(a) If $x\cos(p+y) + \cos p\,\sin(p+y) = 0$, prove that $\cos p\,\dfrac{dy}{dx} = -\cos^{2}(p+y)$, where $p$ is a constant. OR (b) Find the values of $a$ and $b$ so that the function $f(x)=\begin{cases}\dfrac{x-2}{|x-2|}+a, & x 2\end{cases}$ is continuous at $x=2$.

(a) If xcos(p+y) + cos psin(p+y) = 0, prove that cos p(dy)/(dx) = -cos²(p+y), where p is a constant. OR (b) Fi

Question

x\cos(p+y) + \cos p\,\sin(p+y) = 0

(OR)

(b) Find the values of

a

and

b

so that the function

f(x)=\begin{cases}\dfrac{x-2}{|x-2|}+a, & x<2\\ a+b, & x=2\\ \dfrac{x-2}{|x-2|}+b, & x>2\end{cases}

is continuous at

x=2

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★

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Part (a)Concept understanding — Implicit Differentiation

Implicit Differentiation

When $y$ isn't alone

You can differentiate $y=x^2+3x$ term by term because $y$ is written explicitly in terms of $x$ . But an equation like $x^2+y^2=25$ , or $x^3+y^3=6xy$ , does not give $y$ by itself — solving for $y$ is messy or downright impossible.

Implicit differentiation finds $\dfrac{dy}{dx}$ without isolating $y$ : treat $y$ as an unknown function of $x$ , differentiate the whole equation as it stands, then solve for $\dfrac{dy}{dx}$ .

The one key move: $y$ is really $y(x)$

Wherever $y$ appears, picture $y(x)$ hiding inside. Differentiating a $y$ -term therefore needs the chain rule, which tacks on a factor of $\dfrac{dy}{dx}$ :

\frac{d}{dx}\big(y^2\big)=2y\,\frac{dy}{dx}.

That extra $\dfrac{dy}{dx}$ on every $y$ -term is the whole trick.

The procedure

x

Worked example

For $x^2+y^2=25$ :

2x+2y\frac{dy}{dx}=0 \quad\Rightarrow\quad \frac{dy}{dx}=-\frac{x}{y}.

The answer naturally contains both $x$ and $y$ — that is normal here. To get the slope at a point on the curve, substitute the coordinates after differentiating; there is no need to solve for $y$ first. …

Part (b)Concept understanding — Continuity At A Point

Continuity at a Point

Imagine drawing the graph of a function and putting your pen down at $x = a$ . If the function is continuous there, you can draw straight through that point without lifting your pen — no jump, no hole, no break. That is the intuition; here is the precision.

The Three-Condition Test

For $f(x)$ to be continuous at $x = a$ , all three must hold. If even one fails, $f$ is discontinuous there.

Important

Continuity at $x = a$ requires:

$f(a)$ is defined,
$\displaystyle \lim_{x \to a} f(x)$ exists (left- and right-hand limits are equal),
$\displaystyle \lim_{x \to a} f(x) = f(a)$ .

Condition 1 says $a$ is in the domain — the pen must have somewhere to land. Condition 2 says the curve approaches a single value from both sides — no jump. Condition 3 says that common approach value actually matches the function's value at $a$ — no misplaced point.

Why All Three Are Needed

$f(x) = \dfrac{x^2 - 1}{x - 1}$ has $\lim_{x \to 1} f(x) = 2$ , yet $f(1)$ is undefined (zero denominator). Condition 1 fails, leaving a hole at $(1, 2)$ .

A piecewise function shows the opposite can be fine:

f(x) = \begin{cases} x + 1 & x < 2 \\ 3 & x = 2 \\ x + 1 & x > 2 \end{cases}

Here $f(2) = 3$ , both one-sided limits equal $3$ , and they match $f(2)$ — so all three hold and $f$ is continuous at $x = 2$ .

Common Pitfalls

Watch out

"Limit exists" does not mean "continuous." The hole example has a limit but no continuity — the limit must equal the function value.

Watch out

"Defined everywhere" does not mean "continuous." A piecewise function can have a value at every point and still jump. Always check the one-sided limits.

A Quick Check …

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Part (a): rewriting the relation as $x=-\cos p\,\tan(p+y)$ and differentiating gives $\cos p\,\dfrac{dy}{dx}=-\cos^2(p+y)$ . Part (b): matching LHL, RHL and $f(2)$ at $x=2$ gives $a=1$ , $b=-1$ .

Part (a)

1. Isolate the trig ratio. From

x\cos(p+y)+\cos p\,\sin(p+y)=0,

move the second term and divide by $\cos(p+y)$ :

x=-\cos p\,\frac{\sin(p+y)}{\cos(p+y)}=-\cos p\,\tan(p+y).

2. Differentiate both sides with respect to $x$ ( $p$ is a constant, so $\cos p$ is a constant and $\dfrac{d}{dx}\tan(p+y)=\sec^2(p+y)\,\dfrac{dy}{dx}$ ):

1=-\cos p\,\sec^2(p+y)\,\frac{dy}{dx}.

3. Rearrange: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 47 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$e^{-x} + e^{-y} = 2$
›Reveal solutionSolution
To find $\frac{dy}{dx}$ for an implicitly defined function, we differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ and applying the chain rule. The result is $\boxed{-e^{y-x}}$ .

When an equation relates $x$ and $y$ but does not explicitly express $y$ as a function of $x$ (like $y = f(x)$ ), we use a technique called implicit differentiation to find $\frac{dy}{dx}$ . The core idea is that even though $y$ isn't isolated, it is still a function of $x$ .

This means that when we differentiate a term involving $y$ with respect to $x$ , we must apply the chain rule. For example, if we differentiate $g(y)$ with respect to $x$ , we get $\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$ . This $\frac{dy}{dx}$ term is crucial and often the source of errors if overlooked.

Let's apply this to the given equation.

Differentiate both sides of the equation with respect to $x$ . The given equation is $e^{-x} + e^{-y} = 2$ . We apply the derivative operator $\frac{d}{dx}$ to every term:

$\frac{d}{dx}(e^{-x}) + \frac{d}{dx}(e^{-y}) = \frac{d}{dx}(2)$

Evaluate each derivative.

For the first term, $\frac{d}{dx}(e^{-x})$ :

Using the chain rule, if $u = -x$ , then $\frac{du}{dx} = -1$ .

So, $\frac{d}{dx}(e^{-x}) = e^{-x} \cdot \frac{d}{dx}(-x) = e^{-x} \cdot (-1) = -e^{-x}$ .

For the second term, $\frac{d}{dx}(e^{-y})$ :

This is where implicit differentiation comes in. We treat $y$ as a function of $x$ .

Using the chain rule, if $v = -y$ , then $\frac{dv}{dx} = \frac{d}{dx}(-y) = -1 \cdot \frac{dy}{dx}$ .

So, $\frac{d}{dx}(e^{-y}) = e^{-y} \cdot \frac{d}{dx}(-y) = e^{-y} \cdot \left(-\frac{dy}{dx}\right) = -e^{-y} \frac{dy}{dx}$ .

Watch out

A common mistake is to forget the $\frac{dy}{dx}$ term when differentiating expressions involving $y$ with respect to $x$ . Remember, $y$ is a function of $x$ .

For the right-hand side, $\frac{d}{dx}(2)$ : …
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CBSE 2026Set 65/3/11 markMCQ
$k$
›Reveal solutionSolution
For continuity at $x = 0$ , the limit of $x^2 \sin \frac{1}{x}$ as $x \to 0$ must equal the function value $k(0+1) = k$ . Since the limit is $0$ , we need $k = 0$ .

A function is continuous at a point when three conditions align: the function is defined there, the limit exists as we approach that point, and crucially, the limit equals the function's value at that point. This problem tests whether you can recognize that continuity at $x = 0$ creates a bridge between two different expressions.

The function behaves as $x^2 \sin \frac{1}{x}$ everywhere except at zero, where it suddenly switches to $k(x+1)$ . At $x = 0$ , this second piece gives us $f(0) = k(0+1) = k$ . For continuity, we need:
$\lim_{x \to 0} f(x) = f(0)$
Since we approach zero from the region where $x \neq 0$ , the relevant limit is:
$\lim_{x \to 0} x^2 \sin \frac{1}{x} = k$
Let me find this limit.

Recognize the bounded oscillation

The sine function satisfies $-1 \leq \sin \frac{1}{x} \leq 1$ for all $x \neq 0$ , no matter how wildly $\frac{1}{x}$ oscillates as $x \to 0$ .

Apply the squeeze theorem

Multiplying the inequality by $x^2$ (which is always non-negative):

$-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$

Evaluate the bounding limits As $x \to 0$ :

$\lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0$

Conclude via the squeeze theorem …
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CBSE 2026Set A1 markMCQ
$y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}}$
(a) $\frac{1}{2y-1}$
(b) $\frac{\cos x}{2y-1}$
(c) $\frac{\sin x}{2y-1}$
(d) $\frac{2y-1}{\cos x}$
›Reveal solutionSolution

$\frac{dy}{dx} = \frac{\cos x}{2y-1}$ .

The infinite nested radical satisfies $y = \sqrt{\sin x + y}$ , so

$y^2 = \sin x + y.$

Differentiate both sides implicitly with respect to $x$ :

$2y\frac{dy}{dx} = \cos x + \frac{dy}{dx}.$

Collect $\frac{dy}{dx}$ : …

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CBSE 2026Set A1 markMCQ
$x^n + y^n = a^n$
(a) $-\frac{x^{n-1}}{y^{n-1}}$
(b) $\frac{x^{n-1}}{y^{n-1}}$
(c) $-\frac{y^{n-1}}{x^{n-1}}$
(d) $nx^{n-1}$
›Reveal solutionSolution

$\frac{dy}{dx} = -\frac{x^{n-1}}{y^{n-1}}$ .

Differentiate $x^n + y^n = a^n$ implicitly ( $a$ constant):

$n x^{n-1} + n y^{n-1}\frac{dy}{dx} = 0.$

Solve: …

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CBSE 2026Set ANNUAL1 markMCQ
$2x + 3y = \sin y$
(a) $\frac{3}{\sin y - 2}$
(b) $\frac{2}{\cos y - 3}$
(c) $\frac{\cos y + 3}{2}$
(d) $\frac{2}{\cos y}$
›Reveal solutionSolution

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ , then solve for $dy/dx$ .

$2x+3y=\sin y$

Differentiating: $2 + 3\dfrac{dy}{dx} = \cos y\,\dfrac{dy}{dx}$

$2 = \dfrac{dy}{dx}(\cos y - 3)$

…

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CBSE 2026Set ANNUAL1 mark
$Q. Prove that the function f(x) = 5x - 3 is continuous at x = -3.$
›Reveal solutionSolution

A function $f$ is continuous at $x=a$ if $\displaystyle\lim_{x\to a} f(x) = f(a)$ ; check this directly for the linear function $f(x)=5x-3$ at $a=-3$ .

Concept: $f$ is continuous at $x=a$ when: (i) $f(a)$ is defined, (ii) $\displaystyle\lim_{x\to a}f(x)$ exists, and (iii) the two are equal.

Working:

$f(-3) = 5(-3) - 3 = -15 - 3 = -18$

$\lim_{x\to -3} f(x) = \lim_{x\to -3}(5x-3) = 5(-3) - 3 = -18$

…

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CBSE 2026Set ANNUAL1 mark
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Differentiate both sides of $2x+3y=\sin y$ with respect to $x$ (using the chain rule for the $y$ -terms), then collect $\dfrac{dy}{dx}$ on one side.

Given: $2x + 3y = \sin y$

Differentiate both sides w.r.t. $x$ :

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$

$2 + 3\frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$

Collect all $\dfrac{dy}{dx}$ terms on one side: …

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CBSE 2025Set 65/4/11 markMCQ
$f$
›Reveal solutionSolution
The function has a jump at $x=1$ because the left-hand limit ( $1$ ) and the right-hand limit ( $5$ ) do not match, so it is discontinuous only at $x=1$ . The correct option is (B).

Continuity at a point means three things must hold: the function is defined there, the limit exists there, and the limit equals the function value. For a piecewise function, the only place where things can go wrong is at the boundary between the pieces — here, at $x=1$ . Everywhere else, the function is just a simple rule (either $x$ or the constant $5$ ), so it's automatically continuous.

Let’s check each candidate point.

At $x=0$

For $x \le 1$ , the rule is $f(x)=x$ . Since $0 \le 1$ , we have $f(0)=0$ .

The left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x = 0$ .

The right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$ (because near $0$ , $x$ is still $\le 1$ ).

So the limit exists and equals $0$ , which matches $f(0)$ . Continuous here.

At $x=1$ — the critical boundary

Left-hand limit: as $x$ approaches $1$ from below, $x \le 1$ , so $f(x)=x$ . Hence

$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1.$

Right-hand limit: as $x$ approaches $1$ from above, $x > 1$ , so $f(x)=5$ . Hence

$\lim_{x \to 1^+} f(x) = 5.$

The left and right limits are different ( $1 \neq 5$ ), so the two-sided limit does not exist.

The function value is $f(1)=1$ (since $1 \le 1$ ). …
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CBSE 2025Set IX1 mark
$f(x) = |x|$
›Reveal solutionSolution
Left limit, right limit and the value all equal $0$ , so $|x|$ is continuous at $0$ .

Concept. $f$ is continuous at $x=a$ iff $\displaystyle\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$ .

Here $f(x)=|x|=\begin{cases}-x,&x<0\\ x,&x\ge0\end{cases}$

Left-hand limit: $\displaystyle\lim_{x\to0^-}f(x)=\lim_{x\to0^-}(-x)=0.$

Right-hand limit: $\displaystyle\lim_{x\to0^+}f(x)=\lim_{x\to0^+}x=0.$ …
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CBSE 2025Set ANNUAL1 markMCQ
$x^2+y^2=2$
(a) $\dfrac{1-2x}{2y}$
(b) $\dfrac{2y}{1-2x}$
(c) $-\dfrac{x}{y}$
(d) $-\dfrac{y}{x}$
›Reveal solutionSolution

Differentiate $x^2+y^2=2$ implicitly with respect to $x$ , treating $y$ as a function of $x$ .

$\dfrac{d}{dx}(x^2+y^2)=\dfrac{d}{dx}(2)$

$2x+2y\dfrac{dy}{dx}=0$ …

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CBSE 2025Set ANNUAL1 markMCQ
$f(x) = |x| - |x + 1|$
(a) continuous at $x = 0$ as well as at $x = -1$
(b) continuous at $x = -1$ but not at $x = 0$
(c) discontinuous at $x = 0$ as well as at $x = -1$
(d) continuous at $x = 0$ but not at $x = -1$
›Reveal solutionSolution

$|x|$ and $|x+1|$ are each continuous everywhere, and the difference of two continuous functions is continuous.

$g(x)=|x|$ is continuous on all of $\mathbb{R}$ , and $h(x)=|x+1|$ (a shifted absolute value) is also continuous on all of $\mathbb{R}$ . Since $f(x)=g(x)-h(x)$ is a difference of two functions continuous eve …

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CBSE 2025Set ANNUAL1 mark
$f$
›Reveal solutionSolution

A function is continuous at a point when its limit there equals its value; check both.

Here $f(x)=2x+3$ (a polynomial), and we test $x=1$ .

Value: $f(1)=2(1)+3=5.$

Limit: $\displaystyle\lim_{x\to1}(2x+3)=2(1)+3=5.$

Since $\displaystyle\lim_{x\to1}f(x)=5=f(1)$ , the function is continuous at $x=1$ .

…

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Q.(a) If xcos⁡(p+y)+cos⁡p sin⁡(p+y)=0x\cos(p+y) + \cos p\,\sin(p+y) = 0xcos(p+y)+cospsin(p+y)=0, prove that cos⁡p dydx=−cos⁡2(p+y)\cos p\,\dfrac{dy}{dx} = -\cos^{2}(p+y)cospdxdy​=−cos2(p+y), where ppp is a constant.

Part (a)

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Q.(a) If $x\cos(p+y) + \cos p\,\sin(p+y) = 0$ , prove that $\cos p\,\dfrac{dy}{dx} = -\cos^{2}(p+y)$ , where $p$ is a constant.

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