Q.At $x = \dfrac{5\pi}{6}$, $f(x) = 2\sin 3x + 3\cos 3x$ is:
(A) maximum
(B) minimum
(C) zero
(D) neither maximum nor minimum

Showing the 12 most recent of 21 on this concept.

1. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.An extreme value of $f(x)=\dfrac{4}{\sin x}+\dfrac{1}{1-\sin x}$ in $(0,\pi/2)$ is (A) 9 (B) 8 (C) $2/3$ (D) $-7/2$

   ›Reveal solutionSolution

   Substituting $s=\sin x$ turns this into a single-variable optimization; the unique interior critical point at $s=2/3$ gives the extreme (minimum) value $9$.

   Concept and Intuition

   Since $f$ depends on $x$ only through $s=\sin x$, and $\sin x$ ranges over $(0,1)$ for $x\in(0,\pi/2)$, we can optimize the simpler function $g(s)=4/s+1/(1-s)$ on $(0,1)$ directly.

   Step-by-Step Solution

   1. Let $s=\sin x\in(0,1)$, $g(s)=\dfrac4s+\dfrac1{1-s}$.
   2. $g'(s)=-\dfrac4{s^2}+\dfrac1{(1-s)^2}$. Setting $g'(s)=0$: $\dfrac1{(1-s)^2}=\dfrac4{s^2}\Rightarrow s^2=4(1-s)^2\Rightarrow s=\pm2(1-s)$.
   3. Taking $s=2(1-s)$ gives $3s=2\Rightarrow s=2/3$ (in range); the other sign gives $s=2$, rejected.
   4. As $s\to0^+$ or $s\to1^-$, $g(s)\to\infty$, so the single interior critical point $s=2/3$ must be a minimum. …

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2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $A$ and $B$ are the minimum and maximum values of $\sin^6 x + \cos^6 x$, then $A + B =$ (A) $1$ (B) $-1$ (C) $\dfrac{5}{4}$ (D) $\dfrac{7}{4}$

   ›Reveal solutionSolution

   Reduce $\sin^6x+\cos^6x$ to a single term in $\sin 2x$ to read off its min and max directly.

   Concept and Intuition

   $\sin^6x+\cos^6x$ looks like a hard sixth-degree expression, but it is a sum of cubes: $(\sin^2x)^3+(\cos^2x)^3$. Using $u^3+v^3=(u+v)^3-3uv(u+v)$ with $u=\sin^2x,\,v=\cos^2x$ (so $u+v=1$) collapses it to something only in $\sin x\cos x$, which is itself periodic and bounded — the whole problem becomes a one-variable range question.

   Step-by-Step Solution

   1. Write $\sin^6x+\cos^6x = (\sin^2x+\cos^2x)^3 - 3\sin^2x\cos^2x(\sin^2x+\cos^2x)$.
   2. Since $\sin^2x+\cos^2x=1$, this simplifies to $1-3\sin^2x\cos^2x$.
   3. Use $\sin x\cos x=\dfrac{1}{2}\sin 2x$, so $3\sin^2x\cos^2x = 3\cdot\dfrac14\sin^2 2x = \dfrac34\sin^2 2x$.
   4. So the expression is $f(x)=1-\dfrac34\sin^2 2x$.
   5. As $\sin^2 2x$ ranges over $[0,1]$, $f(x)$ ranges over $\left[1-\dfrac34,\ 1\right] = \left[\dfrac14,\,1\right]$. …

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3. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The maximum value of $12\sin x - 5\cos x + 3$ is (A) 18 (B) 13 (C) 16 (D) 10

   ›Reveal solutionSolution

   The maximum value of $12\sin x-5\cos x+3$ is 16, using the amplitude formula for $a\sin x+b\cos x$.

   Concept and Intuition

   Any expression of the form $a\sin x+b\cos x$ can be written as $R\sin(x+\phi)$ where $R=\sqrt{a^2+b^2}$, so its maximum value is $R$ and minimum is $-R$. Adding a constant just shifts this range.

   Step-by-Step Solution

   1. Write $12\sin x - 5\cos x = R\sin(x-\phi)$ where $R=\sqrt{12^2+(-5)^2}=\sqrt{144+25}=\sqrt{169}=13$.
   2. So $12\sin x-5\cos x$ ranges over $[-13,13]$, with maximum $13$.
   3. Adding the constant $+3$: the maximum of $12\sin x-5\cos x+3$ is $13+3=16$. …

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4. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.The number of all the values of $x$ for which the function $f(x) = \sin x + \dfrac{1-\tan^2 x}{1+\tan^2 x}$ attains its maximum value on $[0, 2\pi]$ is (A) $4$ (B) $1$ (C) $2$ (D) infinite

   ›Reveal solutionSolution

   Rewrite $f(x)$ purely in terms of $\sin x$, maximise the resulting quadratic, and count how many $x$ in $[0,2\pi]$ give that value of $\sin x$.

   Concept and Intuition

   The expression $\dfrac{1-\tan^2x}{1+\tan^2x}$ is exactly the double-angle identity $\cos2x$. Substituting $\cos2x=1-2\sin^2x$ turns $f$ into a simple quadratic in $s=\sin x$, which is easy to maximise using calculus/vertex formula.

   Step-by-Step Solution

   1. $\dfrac{1-\tan^2x}{1+\tan^2x}=\cos2x$ (standard identity, valid wherever $\tan x$ is defined, i.e. $\cos x\ne0$).
   2. $f(x)=\sin x+\cos2x=\sin x+(1-2\sin^2x)=-2\sin^2x+\sin x+1$.
   3. Let $s=\sin x\in[-1,1]$, $g(s)=-2s^2+s+1$. $g'(s)=-4s+1=0\Rightarrow s=\dfrac14$; since $g''(s)=-4<0$, this is a maximum.
   4. $g(1/4)=-2\left(\dfrac1{16}\right)+\dfrac14+1=-\dfrac18+\dfrac14+1=\dfrac98$. …

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5. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.The maximum value of $f(x)=\sin(x)$ in the interval $\left[\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right]$ is ______. (A) $0$ (B) $-1$ (C) $1$ (D) $\sqrt{2}$

   ›Reveal solutionSolution

   $\sin x$ is monotonically increasing on $[-\pi/2,\pi/2]$, so its max is at $x=\pi/2$. Answer: $1$.

   Concept and Intuition

   On $\left[-\frac\pi2,\frac\pi2\right]$, $\cos x\ge0$ so $\sin x$ has non-negative derivative throughout, meaning $\sin x$ is increasing on the whole interval — its maximum is simply its value at the right endpoint.

   Step-by-Step Solution

   1. $f'(x)=\cos x\ge0$ for all $x\in\left[-\frac\pi2,\frac\pi2\right]$, so $f$ is increasing (non-decreasing) throughout.
   2. Maximum occurs at $x=\frac\pi2$ (the right endpoint). …

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6. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.The value of $5\cos\theta + 3\cos\left(\theta + \dfrac{\pi}{3}\right) + 3$ lies between (A) $-2$ and 5 (B) $-1$ and 8 (C) $-3$ and 6 (D) $-4$ and 10

   ›Reveal solutionSolution

   This tests reducing $a\cos\theta+b\cos(\theta+\alpha)$ to a single sinusoid $R\cos(\theta+\phi)$ to read off its range; the range is $[-4,10]$.

   Concept and Intuition

   Any expression of the form $p\cos\theta+q\sin\theta$ has range $[-R,R]$ where $R=\sqrt{p^2+q^2}$, regardless of $\phi$. Expanding the shifted cosine term first converts the whole expression into this standard form.

   Step-by-Step Solution

   1. Expand: $3\cos\left(\theta+\dfrac{\pi}{3}\right)=3\left[\cos\theta\cos\dfrac{\pi}{3}-\sin\theta\sin\dfrac{\pi}{3}\right]=\dfrac32\cos\theta-\dfrac{3\sqrt3}{2}\sin\theta$.
   2. Add to $5\cos\theta$: total cosine coefficient $=5+\dfrac32=\dfrac{13}{2}$, sine coefficient $=-\dfrac{3\sqrt3}{2}$.
   3. So expression $=\dfrac{13}{2}\cos\theta-\dfrac{3\sqrt3}{2}\sin\theta+3$.
   4. Amplitude $R=\sqrt{\left(\dfrac{13}{2}\right)^2+\left(\dfrac{3\sqrt3}{2}\right)^2}=\sqrt{\dfrac{169}{4}+\dfrac{27}{4}}=\sqrt{\dfrac{196}{4}}=\sqrt{49}=7$. …

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7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\sqrt{3}\cos\theta+\sin\theta>0$ then (A) $-\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}$ (B) $-\dfrac{\pi}{3}<\theta<\dfrac{2\pi}{3}$ (C) $-\dfrac{2\pi}{3}<\theta<\dfrac{\pi}{3}$ (D) $-\dfrac{\pi}{6}<\theta<\dfrac{5\pi}{6}$

   ›Reveal solutionSolution

   This tests converting a linear combination of sine and cosine into a single sinusoid to solve an inequality. Answer: $-\frac{\pi}{3}<\theta<\frac{2\pi}{3}$.

   Concept and Intuition

   Any expression $a\cos\theta+b\sin\theta$ can be written as $R\sin(\theta+\phi)$ where $R=\sqrt{a^2+b^2}$ and $\phi$ is chosen so $R\cos\phi=1$ (coefficient of $\sin\theta$ inside becomes 1... more precisely matching the expansion $R\sin(\theta+\phi)=R\sin\theta\cos\phi+R\cos\theta\sin\phi$). This lets us reduce a two-term trig inequality to a single sine inequality with a known solution interval.

   Step-by-Step Solution

   1. Write $\sin\theta+\sqrt3\cos\theta = R\sin(\theta+\phi)=R\sin\theta\cos\phi+R\cos\theta\sin\phi$.
   2. Match: $R\cos\phi=1$, $R\sin\phi=\sqrt3$. So $R=\sqrt{1+3}=2$ and $\tan\phi=\sqrt3\Rightarrow\phi=\dfrac{\pi}{3}$.
   3. So the expression equals $2\sin\left(\theta+\dfrac{\pi}{3}\right)$. …

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8. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.The range of the real valued function $f(x) = \dfrac{15}{3\sin x+4\cos x+10}$ is (A) $[0,3]$ (B) $[-1,3]$ (C) $[1,3]$ (D) $[-1,1]$

   ›Reveal solutionSolution

   Bound the linear combination $3\sin x+4\cos x$ by its amplitude $5$, then invert the resulting bounded, always-positive denominator to get the range of $f$.

   Concept and Intuition

   Any expression of the form $a\sin x+b\cos x$ can be written as $R\sin(x+\varphi)$ with $R=\sqrt{a^2+b^2}$, so it continuously sweeps the entire interval $[-R,R]$ as $x$ varies over all reals. Here $R=\sqrt{3^2+4^2}=5$. Once we know the denominator's exact range (and that it never touches zero, so $f$ is defined for all real $x$), inverting a positive quantity that ranges over $[m,M]$ (with $0<m\le M$) gives a reciprocal-type function whose range is $[c/M, c/m]$ for a positive constant $c$ — reciprocation reverses order but keeps the interval closed and connected because the denominator itself is continuous.

   Step-by-Step Solution

   1. $3\sin x+4\cos x$ has amplitude $R=\sqrt{9+16}=\sqrt{25}=5$, so its range is $[-5,5]$.
   2. Denominator: $D(x)=3\sin x+4\cos x+10 \in [10-5,\,10+5]=[5,15]$.
   3. Since $D(x)>0$ always, $f(x)=15/D(x)$ is well-defined and continuous everywhere.
   4. $f$ is a strictly decreasing function of $D$ (as $D$ increases from 5 to 15, $15/D$ decreases from $15/5=3$ to $15/15=1$). …

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9. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If $\alpha$ is the maximum value and $\beta$ is the minimum value of $\cos^2\dfrac{x}{4} + \sin\dfrac{x}{4}$, $x \in R$, then $\alpha - \beta =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{9}{4}$ (C) $2$ (D) $3$

   ›Reveal solutionSolution

   This tests converting a trig expression into a quadratic in $\sin(x/4)$ and finding its max/min over the valid range. The answer is $\alpha-\beta=9/4$.

   Concept and Intuition

   $\cos^2\theta+\sin\theta$ is a quadratic in $t=\sin\theta$ once we replace $\cos^2\theta=1-\sin^2\theta$. Since $t$ is restricted to $[-1,1]$, the max/min of the quadratic must be found over that closed interval — checking both the vertex (if it lies inside the interval) and the endpoints.

   Step-by-Step Solution

   1. Let $t=\sin(x/4)$, so $t\in[-1,1]$ as $x$ ranges over $\mathbb R$.
   2. $\cos^2(x/4)+\sin(x/4)=(1-t^2)+t=-t^2+t+1=f(t)$.
   3. $f(t)$ is a downward parabola; its vertex is at $t=\dfrac{-1}{2(-1)}=\dfrac12$, which lies in $[-1,1]$.
   4. $f(1/2)=-\dfrac14+\dfrac12+1=\dfrac54$. Since the parabola opens downward, this is the maximum: $\alpha=\dfrac54$. …

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10. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.If $M_1$ and $M_2$ are the maximum values of $\dfrac{1}{11\cos 2x + 60\sin 2x + 69}$ and $3\cos^2 5x + 4\sin^2 5x$ respectively, then $\dfrac{M_1}{M_2} =$ (A) $\dfrac{65}{2}$ (B) $\dfrac{1}{32}$ (C) $\dfrac{8}{3}$ (D) $2$

    ›Reveal solutionSolution

    $M_1=1/8$ (reciprocal of the denominator's minimum, using the amplitude $\sqrt{11^2+60^2}=61$) and $M_2=4$; the ratio is $\dfrac{1}{32}$.

    Concept and Intuition

    Maximizing $\dfrac1{D(x)}$ (with $D>0$) is the same as minimizing $D(x)$; and $a\cos\theta+b\sin\theta$ has minimum $-\sqrt{a^2+b^2}$. For $M_2$, use $\cos^2+\sin^2=1$ to rewrite the expression with a single trig-squared term.

    Step-by-Step Solution

    1. $11\cos2x+60\sin2x$ has amplitude $\sqrt{11^2+60^2}=\sqrt{121+3600}=\sqrt{3721}=61$, so its minimum value is $-61$.
    2. Minimum of the denominator $=-61+69=8$ (positive, so this indeed gives the maximum of the reciprocal).
    3. $M_1=\dfrac18$. …

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11. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.The range of $\dfrac{1}{\sin^2 x + 3\sin x \cos x + 5\cos^2 x}$ is (A) $\left[2, \dfrac{11}{2}\right]$ (B) $\left[\dfrac{1}{2}, \dfrac{11}{2}\right]$ (C) $\left[\dfrac{2}{11}, \dfrac{1}{2}\right]$ (D) $\left[\dfrac{2}{11}, 2\right]$

    ›Reveal solutionSolution

    Converting the denominator into double-angle form $3+2\cos2x+1.5\sin2x$ and using the amplitude bound gives $f\in[0.5,5.5]$, so $1/f\in[2/11,2]$. Answer: (D).

    Concept and Intuition

    $A+B\cos\theta+C\sin\theta$ oscillates between $A-\sqrt{B^2+C^2}$ and $A+\sqrt{B^2+C^2}$.

    Step-by-Step Solution

    1. $f(x) = \sin^2x+3\sin x\cos x+5\cos^2x = 1 + 4\cos^2x + 3\sin x\cos x$.
    2. Use double-angle forms: $f(x) = 1+2(1+\cos2x)+1.5\sin2x = 3+2\cos2x+1.5\sin2x$.
    3. Amplitude of oscillating part: $\sqrt{2^2+1.5^2}=\sqrt{6.25}=2.5$.
    4. $f(x) \in [0.5, 5.5]$. …

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12. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.The sides of a triangle inscribed in a given circle subtend angles $\alpha, \beta, \gamma$ at the center. The minimum value of the A.M. of $\cos\left(\alpha + \dfrac{\pi}{2}\right)$, $\cos\left(\beta + \dfrac{\pi}{2}\right)$ and $\cos\left(\gamma + \dfrac{\pi}{2}\right)$ is equal to (A) $\dfrac{\sqrt{3}}{2}$ (B) $\dfrac{-\sqrt{3}}{2}$ (C) $\dfrac{-2}{\sqrt{3}}$ (D) $\sqrt{2}$

    ›Reveal solutionSolution

    The A.M. simplifies to $-\frac13(\sin\alpha+\sin\beta+\sin\gamma)$; its minimum (most negative) value occurs when the sum of sines is maximum, which by symmetry/Jensen happens at $\alpha=\beta=\gamma=\frac{2\pi}{3}$, giving $-\frac{\sqrt3}{2}$.

    Concept and Intuition

    The central angles subtended by the three sides of a triangle inscribed in a circle always add up to the full angle $2\pi$ (they partition the circle). The quantity asked for is a co-function of these angles' sines, so the problem reduces to optimizing $\sin\alpha+\sin\beta+\sin\gamma$ under the fixed-sum constraint $\alpha+\beta+\gamma=2\pi$ — a textbook use of concavity/Jensen's inequality (or Lagrange multipliers), which is maximized at equal angles for a concave function.

    Step-by-Step Solution

    1. Use the identity $\cos\left(x+\dfrac{\pi}2\right)=-\sin x$ to rewrite each term.
    2. A.M. $= \dfrac{-\sin\alpha-\sin\beta-\sin\gamma}{3} = -\dfrac{\sin\alpha+\sin\beta+\sin\gamma}{3}$.
    3. We want the minimum of this A.M., which (because of the leading minus sign) corresponds to the maximum of $S=\sin\alpha+\sin\beta+\sin\gamma$ subject to $\alpha+\beta+\gamma=2\pi$, $\alpha,\beta,\gamma>0$. …

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