Q.What is the maximum value of the function $\sin x+\cos x$?

Showing the 12 most recent of 21 on this concept.

1. 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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2. 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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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 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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5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $7\sin x + 15\sin y = 17$, then the maximum value of $7\cos x + 15\cos y$ is (A) $\sqrt{190}$ (B) $\sqrt{195}$ (C) $\sqrt{200}$ (D) $\sqrt{205}$

   ›Reveal solutionSolution

   Treat the two constraints as components of a vector sum of fixed-length vectors; the maximum possible resultant length is $7+15=22$, giving a maximum $x$-component of $\sqrt{22^2-17^2}=\sqrt{195}$.

   Concept and Intuition

   Expressions like $7\cos x+15\cos y$ and $7\sin x+15\sin y$ are the components of the vector sum of two vectors with fixed magnitudes $7$ and $15$ but free, independent directions $x$ and $y$. As $x,y$ vary independently over all angles, this resultant vector can point in any direction with any magnitude between $|15-7|=8$ and $15+7=22$ (the usual triangle-inequality range for summing two vectors). This converts a trigonometric optimization into simple 2D geometry.

   Step-by-Step Solution

   1. Let $u=(7\cos x,\,7\sin x)$ and $v=(15\cos y,\,15\sin y)$, so $|u|=7,\,|v|=15$ always, regardless of $x,y$.
   2. Their sum $S=u+v=(7\cos x+15\cos y,\ 7\sin x+15\sin y)$ has $y$-component fixed at $17$ by the given condition, and we want to maximize its $x$-component.
   3. For any target direction, $|S|$ can be made anywhere in $[8,22]$ by choosing the angle between $u$ and $v$ appropriately (rotating both $x,y$ together lets $S$ point any direction for a given magnitude).
   4. Since $S_x = \sqrt{|S|^2-S_y^2}$ (when $S_x\ge0$), and $S_y=17$ is fixed, $S_x$ is maximized by taking $|S|$ as large as possible, i.e., $|S|=22$ (achieved when $u,v$ are parallel, $x=y$). …

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6. 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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7. 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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8. 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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9. 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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10. 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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11. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.If $u = \sqrt{a^2\cos^2\theta + b^2\sin^2\theta} + \sqrt{a^2\sin^2\theta + b^2\cos^2\theta}$, then the difference between the maximum and minimum values of $u^2$ is (A) $(a + b)^2$ (B) $(a - b)^2$ (C) $2\sqrt{a^2 + b^2}$ (D) $2\sqrt{a^2 - b^2}$

    ›Reveal solutionSolution

    Express $u^2$ in terms of $\sin^2 2\theta$, find its max and min, and subtract; the difference is $(a-b)^2$.

    Concept and Intuition

    $u$ is a sum of two square roots whose radicands add to the constant $a^2+b^2$. So $u^2=(a^2+b^2)+2\sqrt{fg}$, and everything about the variation of $u^2$ with $\theta$ is carried by the product $fg$ of the two radicands — we just need its range.

    Step-by-Step Solution

    1. Let $f=a^2\cos^2\theta+b^2\sin^2\theta$ and $g=a^2\sin^2\theta+b^2\cos^2\theta$. Then $f+g=a^2+b^2$ (constant), and

    $u^2=f+g+2\sqrt{fg}=(a^2+b^2)+2\sqrt{fg}.$

    2. Compute $fg$:

    $fg=a^4\sin^2\theta\cos^2\theta+a^2b^2\cos^4\theta+a^2b^2\sin^4\theta+b^4\sin^2\theta\cos^2\theta.$

    Using $\sin^4\theta+\cos^4\theta=1-\tfrac12\sin^2 2\theta$ and $\sin^2\theta\cos^2\theta=\tfrac14\sin^2 2\theta$:

    $fg=a^2b^2+\frac{(a^2-b^2)^2}{4}\sin^2 2\theta.$

    3. So $fg$ ranges from $a^2b^2$ (when $\sin^2 2\theta=0$) to $a^2b^2+\dfrac{(a^2-b^2)^2}{4}$ (when $\sin^2 2\theta=1$).
    4. Hence

    $u^2_{\max}=(a^2+b^2)+2\sqrt{a^2b^2+\frac{(a^2-b^2)^2}{4}},\qquad u^2_{\min}=(a^2+b^2)+2ab.$

    5. The difference is $u^2_{\max}-u^2_{\min}=2\left[\sqrt{a^2b^2+\frac{(a^2-b^2)^2}{4}}-ab\right].$ …

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12. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.The number of solutions of $\sin 2x + \cos 4x = 2$ in the interval $[-\pi, \pi]$ is (A) 3 (B) 2 (C) 0 (D) 1

    ›Reveal solutionSolution

    Since both $\sin2x$ and $\cos4x$ are individually bounded above by $1$, their sum can reach $2$ only if each equals $1$ at the same $x$ — but that is algebraically impossible here, so there are $0$ solutions.

    Concept and Intuition

    When an equation is a sum of two bounded quantities set equal to the sum of their individual maximums, the only way to satisfy it is for both to hit their maximum simultaneously. This is a much faster route than trying to solve the trigonometric equation directly.

    Step-by-Step Solution

    1. For all real $x$: $\sin2x\in[-1,1]$ and $\cos4x\in[-1,1]$, so $\sin2x+\cos4x\le 2$, with equality only if $\sin2x=1$ AND $\cos4x=1$ simultaneously.
    2. Suppose $\sin2x=1$. Then using $\cos4x=1-2\sin^22x$, we get $\cos4x=1-2(1)^2=1-2=-1$.
    3. But we need $\cos4x=1$, and we just showed $\cos4x=-1$ whenever $\sin2x=1$ — a direct contradiction. …

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