Q.Prove that the function $f : \mathbf{R} \to \mathbf{R}$ defined by $f(x) = 2x + 5$ is one-one, using the contrapositive.

Showing the 12 most recent of 23 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.A real valued function $f$ defined by $f(x) = |x| - x$ is (A) an injection but not surjection, if $[0, \infty)$ is its domain and $(-\infty, 0]$ is its codomain (B) a bijection, if $(-\infty, 0]$ is its domain and also codomain (C) a bijection, if $[0, \infty)$ is its domain and also codomain (D) a surjection but not injection, if $\mathbb{R}$ is its domain and $[0, \infty)$ is its codomain

   ›Reveal solutionSolution

   The function $f(x) = |x| - x$ simplifies to $0$ for $x \ge 0$ and $-2x$ for $x < 0$, so its behaviour depends entirely on the chosen domain and codomain. The correct option is (D).

   The key to this problem is understanding that $|x| - x$ is not a single, fixed mapping — it changes its rule depending on whether $x$ is negative or non‑negative. Before checking each option, let's simplify the function piecewise.

   For $x \ge 0$, $|x| = x$, so $f(x) = x - x = 0$.

   For $x < 0$, $|x| = -x$, so $f(x) = -x - x = -2x$.

   So $f$ maps every non‑negative input to $0$, and every negative input to a positive number (since $-2x > 0$ when $x < 0$). This immediately tells us that $f$ is not injective on any domain that contains two or more non‑negative numbers (they all go to $0$), and its range is always a subset of $[0, \infty)$.

   Now examine each option.

   1. Option (A): Domain $[0, \infty)$, codomain $(-\infty, 0]$.

      On this domain, $f(x) = 0$ for every $x$. Since $0 \in (-\infty, 0]$, the function is well-defined into this codomain. But it maps every input to the same output $0$, so it is not injective, and it is not surjective either since no negative value in the codomain is ever attained. So (A) is false.

   2. Option (B): Domain $(-\infty, 0]$, codomain $(-\infty, 0]$.

      For $x \le 0$, $f(x) = -2x$. Since $x \le 0$, $-2x \ge 0$. But the codomain is $(-\infty, 0]$, which contains only non‑positive numbers. The output $-2x$ is non‑negative, so the only way it can belong to $(-\infty, 0]$ is if it equals $0$. That happens only when $x = 0$. For any $x < 0$, $-2x > 0$, which is not in $(-\infty, 0]$. So the function is not even well‑defined for most of its domain — it fails to map into the given codomain. Hence (B) is false.

      Watch out

      A common mistake is to assume $f(x) = -2x$ automatically lands in $(-\infty, 0]$ because the domain is negative. But $-2x$ for a negative $x$ is positive — always check the sign of the output. …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $z = i^i$, then $z^i =$ (A) $-i$ (B) $i$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   $i^i = e^{-\pi/2}$ is real, so $z^i = e^{-i\pi/2} = -i$.

   Step 1 — evaluate $z = i^i$. Using the principal value $\log i = i\tfrac{\pi}{2}$,

   $z = i^i = e^{i \log i} = e^{i \cdot i\pi/2} = e^{-\pi/2},$

   a positive real number.

   Step 2 — raise to the power $i$. …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $f(x) = xe^{x^2 - 2x - 3}$ is a real valued function, then $f$ is (A) a monotonically decreasing function (B) an increasing function in the interval $(-1,3)$ and decreasing function in the interval $(-\infty,-1) \cup (3,\infty)$ (C) a monotonically increasing function (D) a decreasing function in the interval $(-1,3)$ and increasing function in the interval $(-\infty,-1) \cup (3,\infty)$

   ›Reveal solutionSolution

   To determine the monotonicity of $f(x)$, we find its first derivative $f'(x)$. Since $f'(x)$ is positive for all real $x$, the function $f(x)$ is monotonically increasing. The correct option is (C).

   To understand whether a function is increasing or decreasing, we examine the sign of its first derivative.

   • If $f'(x) > 0$ for all $x$ in an interval, the function $f(x)$ is increasing in that interval.
   • If $f'(x) < 0$ for all $x$ in an interval, the function $f(x)$ is decreasing in that interval.
   • If $f'(x) = 0$ at isolated points, but maintains its sign across these points, the function is still monotonic. If $f'(x)$ changes sign, the function changes its monotonicity.

   In this problem, we need to find the first derivative of $f(x)$ and then analyze its sign.

   1. Find the first derivative $f'(x)$:

      The given function is $f(x) = xe^{x^2 - 2x - 3}$.

      We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

      Let $u(x) = x$ and $v(x) = e^{x^2 - 2x - 3}$.

      Then, $u'(x) = \frac{d}{dx}(x) = 1$.

      To find $v'(x)$, we use the chain rule. If $v(x) = e^{g(x)}$, then $v'(x) = e^{g(x)} \cdot g'(x)$.

      Here, $g(x) = x^2 - 2x - 3$.

      So, $g'(x) = \frac{d}{dx}(x^2 - 2x - 3) = 2x - 2$.

      Therefore, $v'(x) = e^{x^2 - 2x - 3} (2x - 2)$.

      Now, substitute $u, u', v, v'$ into the product rule formula:

      $f'(x) = (1) \cdot e^{x^2 - 2x - 3} + x \cdot e^{x^2 - 2x - 3} (2x - 2)$

      Factor out the common term $e^{x^2 - 2x - 3}$:

      $f'(x) = e^{x^2 - 2x - 3} [1 + x(2x - 2)]$

      $f'(x) = e^{x^2 - 2x - 3} [1 + 2x^2 - 2x]$

      $f'(x) = e^{x^2 - 2x - 3} (2x^2 - 2x + 1)$

   2. Analyze the sign of $f'(x)$:

      We need to determine when $f'(x) > 0$ or $f'(x) < 0$.

      The expression for $f'(x)$ has two factors: $e^{x^2 - 2x - 3}$ and $(2x^2 - 2x + 1)$.

      • Factor 1: $e^{x^2 - 2x - 3}$

        The exponential function $e^A$ is always positive for any real value of $A$.

        Thus, $e^{x^2 - 2x - 3} > 0$ for all $x \in \mathbb{R}$.

      • Factor 2: $2x^2 - 2x + 1$

        This is a quadratic expression. To determine its sign, we can look at its discriminant ($\Delta = b^2 - 4ac$) and the sign of its leading coefficient.

        For $2x^2 - 2x + 1$, we have $a=2$, $b=-2$, $c=1$.

        The discriminant is $\Delta = (-2)^2 - 4(2)(1) = 4 - 8 = -4$. …

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4. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The number of values of $x$ satisfying the equation

   [!FORMULA] $\tan^{-1}\left(x + \frac{\sqrt{2}}{x}\right) + \tan^{-1}\left(x - \frac{\sqrt{2}}{x}\right) = \tan^{-1}(x)$

   is (A) $0$ (B) $1$ (C) $2$ (D) $3$

   ›Reveal solutionSolution

   The key idea is to apply the tangent addition formula to the sum of two inverse tangents, simplify the resulting algebraic equation, and then check for extraneous solutions introduced by the domain restrictions of inverse trigonometric functions. The equation reduces to a cubic with only one valid solution, so the answer is (B) 1.

   We start with the equation

   $\tan^{-1}\left(x + \frac{\sqrt{2}}{x}\right) + \tan^{-1}\left(x - \frac{\sqrt{2}}{x}\right) = \tan^{-1}(x).$

   Concept and intuition:

   When we have an equation involving a sum of two inverse tangents, a natural first step is to take the tangent of both sides. This uses the identity

   $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B},$

   which converts the inverse-trig equation into an ordinary algebraic equation. However, we must be careful: taking the tangent can introduce extraneous solutions because the inverse tangent function is not one-to-one over the whole real line — its range is $(-\pi/2, \pi/2)$. So after solving the algebra, we must check each candidate against the original equation’s domain and range.

   ---

   Step-by-step solution:

   1. Set up the tangent addition. Let

   $A = \tan^{-1}\left(x + \frac{\sqrt{2}}{x}\right), \quad B = \tan^{-1}\left(x - \frac{\sqrt{2}}{x}\right), \quad C = \tan^{-1}(x).$

   Then the equation is $A + B = C$. Taking the tangent of both sides gives

   $\tan(A+B) = \tan C.$

   2. Apply the tangent addition formula.

   $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}.$

   Since $\tan A = x + \frac{\sqrt{2}}{x}$ and $\tan B = x - \frac{\sqrt{2}}{x}$, we have

   $\tan A + \tan B = \left(x + \frac{\sqrt{2}}{x}\right) + \left(x - \frac{\sqrt{2}}{x}\right) = 2x,$

   and

   $\tan A \tan B = \left(x + \frac{\sqrt{2}}{x}\right)\left(x - \frac{\sqrt{2}}{x}\right) = x^2 - \frac{2}{x^2}.$

   Thus

   $\tan(A+B) = \frac{2x}{1 - \left(x^2 - \frac{2}{x^2}\right)} = \frac{2x}{1 - x^2 + \frac{2}{x^2}}.$

   3. Simplify the denominator. Write $1 - x^2 + \frac{2}{x^2}$ as a single fraction:

   $1 - x^2 + \frac{2}{x^2} = \frac{x^2 - x^4 + 2}{x^2} = \frac{-(x^4 - x^2 - 2)}{x^2}.$

   So

   $\tan(A+B) = \frac{2x}{\frac{-(x^4 - x^2 - 2)}{x^2}} = -\frac{2x \cdot x^2}{x^4 - x^2 - 2} = -\frac{2x^3}{x^4 - x^2 - 2}.$

   4. Set equal to $\tan C = x$. The equation becomes

   $-\frac{2x^3}{x^4 - x^2 - 2} = x.$

   Multiply both sides by the denominator (noting that $x=0$ is a special case we’ll check later):

   $-2x^3 = x(x^4 - x^2 - 2).$

   If $x \neq 0$, we can divide both sides by $x$:

   $-2x^2 = x^4 - x^2 - 2.$

   Rearranging:

   $0 = x^4 - x^2 - 2 + 2x^2 = x^4 + x^2 - 2.$

   5. Solve the quartic. Let $u = x^2 \ge 0$. Then $u^2 + u - 2 = 0$, so

   $(u+2)(u-1) = 0.$

   Hence $u = 1$ or $u = -2$ (impossible since $u \ge 0$). So $x^2 = 1$, giving $x = 1$ or $x = -1$.

   6. Check the case $x = 0$.

      If $x = 0$, the original equation contains terms $\frac{\sqrt{2}}{x}$, which are undefined. So $x=0$ is not in the domain. Thus only $x = 1$ and $x = -1$ are candidates.

   7. Check for extraneous solutions. …

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5. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.For a real number ‘a’, if a real valued function $f(x) = 4x^3 + ax^2 + 3x - 2$ is monotonic in its domain, then the range of ‘a’ is (A) $(-6, 6)$ (B) Empty set (C) $(-2, 2)$ (D) $(2, 4)$

   ›Reveal solutionSolution

   A cubic is monotonic (always increasing or always decreasing) if its derivative never changes sign — i.e., the quadratic derivative has no real roots or is always non‑negative/non‑positive. For $f(x)=4x^3+ax^2+3x-2$, the derivative is $12x^2+2ax+3$; its discriminant must be $\le 0$, giving $a^2 \le 36$, so $a \in [-6,6]$. Since the question asks for the range of $a$ for monotonicity, the correct choice is (A) $(-6,6)$.

   ---

   Concept & Intuition

   A function is monotonic on its domain if it is either entirely non‑decreasing or entirely non‑increasing — no turning points, no local maxima or minima. For a polynomial, this means its derivative never changes sign.

   Here $f$ is a cubic; its derivative is a quadratic. A quadratic changes sign only if it has two distinct real roots (i.e., it crosses zero twice). If it has no real roots or a double root, it is always positive or always negative (except possibly zero at one point). So the condition for monotonicity is: the derivative’s discriminant $\le 0$.

   ---

   Step‑by‑step reasoning

   1. Find the derivative

      $f(x) = 4x^3 + a x^2 + 3x - 2$

      $f'(x) = 12x^2 + 2a x + 3$.

   2. Condition for monotonicity

      $f$ is monotonic on $\mathbb{R}$ iff $f'(x) \ge 0$ for all $x$ (always increasing) or $f'(x) \le 0$ for all $x$ (always decreasing). Since the leading coefficient of $f'(x)$ is $12 > 0$, the quadratic opens upward. Thus it can only be $\ge 0$ for all $x$; it can never be $\le 0$ for all $x$ (because as $x \to \pm\infty$, $f'(x) \to +\infty$). So we require $f'(x) \ge 0$ for every real $x$.

   3. Discriminant condition

      A quadratic $Ax^2 + Bx + C$ with $A>0$ is non‑negative for all $x$ iff its discriminant $\Delta = B^2 - 4AC \le 0$.

      Here $A=12$, $B=2a$, $C=3$.

      $\Delta = (2a)^2 - 4 \cdot 12 \cdot 3 = 4a^2 - 144$.

   4. Solve the inequality

      $4a^2 - 144 \le 0 \implies a^2 \le 36 \implies -6 \le a \le 6$.

   5. Interpret the result

      For $a$ strictly between $-6$ and $6$, $\Delta < 0$, so $f'(x) > 0$ everywhere — strictly increasing. …

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.For a real number ‘a’, if a real valued function $f(x) = 4x^3 + ax^2 + 3x - 2$ is monotonic in its domain, then the range of ‘a’ is (A) $(-2, 2)$ (B) $(-6, 6)$ (C) Empty set (D) $(2, 4)$

   ›Reveal solutionSolution

   A cubic is monotonic (always increasing or always decreasing) if its derivative never changes sign — i.e., its derivative quadratic has no real roots or a double root. For $f(x)=4x^3+ax^2+3x-2$, the derivative is $12x^2+2ax+3$, and requiring its discriminant $\le 0$ gives $a^2 \le 36$, so $a \in [-6,6]$. Since the question asks for the range of $a$ for which $f$ is monotonic, the correct option is (B) $(-6,6)$.

   ---

   Concept & Intuition

   A function is monotonic on its domain if it is either entirely non‑decreasing or entirely non‑increasing. For a differentiable function, this means its derivative never changes sign — it is either always $\ge 0$ or always $\le 0$.

   Here $f$ is a cubic polynomial. Its derivative is a quadratic. A quadratic that never changes sign must have no real roots (or a double root), i.e., its discriminant must be $\le 0$. That’s the key: we don’t need to check which sign the derivative has — just that it doesn’t cross zero.

   ---

   Step‑by‑step reasoning

   1. Find the derivative $f(x) = 4x^3 + a x^2 + 3x - 2$ Differentiate term‑by‑term:

   $f'(x) = 12x^2 + 2a x + 3.$

   2. Condition for monotonicity

      For $f$ to be monotonic on $\mathbb{R}$, $f'(x)$ must not change sign. Since $f'(x)$ is a quadratic with positive leading coefficient ($12>0$), it is always $\ge 0$ if it has at most one real root. That happens exactly when its discriminant $\Delta \le 0$.

   3. Compute the discriminant

      For $f'(x)=12x^2+2a x+3$,

   $\Delta = (2a)^2 - 4\cdot 12 \cdot 3 = 4a^2 - 144.$

   4. Set the inequality

   $\Delta \le 0 \quad\Rightarrow\quad 4a^2 - 144 \le 0.$

   Divide by 4:

   $a^2 - 36 \le 0 \quad\Rightarrow\quad a^2 \le 36.$

   5. Solve for $a$

   $|a| \le 6 \quad\Rightarrow\quad -6 \le a \le 6.$

   6. Interpret the options The problem gives intervals: (A) $(-2,2)$ — too narrow …

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7. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The number of values of $x$ satisfying the equation

   [!FORMULA] $\tan^{-1}\left(\frac{x+\sqrt{2}}{x}\right) + \tan^{-1}\left(\frac{x-\sqrt{2}}{x}\right) = \tan^{-1}(x)$

   is (A) 0 (B) 1 (C) 3 (D) 2

   ›Reveal solutionSolution

   The two arctangents combine to $\tan^{-1}(x^2)$ (valid since $1-uv = 2/x^2 > 0$), reducing the equation to $x^2 = x$; excluding $x=0$, only $x=1$ survives — exactly 1 solution, option (B).

   Concept. $\tan^{-1}u + \tan^{-1}v = \tan^{-1}\dfrac{u+v}{1-uv}$ holds exactly (no $\pm\pi$ adjustment) when $uv < 1$. Always verify that condition before collapsing the sum.

   Step 1 — combine the arctangents. With $u = \dfrac{x+\sqrt2}{x}$ and $v = \dfrac{x-\sqrt2}{x}$ (note $x \ne 0$):

   $u + v = \frac{2x}{x} = 2, \qquad uv = \frac{x^2 - 2}{x^2}, \qquad 1 - uv = \frac{2}{x^2} > 0 \ \text{for all } x \ne 0.$

   Since $uv < 1$ always, the addition formula applies with no correction:

   $\tan^{-1}u + \tan^{-1}v = \tan^{-1}\frac{2}{2/x^2} = \tan^{-1}(x^2).$

   Step 2 — solve the reduced equation.

   $\tan^{-1}(x^2) = \tan^{-1}(x) \;\Rightarrow\; x^2 = x \;\Rightarrow\; x = 0 \text{ or } x = 1.$ …

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8. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The domain and range of $f(x) = \dfrac{1}{\sqrt{|x| - x^2}}$ are A and B respectively. Then $A \cup B =$ (A) $\mathbb{R} - \{-1, 0, 1\}$ (B) $(-1, \infty) - \{0, 1\}$ (C) $(-1, 0) \cup (0, 1) \cup [2, \infty)$ (D) $(-1, 1) \cup [2, \infty)$

   ›Reveal solutionSolution

   The domain is $(-1,0)\cup(0,1)$ and the range is $[2,\infty)$; their union is $(-1,0)\cup(0,1)\cup[2,\infty)$, which matches option (C).

   We need the set of all $x$ for which $f(x)$ is defined (domain $A$) and the set of all values $f(x)$ can take (range $B$), then find $A\cup B$.

   Concept & Intuition

   The function involves a square root in the denominator, so the expression inside the root must be positive (strictly, because it’s in the denominator). That gives a condition on $x$. The range is trickier: the denominator’s size controls the output, and because the denominator can approach zero, the function can blow up to infinity. We’ll find the minimum possible value of the denominator to get the smallest output, then see that outputs can be arbitrarily large.

   ---

   1. Find the domain $A$

      We require $|x| - x^2 > 0$ (positive inside the square root, and denominator non-zero).

      Write $|x| - x^2 > 0$ as $|x| > x^2$.

      • If $x \ge 0$, this is $x > x^2 \implies x(1 - x) > 0$. Since $x \ge 0$, we need $x > 0$ and $1 - x > 0$, so $0 < x < 1$.
      • If $x < 0$, then $|x| = -x$, so $-x > x^2 \implies 0 > x^2 + x = x(x+1)$. The product $x(x+1)$ is negative when $-1 < x < 0$. Combining: $x \in (-1,0) \cup (0,1)$. Note $x=0$ is excluded because denominator becomes $0$, and $x = \pm 1$ give $|x|-x^2 = 0$. So $A = (-1,0) \cup (0,1)$.

   2. Find the range $B$

      Let $g(x) = |x| - x^2$. Then $f(x) = 1/\sqrt{g(x)}$.

      On $A$, $g(x) > 0$. We need the set of all possible values of $1/\sqrt{g(x)}$ as $x$ runs over $A$.

      • For $x \in (0,1)$: $g(x) = x - x^2 = x(1-x)$. This is a downward parabola with maximum at $x = 1/2$, value $g(1/2) = 1/4$. As $x \to 0^+$ or $x \to 1^-$, $g(x) \to 0^+$. So $g(x) \in (0, 1/4]$.
      • For $x \in (-1,0)$: $|x| = -x$, so $g(x) = -x - x^2 = -x(1+x)$. Let $t = -x$, then $t \in (0,1)$ and $g = t - t^2$, same as above. So again $g(x) \in (0, 1/4]$. Thus $g(x)$ takes all values in $(0, 1/4]$ on $A$. Then $f(x) = 1/\sqrt{g(x)}$: …

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9. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The number of real solutions of $\tan^{-1}x + \tan^{-1}2x = \dfrac{\pi}{4}$ is (A) 2 (B) 1 (C) 0 (D) infinitely many

   ›Reveal solutionSolution

   The key idea is to apply the tangent addition formula to the inverse-tangent equation, then solve the resulting quadratic and check for extraneous solutions. Only one real solution satisfies the original domain constraints, so the answer is (B) 1.

   We start with the equation

   $\tan^{-1}x + \tan^{-1}2x = \frac{\pi}{4}.$

   The natural instinct is to take the tangent of both sides, because we know a formula for $\tan(A+B)$. But we must be careful: taking the tangent can introduce extraneous solutions if the sum of the angles is outside the principal range of $\tan^{-1}$. So after solving algebraically, we must check each candidate against the original equation.

   Why this approach works:

   The tangent addition formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ lets us convert the inverse-tangent equation into an ordinary algebraic equation. However, the inverse tangent function has a restricted range $(-\pi/2, \pi/2)$, so the sum of two such angles could be outside that range, and the tangent function is not one-to-one there. Hence we must verify each solution.

   ---

   Step-by-step solution:

   1. Apply tangent to both sides Let $A = \tan^{-1}x$ and $B = \tan^{-1}2x$. Then $\tan A = x$, $\tan B = 2x$, and the equation becomes

   $\tan(A+B) = \tan\left(\frac{\pi}{4}\right) = 1.$

   Using the formula:

   $\frac{x + 2x}{1 - x \cdot 2x} = 1 \quad\Rightarrow\quad \frac{3x}{1 - 2x^2} = 1.$

   2. Solve the algebraic equation Multiply both sides by $1 - 2x^2$ (assuming $1 - 2x^2 \neq 0$):

   $3x = 1 - 2x^2 \quad\Rightarrow\quad 2x^2 + 3x - 1 = 0.$

   Solve the quadratic:

   $x = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}.$

   So two candidates:

   $x_1 = \frac{-3 + \sqrt{17}}{4}, \quad x_2 = \frac{-3 - \sqrt{17}}{4}.$

   3. Check domain restrictions

      The original equation involves $\tan^{-1}x$ and $\tan^{-1}2x$. Their sum must equal $\pi/4$. Since each inverse tangent lies in $(-\pi/2, \pi/2)$, their sum lies in $(-\pi, \pi)$. But we need the sum to be exactly $\pi/4$, a positive acute angle.

      • For $x_2 = \frac{-3 - \sqrt{17}}{4} \approx \frac{-3 - 4.123}{4} = -1.7808$, both $x$ and $2x$ are negative. Then $\tan^{-1}x$ and $\tan^{-1}2x$ are both negative, so their sum is negative. It cannot equal $\pi/4 > 0$. Hence $x_2$ is extraneous.

      • For $x_1 = \frac{-3 + \sqrt{17}}{4} \approx \frac{-3 + 4.123}{4} = 0.2808$, both $x$ and $2x$ are positive. Then both inverse tangents are positive, and their sum is positive. Could it be exactly $\pi/4$? We must verify numerically or by reasoning: …

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10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.Consider the following statements Statement-I: $\cosh^{-1}x = \tanh^{-1}x$ has no solution Statement-II: $\cosh^{-1}x = \coth^{-1}x$ has only one solution The correct answer is (A) Both statements I and II are true (B) Both statements I and II are false (C) Statement I is true, but statement II is false (D) Statement I is false, but statement II is true

    ›Reveal solutionSolution

    The key idea is to rewrite each inverse hyperbolic equation in terms of exponentials or algebraic forms, solve for $x$, and check domain restrictions. Statement I is false (there is a solution), and Statement II is false (there are two solutions, not one). So both statements are false.

    We need to decide whether each statement about inverse hyperbolic equations is true or false. The best approach is to convert each inverse hyperbolic equation into an algebraic equation in $x$ using the known logarithmic definitions, then solve and check the domain.

    ---

    1. Recall the definitions and domains

    • $\cosh^{-1} x$ is defined for $x \ge 1$ and gives non-negative output.
    • $\tanh^{-1} x$ is defined for $|x| < 1$.
    • $\coth^{-1} x$ is defined for $|x| > 1$.

    These domains will immediately tell us whether a solution is possible.

    ---

    2. Analyze Statement I: $\cosh^{-1} x = \tanh^{-1} x$

    Let $y = \cosh^{-1} x = \tanh^{-1} x$. Then:

    $x = \cosh y \quad \text{and} \quad x = \tanh y.$

    So we require $\cosh y = \tanh y$. But $\tanh y = \frac{\sinh y}{\cosh y}$, so:

    $\cosh y = \frac{\sinh y}{\cosh y} \quad\Rightarrow\quad \cosh^2 y = \sinh y.$

    Using $\cosh^2 y = 1 + \sinh^2 y$, we get:

    $1 + \sinh^2 y = \sinh y \quad\Rightarrow\quad \sinh^2 y - \sinh y + 1 = 0.$

    The discriminant is $(-1)^2 - 4(1)(1) = 1 - 4 = -3 < 0$. No real solution for $\sinh y$, hence no real $y$. So it seems no solution — but wait: we must check if the domains even overlap.

    Domain of $\cosh^{-1} x$: $x \ge 1$.

    Domain of $\tanh^{-1} x$: $|x| < 1$.

    These domains are disjoint (no overlap). So indeed no real $x$ can satisfy both.

    Thus Statement I says "has no solution" — that is true.

    Watch out

    The algebraic approach gave no real solution anyway, but the domain argument is faster and more fundamental. Always check domains first for inverse hyperbolic equations.

    ---

    3. Analyze Statement II: $\cosh^{-1} x = \coth^{-1} x$

    Let $y = \cosh^{-1} x = \coth^{-1} x$. Then:

    $x = \cosh y \quad\text{and}\quad x = \coth y = \frac{\cosh y}{\sinh y}.$

    Equating:

    $\cosh y = \frac{\cosh y}{\sinh y}.$

    If $\cosh y \neq 0$ (it never is, since $\cosh y \ge 1$), we can divide:

    $1 = \frac{1}{\sinh y} \quad\Rightarrow\quad \sinh y = 1.$

    Thus $y = \sinh^{-1} 1 = \log(1 + \sqrt{2})$. Then:

    $x = \cosh y = \cosh(\log(1+\sqrt{2})).$

    We can compute: $\cosh(\log a) = \frac{a + 1/a}{2}$. With $a = 1+\sqrt{2}$, we have $1/a = \sqrt{2} - 1$, so:

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

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11. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.Consider the following statements Statement-I: $\cosh^{-1}x = \tanh^{-1}x$ has no solution Statement-II: $\cosh^{-1}x = \coth^{-1}x$ has only one solution The correct answer is (A) Statement I is false, but statement II is true (B) Both statements I and II are true (C) Both statements I and II are false (D) Statement I is true, but statement II is false

    ›Reveal solutionSolution

    $\cosh^{-1}x=\tanh^{-1}x$ has no real solution (Statement-I true), and $\cosh^{-1}x=\coth^{-1}x$ has exactly one solution $x=\sqrt2$ (Statement-II true). Both statements are true — option (B).

    Concept

    Use the logarithmic forms and the disjoint domains: $\cosh^{-1}x$ needs $x\ge1$, $\tanh^{-1}x$ needs $|x|<1$, and $\coth^{-1}x$ needs $|x|>1$.

    Statement-I: $\cosh^{-1}x=\tanh^{-1}x$

    Applying $\cosh$ to both sides requires $|x|<1$, where $\cosh(\tanh^{-1}x)=\dfrac{1}{\sqrt{1-x^2}}$:

    $x=\frac{1}{\sqrt{1-x^2}}\ \Rightarrow\ x^2(1-x^2)=1\ \Rightarrow\ x^4-x^2+1=0.$

    Setting $t=x^2$: $t^2-t+1=0$ has discriminant $1-4=-3<0$, so there is no real $x$. Statement-I ("has no solution") is true.

    Statement-II: $\cosh^{-1}x=\coth^{-1}x$

    Here $x>1$. Using $\coth^{-1}x=\tanh^{-1}\!\left(\tfrac1x\right)$ and applying $\cosh$, with $\cosh(\tanh^{-1}u)=\dfrac{1}{\sqrt{1-u^2}}$ at $u=\tfrac1x$: …

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12. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.The number of real solutions of $\tan^{-1}x + \tan^{-1}2x = \frac{\pi}{4}$ is (A) 0 (B) infinitely many (C) 1 (D) 2

    ›Reveal solutionSolution

    The key idea is to use the identity for the sum of two inverse tangents and then check each candidate solution against the domain restrictions of $\tan^{-1}$. The equation has exactly one real solution.

    The problem asks for the number of real solutions to $\tan^{-1}x + \tan^{-1}2x = \frac{\pi}{4}$. This is not a simple algebraic equation — it involves inverse trigonometric functions, which have restricted ranges. The standard approach is to take the tangent of both sides, which converts the problem into an algebraic equation, but we must be careful: taking the tangent can introduce extraneous solutions that don't satisfy the original equation because the sum of the two angles might lie outside the principal range of $\tan^{-1}$.

    The range of $\tan^{-1}$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. So each term individually lies in this open interval. Their sum could be anywhere between $-\pi$ and $\pi$, but the equation says it equals $\frac{\pi}{4}$, which is well within the range where the tangent function is one-to-one. However, the identity for $\tan^{-1}a + \tan^{-1}b$ has a condition: it equals $\tan^{-1}\left(\frac{a+b}{1-ab}\right)$ only when $ab < 1$; otherwise, we need to add or subtract $\pi$ to adjust the angle to the principal range. This is the subtle point that most students miss.

    Let's work through it step by step.

    1. Take the tangent of both sides. Since $\tan\left(\frac{\pi}{4}\right) = 1$, we get:

    $\tan\left(\tan^{-1}x + \tan^{-1}2x\right) = 1$

    Using the tangent addition formula:

    $\frac{x + 2x}{1 - x \cdot 2x} = 1$

    which simplifies to:

    $\frac{3x}{1 - 2x^2} = 1$

    2. Solve the algebraic equation. Multiply both sides by $1 - 2x^2$ (note: $1 - 2x^2 \neq 0$ for now):

    $3x = 1 - 2x^2$

    Rearranging:

    $2x^2 + 3x - 1 = 0$

    This is a quadratic. Its discriminant is:

    $\Delta = 3^2 - 4(2)(-1) = 9 + 8 = 17$

    So the two candidate solutions are:

    $x = \frac{-3 \pm \sqrt{17}}{4}$

    3. Check the domain condition for the identity. The identity $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$ holds only when $ab < 1$. Here $a = x$, $b = 2x$, so $ab = 2x^2$. For each candidate, compute $2x^2$:
       • For $x = \frac{-3 + \sqrt{17}}{4}$: $\sqrt{17} \approx 4.123$, so $x \approx \frac{1.123}{4} \approx 0.2808$. Then $2x^2 \approx 2(0.0788) = 0.1576 < 1$. So the identity applies directly, and the sum is indeed $\frac{\pi}{4}$.
       • For $x = \frac{-3 - \sqrt{17}}{4}$: …

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