Q.Find the maximum and minimum values, if any, of the following functions given by

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

   $f = 3a^6 + 4b^8.$

   Since $a, b \le 1$, raising to powers reduces them: $a^6 \le a$ and $b^8 \le b$.

   Therefore

   $f \le 3a + 4b.$

   Using $b = 1 - a$,

   $f \le 3a + 4(1 - a) = 4 - a.$

   Since $a \ge 0$, we get $f \le 4$, with equality only when $a = 0$ (i.e., $\sin^2 x = 0$). That’s exactly the case we already found.

   4. Conclusion The maximum possible value is 4, achieved when $\cos^2 x = 1$ and $\sin^2 x = 0$, e.g., at $x = 0$.

   Watch out

   A common mistake is to think that because each term can be at most 3 and 4 respectively, the sum can be 7. But the conditions for each maximum are mutually exclusive — they cannot happen at the same $x$.

   Tip

   For sums of powers of sine and cosine with positive coefficients, the maximum often occurs at an endpoint where one of them is 0 and the other is 1, because the exponents make the functions concave.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

2. 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. So the inequality holds for $y \leq -1$ or $y \geq \frac{7}{9}$.

   6. Check if the endpoints are actually attained.

      • At $y = -1$: plug into the quadratic: $(-1 - 1)x^2 + (-1 + 1)x - 2(-1 + 1) = -2x^2 = 0 \Rightarrow x = 0$, which is in the domain. So $y = -1$ is in the range.
      • At $y = \frac{7}{9}$: plug in and solve — you’ll get a real $x$ (specifically $x = -5$), so it’s attained.
      • Also recall $y = 1$ is attained (from step 3), and $1$ lies in $[\frac{7}{9}, \infty)$, so it’s already covered.

   7. Combine the intervals.

      The range is $(-\infty, -1] \cup \left[\frac{7}{9}, \infty\right)$.

   Watch out

   A common mistake is to forget to check the $y = 1$ case separately, since the quadratic degenerates. Also, always verify that the endpoints are actually reachable with some $x$ in the domain — sometimes they aren’t, and that changes the answer.

   ✓Final answer

   The correct option is (C) $(-\infty, -1] \cup \left[\frac{7}{9}, \infty\right)$.

3. 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 $f'(c)=0$.

   $c=\frac{28\pm\sqrt{28^2+4\cdot72\cdot11}}{2\cdot72}=\frac{28\pm\sqrt{784+3168}}{144}=\frac{28\pm\sqrt{3952}}{144}.$

   Since $\sqrt{3952}=4\sqrt{247}$,

   $c=\frac{28\pm4\sqrt{247}}{144}=\frac{7\pm\sqrt{247}}{36}.$

   4. Select the root inside $\left(\tfrac12,\tfrac34\right)$. With $\sqrt{247}\approx15.72$,

   $\frac{7+\sqrt{247}}{36}\approx0.63\in(0.5,0.75),\qquad \frac{7-\sqrt{247}}{36}\approx-0.24\notin(0.5,0.75).$

   Only the positive-radical root qualifies.

   ✓Final answer

   $c=\dfrac{7+\sqrt{247}}{36}$. Correct option: (D).

   ANSWER: D

4. 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 $g$: $1+2+a=0$).

   Now form $f(x)+g(x)$ with $a=-3$:

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

   Sum of its roots:

   $-\frac{\text{coeff of }x}{\text{coeff of }x^2} = -\frac{-1}{2} = \frac{1}{2}.$

   ✓Final answer

   Sum of the roots of $f(x)+g(x)=0$ is $\dfrac{1}{2}$ — option (C).

5. 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 $mx - y + \frac{2}{m} = 0$. Multiply through by $m$ to match the constant term:

   $m^2 x - m y + 2 = 0.$

   So we have $a = m^2$, $b = -m$.

   Then $-\frac{a}{b} = -\frac{m^2}{-m} = m$.

   The condition $-\frac{a}{b} > 0$ means $m > 0$, so $m = 1$.

   5. Find $a$ and $b$ With $m = 1$:

   $a = m^2 = 1,\quad b = -m = -1.$

   The tangent line is $x - y + 2 = 0$.

   6. Compute the required expression

   $3a^2 + 2b + 1 = 3(1)^2 + 2(-1) + 1 = 3 - 2 + 1 = 2.$

   Watch out

   A common mistake is to forget that the constant term in the given line is $+2$, so when converting $y = mx + \frac{2}{m}$ to the form $ax + by + 2 = 0$, you must multiply through by $m$ to get the constant exactly $2$, not leave it as $\frac{2}{m}$.

   Tip

   The condition $-\frac{a}{b} > 0$ directly gives the sign of the slope $m$, saving you from checking both possibilities later.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

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. A quadratic with roots $r_1, r_2$ is $x^2 - (r_1 + r_2)x + r_1 r_2 = 0$. Substituting:

   $x^2 - \frac{3}{4}x + \frac{1}{16} = 0.$

   Multiply through by 16 to clear denominators:

   $16x^2 - 12x + 1 = 0.$

   Tip

   Notice that $\sin^2 18^\circ$ and $\cos^2 36^\circ$ are conjugates of the form $\frac{3 \pm \sqrt{5}}{8}$, so their sum and product are rational — that’s the elegant shortcut.

   Watch out

   A common mistake is to confuse $\cos 36^\circ$ with $\sin 54^\circ$ and mis-square. Always double-check the exact forms: $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$, $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

7. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.In the Binomial expansion of $(1+x)^{2k}$, if its middle term is the only numerically greatest term, then $x$ lies in the interval (A) $(-2k\ \ 2k)$ (B) $\left(-\dfrac{k+1}{k}\ \ \dfrac{k+1}{k}\right)$ (C) $(-k\ \ k)$ (D) \left(- (k+1)\ \ (k+1)\right)

   ›Reveal solutionSolution

   The middle term $T_{k+1}$ is the numerically greatest only when $|x| < \dfrac{k+1}{k}$ — option (B).

   For $(1+x)^{2k}$ the general term is $T_{r+1} = \binom{2k}{r}x^r$, and the middle term (of the $2k+1$ terms) is $T_{k+1}$ (i.e. $r=k$).

   The middle term is numerically greatest precisely when it dominates its neighbours. The binding condition is that it exceeds the following term:

   $\left|\frac{T_{k+2}}{T_{k+1}}\right| = \frac{2k-k}{k+1}\,|x| = \frac{k}{k+1}\,|x| < 1 \implies |x| < \frac{k+1}{k}.$

   Hence $x \in \left(-\dfrac{k+1}{k},\ \dfrac{k+1}{k}\right)$.

   ✓Final answer

   $x \in \left(-\dfrac{k+1}{k},\ \dfrac{k+1}{k}\right)$ — option (B).

8. 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

   $r^2 = 144 - \frac{(8\sqrt{3})^2}{4} = 144 - \frac{192}{4} = 144 - 48 = 96.$

   So $r = \sqrt{96} = 4\sqrt{6}$.

   5. Compute the maximum volume

   $V_{\text{max}} = \pi r^2 h = \pi \cdot 96 \cdot 8\sqrt{3} = 768\sqrt{3}\pi.$

   Tip

   Notice that the optimal height is $h = \frac{2R}{\sqrt{3}}$ for a sphere of radius $R$. Here $R=12$ gives $h = \frac{24}{\sqrt{3}} = 8\sqrt{3}$, and the volume formula becomes $V_{\text{max}} = \frac{4\pi R^3}{3\sqrt{3}}$. Plugging $R=12$ yields $768\sqrt{3}\pi$ directly.

   Watch out

   A common mistake is to forget that the cylinder’s half-height appears in the Pythagorean relation, leading to an incorrect link between $r$ and $h$. Always draw the cross-section to avoid this.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. 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:

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

   4. Expand and simplify the equation:

      Expand both sides of the equation:

      Left Hand Side (LHS): $(y^2 - 21)^2 = (y^2)^2 - 2(y^2)(21) + (21)^2 = y^4 - 42y^2 + 441$.

      Right Hand Side (RHS): $y (2y - 6)^2 = y [(2y)^2 - 2(2y)(6) + 6^2] = y (4y^2 - 24y + 36)$.

      RHS: $y (4y^2 - 24y + 36) = 4y^3 - 24y^2 + 36y$.

      Equate the expanded LHS and RHS:

      $y^4 - 42y^2 + 441 = 4y^3 - 24y^2 + 36y$

      Move all terms to one side to form a standard polynomial equation:

      $y^4 - 4y^3 - 42y^2 + 24y^2 - 36y + 441 = 0$

      Combine the like terms:

      $y^4 - 4y^3 + (-42 + 24)y^2 - 36y + 441 = 0$

      $y^4 - 4y^3 - 18y^2 - 36y + 441 = 0$

   5. Compare with the given options:

      The derived equation is $x^4 - 4x^3 - 18x^2 - 36x + 441 = 0$ (using $x$ as the variable for the new polynomial, as is common in options). This matches option (A).

   ✓Final answer

   The equation whose roots are squares of the roots of $x^4 - 2x^3 + 6x - 21 = 0$ is $\boxed{x^4 - 4x^3 - 18x^2 - 36x + 441 = 0}$.

10. 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:

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

    So the roots are $3,\,-3,\,-4$. The condition $\alpha+\beta=0$ means $\alpha$ and $\beta$ are the opposite pair $3$ and $-3$ (either order), leaving $\gamma=-4$.

    Compute the required expression. Since $\alpha^2=\beta^2=9$ and $\gamma^2=16$:

    $\alpha^2+2\beta^2+3\gamma^2 = 9+2(9)+3(16)=9+18+48=75.$

    ✓Final answer

    $\alpha^2+2\beta^2+3\gamma^2=75$ — option (A).

11. 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, $e = \sqrt{1 - \frac{b^2}{a^2}}$.

       Here $b^2 = 16$, $a^2 = 64$, so $\frac{b^2}{a^2} = \frac{16}{64} = \frac{1}{4}$.

       Hence $e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

    Watch out

    A common mistake is to forget that the maximum-area rectangle occurs at $\theta = 45^\circ$ only when sides are parallel to the axes. If the rectangle is rotated, the problem becomes far more complex — but here, the standard result applies directly.

    Tip

    Notice that the given dimensions $8\sqrt{2}$ and $4\sqrt{2}$ already contain the factor $\sqrt{2}$, which hints at the $\theta = 45^\circ$ optimum. This can be a quick sanity check.

    ✓Final answer

    The eccentricity of the ellipse is $\boxed{\frac{\sqrt{3}}{2}}$, which corresponds to option (B).

12. 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 $\beta$.

       Since $\alpha$ is the smaller endpoint and $\beta$ the larger, we have $\alpha = -2$ and $\beta = 7$. The interval $[\alpha, \beta]$ is $[-2, 7]$, and indeed for $x$ outside this interval the inequality $x^2 - 5x - 14 > 0$ holds.

    3. Compute the required ratio.

    $\frac{\alpha}{\beta} = \frac{-2}{7}$

    Watch out

    A common mistake is to mix up which root is $\alpha$ and which is $\beta$. Remember: $\alpha$ is the smaller number, so here $\alpha = -2$, not $7$. Swapping them gives $\frac{7}{-2} = -\frac{7}{2}$, which is option (B) — a tempting distractor.

    ✓Final answer

    The value is $\boxed{-\frac{2}{7}}$, which corresponds to option (A).