CBSE Sample Paper 2026-27 · Section B · 2 marks

Q22: If $y = \sin(x + y)$ and $\frac{dy}{dx} = 0$ ⁠, prove that $y = \pm 1$

Sample 2026-27Ch 5 · Continuity and Differentiability

The question

If $y = \sin(x + y)$ and $\dfrac{dy}{dx} = 0$ ⁠, then prove that $y = \pm 1$ ⁠.

Exam answer

Differentiating $y = \sin(x + y)$ w.r.t. $x$ ⁠,

\frac{dy}{dx} = \cos(x + y)\left(1 + \frac{dy}{dx}\right)

Why?

Use the chain rule, with $u = x + y$ inside the sine:

\frac{d}{dx}\sin u = \cos u\cdot\frac{du}{dx}

Here $\frac{du}{dx} = 1 + \frac{dy}{dx}$ ⁠.

Watch out: $y$ depends on $x$ ⁠, so the derivative of $y$ is $\frac{dy}{dx}$ ⁠, not 0.

\begin{aligned} \Rightarrow\ &\frac{dy}{dx} - \cos(x + y)\frac{dy}{dx} \\ &\quad = \cos(x + y) \end{aligned}

\begin{aligned} \Rightarrow\ &\frac{dy}{dx}\big(1 - \cos(x + y)\big) \\ &\quad = \cos(x + y) \end{aligned}

Why?

Open the bracket, move the $\frac{dy}{dx}$ term to the left, and take $\frac{dy}{dx}$ out as a common factor.

Write $c$ for $\cos(x + y)$ to keep the lines short:

\begin{aligned} \frac{dy}{dx} &= c\left(1 + \frac{dy}{dx}\right) \\ \frac{dy}{dx} &= c + c\frac{dy}{dx} \\ \frac{dy}{dx} - c\frac{dy}{dx} &= c \\ \frac{dy}{dx}(1 - c) &= c \end{aligned}

As $\frac{dy}{dx} = 0$ ⁠, this gives $\cos(x + y) = 0$ ⁠.

Why?

Put $\frac{dy}{dx} = 0$ ⁠: the left side is $0 \times \big(1 - \cos(x + y)\big) = 0$ ⁠. So the right side, $\cos(x + y)$ ⁠, must also be 0.

Since $\sin^2(x + y) + \cos^2(x + y) = 1$ ⁠,

\begin{aligned} &\sin^2(x + y) = 1 \\ \Rightarrow\ &\sin(x + y) = \pm 1 \end{aligned}

Why?

With $\cos(x + y) = 0$ ⁠, the identity gives $\sin^2(x + y) = 1 - 0 = 1$ ⁠.

\begin{aligned} \sin^2(x + y) &= 1 - \cos^2(x + y) \\ &= 1 - (0)^2 \\ &= 1 - 0 = 1 \end{aligned}

A square equal to 1 has two square roots, $+1$ and $-1$ ⁠. Keep both.

Finally $\sin(x + y)$ is $y$ itself, by the given equation.

$\therefore\ y = \sin(x + y) = \pm 1$ ⁠. Hence proved.

Where this comes from

Chapter 5, Continuity and Differentiability: Implicit differentiation.

Question type: Implicit

How to spot it: click to see

Type 5. Implicit: $x$ and $y$ mixed in one equation

Spot it: an equation such as $x^2 + xy + y^2 = 100$ that you cannot easily solve for $y$ ⁠.

Here: $y$ appears on both sides of $y = \sin(x + y)$ ⁠: differentiate, and the inside $x + y$ gives $1 + \frac{dy}{dx}$ ⁠. With $\frac{dy}{dx} = 0$ ⁠, $\cos(x + y) = 0$ ⁠, so $\sin(x + y) = \pm1$ ⁠.

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Continuity and Differentiability in past papers: question 16 of 35 · see them all

Q22: If y=sin⁡(x+y)y = \sin(x + y)y=sin(x+y) and dydx=0\frac{dy}{dx} = 0dxdy​=0⁠, prove that y=±1y = \pm 1y=±1

Q22: If $y = \sin(x + y)$ and $\frac{dy}{dx} = 0$ ⁠, prove that $y = \pm 1$