Q.Show that $f(x) = \tan^{-1}(\sin x + \cos x)$ is an increasing function in $\left(0, \dfrac{\pi}{4}\right)$.

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The range of the real valued function $f(x) = \sin^{-1}\left(\sqrt{x^2 + x + 1}\right)$ is (A) $\left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]$ (B) $\left[ 0, \dfrac{\pi}{2} \right]$ (C) $\left[ \dfrac{\pi}{6}, \dfrac{\pi}{2} \right]$ (D) $\left[ \dfrac{\pi}{3}, \dfrac{\pi}{2} \right]$

   ›Reveal solutionSolution

   The key is that the argument of $\sin^{-1}$ must lie in $[-1,1]$, and the quadratic inside the square root has a minimum of $\frac34$, so the range of $\sqrt{x^2+x+1}$ is $[\frac{\sqrt3}{2}, \infty)$, but only $[\frac{\sqrt3}{2}, 1]$ is valid for $\sin^{-1}$. Hence the output range is $[\frac{\pi}{3}, \frac{\pi}{2}]$.

   Concept and intuition:

   The function $f(x) = \sin^{-1}\bigl(\sqrt{x^2 + x + 1}\bigr)$ is an inverse sine of a square root. Inverse sine only accepts inputs in $[-1, 1]$, and its output is in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. But here the input is a square root, so it’s always non‑negative. That already restricts the output to $[0, \frac{\pi}{2}]$. The real question is: what are the actual possible values of $\sqrt{x^2 + x + 1}$ as $x$ runs over all real numbers? That will determine the exact subinterval of $[0, \frac{\pi}{2}]$ that $f$ can hit.

   1. Find the range of the quadratic inside the square root. The expression $x^2 + x + 1$ is a quadratic with positive leading coefficient. Its minimum occurs at $x = -\frac{1}{2}$:

   $\left(-\frac12\right)^2 + \left(-\frac12\right) + 1 = \frac14 - \frac12 + 1 = \frac34.$

   Since the quadratic opens upward, its range is $\left[\frac34, \infty\right)$.

   2. Apply the square root. Taking the square root preserves order for non‑negative numbers, so

   $\sqrt{x^2 + x + 1} \in \left[\sqrt{\frac34}, \infty\right) = \left[\frac{\sqrt3}{2}, \infty\right).$

   3. Intersect with the domain of $\sin^{-1}$. The inverse sine function $\sin^{-1}(t)$ is defined only for $t \in [-1, 1]$. Since our $t = \sqrt{x^2 + x + 1}$ is always $\ge \frac{\sqrt3}{2} \approx 0.866$, the only values that actually appear in the domain of $\sin^{-1}$ are those in the intersection: $\left[\frac{\sqrt3}{2}, \infty\right) \cap [-1, 1] = \left[\frac{\sqrt3}{2}, 1\right].$ …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $f(x) = \log(2x-3) - 2x^2 + 6x - 4$ is a real valued function then the interval in which $f$ is an increasing function is (A) $(-\infty, 2)$ (B) $\left(\dfrac{3}{2}, 2\right)$ (C) $(2, \infty)$ (D) $\left(\dfrac{3}{2}, \infty\right)$

   ›Reveal solutionSolution

   For a function to be increasing, its derivative must be positive. After finding the derivative and considering the domain, the function increases only on $\left(\frac{3}{2}, 2\right)$.

   The key idea is simple: a function is increasing where its derivative is positive. But we must also respect the domain — the logarithm $\log(2x-3)$ is only defined when its argument is positive, so $2x-3 > 0$, i.e., $x > \frac{3}{2}$. Any interval we consider must lie entirely within this domain.

   Now, to find where $f$ increases, we differentiate and solve $f'(x) > 0$.

   1. Differentiate $f(x)$ $f(x) = \log(2x-3) - 2x^2 + 6x - 4$ Using the chain rule: derivative of $\log(2x-3)$ is $\frac{2}{2x-3}$. So

   $f'(x) = \frac{2}{2x-3} - 4x + 6.$

   2. Set up the inequality for increasing We need $f'(x) > 0$:

   $\frac{2}{2x-3} - 4x + 6 > 0.$

   3. Combine into a single fraction Write $-4x + 6$ as $\frac{(-4x+6)(2x-3)}{2x-3}$:

   $\frac{2 + (-4x+6)(2x-3)}{2x-3} > 0.$

   Expand the numerator:

   $(-4x+6)(2x-3) = -8x^2 + 12x + 12x - 18 = -8x^2 + 24x - 18$.

   Add the $2$:

   $2 - 8x^2 + 24x - 18 = -8x^2 + 24x - 16.$

   Factor out $-8$:

   $-8(x^2 - 3x + 2) = -8(x-1)(x-2).$

   So the inequality becomes:

   $\frac{-8(x-1)(x-2)}{2x-3} > 0.$

   4. Simplify the sign analysis Since $-8$ is a negative constant, multiplying both sides by $-1$ (which flips the inequality) gives:

   $\frac{(x-1)(x-2)}{2x-3} < 0.$

   Now we only need to find where this rational expression is negative.

   5. Find critical points The numerator $(x-1)(x-2) = 0$ at $x=1$ and $x=2$. The denominator $2x-3 = 0$ at $x = \frac{3}{2}$. These three points divide the real line into intervals. But remember: the domain of $f$ is $x > \frac{3}{2}$, so we only care about intervals to the right of $\frac{3}{2}$. …

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3. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $f(x) = x + \log\left(\frac{x-1}{x+1}\right)$ is a well-defined real valued function then $f$ is (A) monotonically decreasing function (B) monotonically increasing function (C) increasing in $(1, \infty)$ and decreasing in $(-\infty, -1)$ (D) decreasing in $(1, \infty)$ and increasing in $(-\infty, -1)$

   ›Reveal solutionSolution

   The function is defined only for $|x|>1$, and its derivative $f'(x)=\frac{x^2+1}{x^2-1}$ is always positive on both intervals, so $f$ is monotonically increasing on each piece. The correct option is (B).

   We need to decide the monotonicity of

   $f(x)=x+\log\left(\frac{x-1}{x+1}\right)$

   where it is a well-defined real-valued function. The logarithm requires its argument to be positive:

   $\frac{x-1}{x+1}>0.$

   Solving this inequality gives $x<-1$ or $x>1$. So the domain is $(-\infty,-1)\cup(1,\infty)$. The function is not defined between $-1$ and $1$, so we examine monotonicity separately on each interval.

   The natural tool is the derivative. If $f'(x)>0$ on an interval, $f$ is increasing there; if $f'(x)<0$, it is decreasing.

   1. Compute the derivative

   $f'(x)=1+\frac{d}{dx}\left[\log(x-1)-\log(x+1)\right].$

   Using $\frac{d}{dx}\log u = \frac{u'}{u}$,

   $f'(x)=1+\frac{1}{x-1}-\frac{1}{x+1}.$

   2. Simplify Combine the fractions:

   $\frac{1}{x-1}-\frac{1}{x+1}=\frac{(x+1)-(x-1)}{(x-1)(x+1)}=\frac{2}{x^2-1}.$

   Hence

   $f'(x)=1+\frac{2}{x^2-1}=\frac{x^2-1+2}{x^2-1}=\frac{x^2+1}{x^2-1}.$

   3. Sign of $f'(x)$ on the domain

      • Numerator: $x^2+1>0$ for all real $x$.
      • Denominator: $x^2-1>0$ when $|x|>1$, which is exactly our domain.

      Therefore, on both $(-\infty,-1)$ and $(1,\infty)$, we have $f'(x)>0$.

   4. Conclusion about monotonicity …

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4. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Consider the following statements Assertion (A): For $x \in \mathbb{R} - \{1\}$, $\dfrac{d}{dx} \left( \tan^{-1} \left( \dfrac{1+x}{1-x} \right) \right) = \dfrac{d}{dx} \left( \tan^{-1} x \right)$ Reason (R): For $x < 1$, $\tan^{-1} \left( \dfrac{1+x}{1-x} \right) = \dfrac{\pi}{4} + \tan^{-1} x$, for $x > 1$, $\tan^{-1} \left( \dfrac{1+x}{1-x} \right) = -\dfrac{3\pi}{4} + \tan^{-1} x$ The correct answer is (A) Both (A) and (R) are true, (R) is the correct explanation of (A) (B) Both (A) and (R) are true, (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

   ›Reveal solutionSolution

   Both (A) and (R) are true, and (R) is the correct explanation of (A) — option (A).

   Since $\tan\!\left(\dfrac{\pi}{4}+\tan^{-1}x\right)=\dfrac{1+x}{1-x}$, taking principal values gives

   • for $x<1$: $\ \tan^{-1}\!\left(\dfrac{1+x}{1-x}\right)=\dfrac{\pi}{4}+\tan^{-1}x$,
   • for $x>1$: $\ \tan^{-1}\!\left(\dfrac{1+x}{1-x}\right)=-\dfrac{3\pi}{4}+\tan^{-1}x$.

   So (R) is true. In each interval the expression differs from $\tan^{-1}x$ only by a constant, hence …

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5. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.The number of solutions of the equation $\tan \theta + \cot 2\theta = 1$ lying in the interval $(-\pi, \pi)$ is (A) 1 (B) 2 (C) 3 (D) 4

   ›Reveal solutionSolution

   The key is to rewrite $\tan\theta + \cot 2\theta$ in terms of $\tan\theta$ using the double-angle identity for $\cot 2\theta$, simplify to a quadratic in $\tan\theta$, solve, and then count distinct solutions in $(-\pi, \pi)$ — there are 2 solutions.

   The equation mixes $\tan\theta$ and $\cot 2\theta$. The natural instinct is to express everything in terms of a single trigonometric function, and $\tan\theta$ is the most convenient choice because $\cot 2\theta$ has a neat double-angle formula in terms of $\tan\theta$.

   Recall that $\cot 2\theta = \frac{1}{\tan 2\theta}$, and $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$. So

   $\cot 2\theta = \frac{1 - \tan^2\theta}{2\tan\theta}.$

   This substitution will turn the equation into an algebraic one in $t = \tan\theta$, provided we are careful about where $\tan\theta$ is undefined (i.e., $\theta = \pm\frac{\pi}{2}$ in $(-\pi, \pi)$) and where $\cot 2\theta$ is undefined (i.e., $2\theta = n\pi$, or $\theta = \frac{n\pi}{2}$). We'll check those points separately after solving.

   1. Substitute and simplify. Let $t = \tan\theta$. Then the equation becomes

   $t + \frac{1 - t^2}{2t} = 1.$

   Multiply through by $2t$ (assuming $t \neq 0$ for now):

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

   So $t^2 - 2t + 1 = 0$, i.e., $(t - 1)^2 = 0$. Hence $t = 1$.

   2. Solve $\tan\theta = 1$ in $(-\pi, \pi)$.

      The general solution is $\theta = \frac{\pi}{4} + n\pi$, $n \in \mathbb{Z}$.

      In the interval $(-\pi, \pi)$, the values are:

      • For $n = 0$: $\theta = \frac{\pi}{4}$.
      • For $n = -1$: $\theta = \frac{\pi}{4} - \pi = -\frac{3\pi}{4}$.
      • For $n = 1$: $\theta = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$, which is outside $(-\pi, \pi)$ since $\frac{5\pi}{4} > \pi$. So we have two candidate solutions: $\theta = \frac{\pi}{4}$ and $\theta = -\frac{3\pi}{4}$.

   3. Check for excluded points.

      We multiplied by $2t$, so we must check if $t = 0$ (i.e., $\theta = 0, \pm\pi$) could be a solution. Plug $\theta = 0$ into the original equation: $\tan 0 + \cot 0$ is undefined because $\cot 0$ blows up. Similarly, $\theta = \pm\pi$ gives $\tan(\pm\pi) = 0$ but $\cot(2\pi)$ or $\cot(-2\pi)$ is undefined. So no extra solutions there.

      Also check points where $\tan\theta$ is undefined: $\theta = \pm\frac{\pi}{2}$. At $\theta = \frac{\pi}{2}$, $\tan\theta$ is undefined, so it cannot satisfy. At $\theta = -\frac{\pi}{2}$, same issue. …

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6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $\alpha = \tan \left( 2\sin^{-1} \left( \frac{2}{3} \right) \right)$ and $\beta = \sin \left( 2\tan^{-1} \left( \frac{1}{3} \right) \right)$, then the maximum value of $\alpha \sin \theta + \beta \cos \theta =$ (A) $1$ (B) $\dfrac{\sqrt{2009}}{5}$ (C) $\dfrac{\sqrt{2024}}{3}$ (D) $4\sqrt{5} + \dfrac{3}{5}$

   ›Reveal solutionSolution

   $\alpha=\tan(2\sin^{-1}\tfrac23)=4\sqrt5$ and $\beta=\sin(2\tan^{-1}\tfrac13)=\tfrac35$; the maximum is $\sqrt{\alpha^2+\beta^2}=\sqrt{2009}/5$.

   $\alpha$: with $\sin^{-1}\tfrac23=\phi$, $\tan\phi=\dfrac{2}{\sqrt5}$, so

   $\alpha=\tan2\phi=\frac{2\cdot\frac{2}{\sqrt5}}{1-\frac{4}{5}}=\frac{4/\sqrt5}{1/5}=4\sqrt5,\qquad \alpha^2=80.$

   $\beta$: with $\tan^{-1}\tfrac13=\psi$, $\tan\psi=\tfrac13$, so …

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7. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.The number of solutions of the equation $\cos 6x + \cos 4x + \cos 2x = -1$ in $[0, \pi]$ is (A) 4 (B) 3 (C) 6 (D) 5

   ›Reveal solutionSolution

   Substituting $t=\cos 2x$ reduces the equation to $2t(2t-1)(t+1)=0$, giving $\cos 2x=0,\tfrac12,-1$ and exactly $5$ solutions in $[0,\pi]$, option (D).

   Let $t=\cos 2x$. Using $\cos 4x=2t^2-1$ and $\cos 6x=4t^3-3t$:

   $(4t^3-3t)+(2t^2-1)+t=-1.$

   $4t^3+2t^2-2t-1=-1\;\Rightarrow\;4t^3+2t^2-2t=0\;\Rightarrow\;2t(2t^2+t-1)=0.$

   Factoring the quadratic, $2t(2t-1)(t+1)=0$, so

   $\cos 2x=0,\quad \cos 2x=\tfrac12,\quad \cos 2x=-1.$

   For $x\in[0,\pi]$ we have $2x\in[0,2\pi]$:

   • $\cos 2x=0$: $2x=\dfrac{\pi}{2},\dfrac{3\pi}{2}\Rightarrow x=\dfrac{\pi}{4},\dfrac{3\pi}{4}$ (2 solutions). …

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8. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.Let $y = 4\sin^2\theta - \cos 2\theta$. If $l$ and $m$ are the minimum and maximum values of $y$ respectively, then (A) $lm = \dfrac{m}{l}$ (B) $lm = \dfrac{l}{m}$ (C) $l + m = \dfrac{l}{m}$ (D) $\dfrac{lm}{l-m} = 1 + m$

   ›Reveal solutionSolution

   Using $\cos 2\theta = 1-2\sin^2\theta$, $y = 6\sin^2\theta - 1 \in [-1,5]$, so $l=-1,\ m=5$ and $lm = m/l = -5$.

   Rewrite $\cos 2\theta = 1 - 2\sin^2\theta$:

   $y = 4\sin^2\theta - (1 - 2\sin^2\theta) = 6\sin^2\theta - 1.$

   Since $\sin^2\theta \in [0,1]$:

   $y_{\min} = 6(0)-1 = -1 = l,\qquad y_{\max} = 6(1)-1 = 5 = m.$

   Test the options: …

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9. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.If $y = \log_2 \sin x$, then the minimum value of $\cosh y$ is (A) $2$ (B) $\dfrac{2}{e}$ (C) $\dfrac{e}{2}$ (D) $1$

   ›Reveal solutionSolution

   The problem reduces to finding the minimum of $\cosh(\log_2 \sin x)$ over $x$ where $\sin x > 0$. Using the definition of $\cosh$ and properties of logs, this becomes a function of $\sin x$ alone; its minimum occurs when $\sin x = 1$, giving the answer $\boxed{1}$.

   The key here is to see that $\cosh y$ is defined as $\frac{e^y + e^{-y}}{2}$, and $y$ itself is $\log_2 \sin x$. So we are really composing a hyperbolic cosine with a logarithmic function of a trigonometric one. The domain is restricted: $\sin x$ must be positive for the log to be defined, so $x \in (0, \pi)$ plus periodic copies.

   Instead of jumping into calculus on a messy composite function, we can simplify algebraically first. That’s the smart move — let the structure of the expression do the work for you.

   1. Write $y = \log_2 \sin x$. Then $\cosh y = \frac{e^y + e^{-y}}{2}$.

   2. Substitute $y$:

      $e^y = e^{\log_2 \sin x} = 2^{\log_2 \sin x \cdot \log_2 e}$? That’s messy. Better: recall $a^{\log_b c} = c^{\log_b a}$. Here, $e^{\log_2 \sin x} = \sin x^{\log_2 e}$. But there’s an even cleaner path.

   3. Use the change-of-base: $\log_2 \sin x = \frac{\ln \sin x}{\ln 2}$. Then

      $e^y = e^{\frac{\ln \sin x}{\ln 2}} = (\sin x)^{1/\ln 2}$.

      Similarly, $e^{-y} = (\sin x)^{-1/\ln 2}$.

   4. So $\cosh y = \frac{1}{2} \left[ (\sin x)^{1/\ln 2} + (\sin x)^{-1/\ln 2} \right]$.

   Let $t = \sin x$. Since $x$ is real and we need $\sin x > 0$, we have $t \in (0, 1]$. The function becomes

   $f(t) = \frac{1}{2} \left( t^{a} + t^{-a} \right)$, where $a = \frac{1}{\ln 2} > 0$. …

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10. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.If $540^\circ < \theta < 630^\circ$ and $\tan \theta = \dfrac{5}{12}$, then $\dfrac{\cos \dfrac{\theta}{2} - 5 \sin \dfrac{\theta}{2}}{\sqrt{-(12 \sec \theta + 5 \csc \theta)}} =$ (A) $-26$ (B) $26$ (C) $1$ (D) $-1$

    ›Reveal solutionSolution

    Fix the quadrant of $\theta$ (third) and of $\theta/2$ (fourth), evaluate numerator and denominator; both equal $\sqrt{26}$, so the expression is $1$.

    Locate $\theta$. Since $540^\circ < \theta < 630^\circ$, subtracting $360^\circ$ gives $180^\circ < \theta < 270^\circ$ — the third quadrant, where $\tan\theta = \dfrac{5}{12} > 0$ is consistent. Hence

    $\sin\theta = -\dfrac{5}{13}, \qquad \cos\theta = -\dfrac{12}{13}.$

    Denominator.

    $12\sec\theta + 5\csc\theta = 12\left(-\dfrac{13}{12}\right) + 5\left(-\dfrac{13}{5}\right) = -13 - 13 = -26,$

    so

    $\sqrt{-(12\sec\theta + 5\csc\theta)} = \sqrt{26}.$

    Locate $\theta/2$. From $540^\circ < \theta < 630^\circ$ we get $270^\circ < \dfrac{\theta}{2} < 315^\circ$ — the fourth quadrant, where $\cos\dfrac{\theta}{2} > 0$ and $\sin\dfrac{\theta}{2} < 0$. Using half-angle formulas with $\cos\theta = -\dfrac{12}{13}$:

    $\cos^2\dfrac{\theta}{2} = \dfrac{1+\cos\theta}{2} = \dfrac{1 - \tfrac{12}{13}}{2} = \dfrac{1}{26} \;\Rightarrow\; \cos\dfrac{\theta}{2} = \dfrac{1}{\sqrt{26}},$ …

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11. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
    Q.If $\cos \theta = -\dfrac{3}{5}$ and $\pi < \theta < 3\pi/2$, then $\tan\left(\dfrac{\theta}{2}\right) =$ (A) $2$ (B) $-2$ (C) $1$ (D) $-1$

    ›Reveal solutionSolution

    Use the half-angle formula for tangent in terms of cosine, and determine the sign of $\tan(\theta/2)$ from the quadrant of $\theta/2$. The value is $\boxed{-2}$.

    The key here is that $\tan(\theta/2)$ can be expressed directly from $\cos\theta$ using a standard identity, but the sign of the result depends on where $\theta/2$ lies. You cannot just plug numbers into a formula and take the positive root — the quadrant decides the sign.

    We are given $\cos\theta = -\frac{3}{5}$ and $\pi < \theta < \frac{3\pi}{2}$. That means $\theta$ is in the third quadrant, where both sine and cosine are negative. Now, what about $\theta/2$? Since $\theta$ is between $\pi$ and $1.5\pi$, dividing by 2 gives $\frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$. That puts $\theta/2$ in the second quadrant, where tangent is negative. So our final answer must be negative — that already eliminates options (A), (C), and (D), leaving only (B) as possible. But let’s verify.

    1. Recall the half-angle formula for tangent. There are several forms; the one that uses only $\cos\theta$ is:

    $\tan\frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}$

    The sign is chosen based on the quadrant of $\theta/2$, not $\theta$.

    2. Substitute the given value.

    $\cos\theta = -\frac{3}{5}$

    So:

    $1 - \cos\theta = 1 - \left(-\frac{3}{5}\right) = 1 + \frac{3}{5} = \frac{8}{5}$

    $1 + \cos\theta = 1 + \left(-\frac{3}{5}\right) = 1 - \frac{3}{5} = \frac{2}{5}$

    Hence:

    $\frac{1 - \cos\theta}{1 + \cos\theta} = \frac{8/5}{2/5} = \frac{8}{2} = 4$

    So:

    $\tan\frac{\theta}{2} = \pm \sqrt{4} = \pm 2$

    3. Apply the sign from the quadrant. …

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12. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.If $a, b$ are real numbers and $\alpha$ is a real root of $x^2 + 12 + 3\sin(a + b x) + 6x = 0$ then the value of $\cos(a + b\alpha)$ for the least positive value of $a + b\alpha$ is (A) $-1$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $\dfrac{1}{2}$ (D) $0$

    ›Reveal solutionSolution

    Completing the square forces $\sin(a+b\alpha)=-1$ and $\alpha=-3$; the least positive $a+b\alpha=\frac{3\pi}{2}$ gives $\cos(a+b\alpha)=0$. Option (D).

    Rewrite the equation as

    $(x+3)^2 + 3 + 3\sin(a+bx) = 0.$

    At a real root $x=\alpha$,

    $3\sin(a+b\alpha) = -\big[(\alpha+3)^2+3\big] \le -3,$

    so $\sin(a+b\alpha)\le -1$. Since $\sin\ge -1$, equality is forced:

    $\sin(a+b\alpha) = -1 \quad\text{and}\quad (\alpha+3)^2 = 0 \Rightarrow \alpha = -3.$ …

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