)$ for some $\theta$ in $(0, \pi/2)$. The side lengths are $2a\cos\theta$ (horizontal) and $2b\sin\theta$ (vertical), so the area is $4ab\sin\theta\cos\theta = 2ab\sin 2\theta$. This is maximized when $\sin 2\theta = 1$, i.e., $\theta = 45^\circ$. At this optimum, the sides become $2a/\sqrt{2}$ and $2b/\sqrt{2}$.

   Now we match these to the given dimensions: the length (longer side) is $8\sqrt{2}$ and the breadth (shorter side) is $4\sqrt{2}$. Since $a > b$ for an ellipse with horizontal major axis, the longer side corresponds to $2a/\sqrt{2}$ and the shorter to $2b/\sqrt{2}$.

   1. Set up the equations.

      Longer side: $\frac{2a}{\sqrt{2}} = 8\sqrt{2}$

      Shorter side: $\frac{2b}{\sqrt{2}} = 4\sqrt{2}$

   2. Solve for $a$ and $b$.

      From the first: $2a = 8\sqrt{2} \cdot \sqrt{2} = 8 \cdot 2 = 16$, so $a = 8$.

      From the second: $2b = 4\sqrt{2} \cdot \sqrt{2} = 4 \cdot 2 = 8$, so $b = 4$.

   3. Find the eccentricity.

      For an ellipse, $e = \sqrt{1 - \frac{b^2}{a^2}}$. …

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8. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the roots of the equation $32x^3 - 48x^2 + 22x - 3 = 0$ are in arithmetic progression, then the square of the common difference of the roots is (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{16}$ (C) $\dfrac{1}{9}$ (D) $\dfrac{1}{25}$

   ›Reveal solutionSolution

   For a cubic with roots in arithmetic progression, the middle root is the average of the three, which equals one-third the sum of the roots. Using Vieta’s formulas, we find the middle root, then the common difference, and square it to get $\frac{1}{16}$.

   We are told the roots of $32x^3 - 48x^2 + 22x - 3 = 0$ are in arithmetic progression. That means they can be written as $a-d$, $a$, $a+d$, where $a$ is the middle term and $d$ is the common difference. The key insight: when roots are equally spaced, the middle root is simply the average of the three roots, which is also one-third of their sum. Vieta’s formulas give us that sum directly from the coefficients, so we can find $a$ immediately. Then we can use another Vieta relation to solve for $d^2$.

   1. Write the roots in AP form

      Let the roots be $p-d$, $p$, $p+d$. Their sum is $(p-d) + p + (p+d) = 3p$.

   2. Use Vieta for the sum of roots

      For $32x^3 - 48x^2 + 22x - 3 = 0$, the sum of roots (with sign) is $-\frac{\text{coefficient of }x^2}{\text{coefficient of }x^3} = -\frac{-48}{32} = \frac{48}{32} = \frac{3}{2}$.

      So $3p = \frac{3}{2}$, giving $p = \frac{1}{2}$.

   3. Use Vieta for the sum of pairwise products

      The sum of products taken two at a time is $\frac{\text{coefficient of }x}{\text{coefficient of }x^3} = \frac{22}{32} = \frac{11}{16}$.

      In terms of $p$ and $d$:

      $(p-d)p + p(p+d) + (p-d)(p+d) = p(p-d) + p(p+d) + (p^2 - d^2)$.

      Simplify:

      $p^2 - pd + p^2 + pd + p^2 - d^2 = 3p^2 - d^2$.

      So $3p^2 - d^2 = \frac{11}{16}$.

   4. Substitute $p = \frac{1}{2}$

      $3\left(\frac{1}{2}\right)^2 - d^2 = \frac{11}{16}$

      $\Rightarrow 3 \cdot \frac{1}{4} - d^2 = \frac{11}{16}$ …

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9. 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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10. 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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11. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.If the focal distance of a point $P(2, y_1)$ on the parabola $y^2 = kx$ is $3$, then the equation of the tangent drawn at $P$ to the given parabola is (A) $x \pm 2\sqrt{2}y + 4 = 0$ (B) $x \pm 2\sqrt{2}y + 2 = 0$ (C) $x \pm \sqrt{2}y + 4 = 0$ (D) $x \pm \sqrt{2}y + 2 = 0$

    ›Reveal solutionSolution

    The focal distance condition gives the point’s coordinates; substituting into the parabola yields the parameter, then the tangent equation is derived and matched to the given options. The correct option is (A).

    Concept & Intuition

    For a parabola $y^2 = kx$, the focus is at $(\frac{k}{4}, 0)$. The focal distance of a point is its distance from the focus. Knowing this distance lets us find the unknown coordinate $y_1$ and the constant $k$. Once we have the point $P$ and the parabola, we can write the tangent equation using the standard formula for a tangent to $y^2 = 4ax$. The multiple-choice options suggest the tangent lines come in a pair (symmetric about the x‑axis), which matches the symmetry of the parabola.

    Step‑by‑step solution

    1. Identify the parabola’s standard form

       The given parabola is $y^2 = kx$. Compare with the standard form $y^2 = 4ax$.

       Hence $4a = k$ so $a = \frac{k}{4}$.

       The focus is at $(a, 0) = \left(\frac{k}{4}, 0\right)$.

    2. Use the focal distance condition

       Point $P$ is $(2, y_1)$. Its distance to the focus is given as 3:

    $\sqrt{\left(2 - \frac{k}{4}\right)^2 + (y_1 - 0)^2} = 3.$

    Square both sides:

    $\left(2 - \frac{k}{4}\right)^2 + y_1^2 = 9. \qquad(1)$

    3. Use that $P$ lies on the parabola Since $P(2, y_1)$ satisfies $y^2 = kx$, we have

    $y_1^2 = k \cdot 2 = 2k. \qquad(2)$

    4. Solve for $k$ Substitute (2) into (1):

    $\left(2 - \frac{k}{4}\right)^2 + 2k = 9.$

    Expand:

    $4 - k + \frac{k^2}{16} + 2k = 9 \quad\Rightarrow\quad 4 + k + \frac{k^2}{16} = 9.$

    Multiply by 16:

    $64 + 16k + k^2 = 144 \quad\Rightarrow\quad k^2 + 16k - 80 = 0.$

    Solve:

    $k = \frac{-16 \pm \sqrt{256 + 320}}{2} = \frac{-16 \pm \sqrt{576}}{2} = \frac{-16 \pm 24}{2}.$

    So $k = 4$ or $k = -20$.

    Since the focal distance is positive and the parabola opens to the right for positive $k$ (and the point has $x=2>0$), we take $k = 4$.

    (If $k = -20$, the parabola opens left and the focus would be at $(-5,0)$; the distance condition could still hold, but the tangent options given are symmetric and positive‑looking, so $k=4$ is the intended case.)

    5. Find $y_1$

       From (2): $y_1^2 = 2 \cdot 4 = 8$ so $y_1 = \pm 2\sqrt{2}$.

       Thus $P$ is $(2, 2\sqrt{2})$ or $(2, -2\sqrt{2})$.

    6. Write the tangent equation …

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12. 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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