CBSE 2026 · Region 5 · Set 1 · Q33 · 5 marks
If $\displaystyle \mathrm{y} \sqrt{\mathrm{x}^{2}+1}=\log \sqrt{\mathrm{x}^{2}+1}-\mathrm{x}$, show that \[\left(\mathrm{x}^{2}+1\right) \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{xy}+1=0 . \]Find the differential of $\displaystyle \mathrm{x}^{\cot \mathrm{x}}+\frac{2 \mathrm{x}^{2}-3}{2 \mathrm{x}^{2}-\mathrm{x}+2}$ with respect to x.
If $\displaystyle \mathrm{y} \sqrt{\mathrm{x}^{2}+1}=\log \sqrt{\mathrm{x}^{2}+1}-\mathrm{x}$, show that \[\left(\mathrm{x}^{2}+1\right) \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{xy}+1=0 . \]
Find the differential of $\displaystyle \mathrm{x}^{\cot \mathrm{x}}+\frac{2 \mathrm{x}^{2}-3}{2 \mathrm{x}^{2}-\mathrm{x}+2}$ with respect to x.
Marking-scheme solution
Differentiating both sides with respect to x, we get
$\displaystyle \sqrt{x^{2}+1} \dfrac{dy}{dx}+y \cdot \dfrac{x}{\sqrt{x^{2}+1}}=\dfrac{1}{\sqrt{x^{2}+1}} \cdot \dfrac{x}{\sqrt{x^{2}+1}}-1$
$\displaystyle \left(x^{2}+1\right) \dfrac{dy}{dx}+y \cdot x=\dfrac{x}{\sqrt{x^{2}+1}}-\sqrt{x^{2}+1}$
In case student, considers $\displaystyle \sqrt{x^{2}+1}-x$ as part of the log:
$\displaystyle \sqrt{x^{2}+1} \dfrac{dy}{dx}+y \cdot \dfrac{x}{\sqrt{x^{2}+1}}=\dfrac{1}{\sqrt{x^{2}+1}-x} \cdot\left(\dfrac{x}{\sqrt{x^{2}+1}}-1\right)$
$\displaystyle \sqrt{x^{2}+1} \dfrac{dy}{dx}+y \cdot \dfrac{x}{\sqrt{x^{2}+1}}=\dfrac{1}{\sqrt{x^{2}+1}-x} \cdot\left(\dfrac{x-\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}\right)$
$\displaystyle \sqrt{x^{2}+1} \dfrac{dy}{dx}+y \cdot \dfrac{x}{\sqrt{x^{2}+1}}=\dfrac{-1}{\sqrt{x^{2}+1}}$
gives $\displaystyle \left(x^{2}+1\right) \dfrac{dy}{dx}+y \cdot x=-1$ or $\displaystyle \left(x^{2}+1\right) \dfrac{dy}{dx}+x y+1=0$
Let $\displaystyle u=x^{\cot x}$ and $\displaystyle v=\dfrac{2 x^{2}-3}{2 x^{2}-x+2}$
$\displaystyle \log u=\cot x \cdot \log x$
$\displaystyle \dfrac{1}{u} \dfrac{du}{dx}=\dfrac{\cot x}{x}+\log x\left(-\operatorname{cosec}^{2} x\right)$
$\displaystyle \dfrac{du}{dx}=x^{\cot x}\left(\dfrac{\cot x}{x}-\operatorname{cosec}^{2} x \cdot \log x\right)$
$\displaystyle \dfrac{dv}{dx}=\dfrac{\left(2 x^{2}-x+2\right)(4 x)-\left(2 x^{2}-3\right)(4 x-1)}{\left(2 x^{2}-x+2\right)^{2}}$
$\displaystyle =\dfrac{-2 x^{2}+20 x-3}{\left(2 x^{2}-x+2\right)^{2}}$
Required derivative $\displaystyle =x^{\cot x}\left(\dfrac{\cot x}{x}-\operatorname{cosec}^{2} x \cdot \log x\right)+\dfrac{-2 x^{2}+20 x-3}{\left(2 x^{2}-x+2\right)^{2}}$
Continuity and DifferentiabilityLogarithmic DifferentiationApplylong_answermedium
More from Continuity and Differentiability
- If f(x)=. is continuous at x=0, then the value of a is:2025 · asked 4×
- The value of k for which function f(x)=. is differentiable at x=0 is:2023 · asked 3×
- If y=( cos x- sin x)/( cos x+ sin x), then (d y)/(d x) is:2023 · asked 3×
- If f(x)=. is continuous at x=0, then the value of k is:2025 · asked 3×
- Differentiate sec^-1( 1√1-x^2) w.r.t. sin^-1(2 x √1-x^2). OR If y= tan x+ sec x, then prove that (d^2 y)/(d…2023 · asked 3×
- If f(x)=. is continuous at x=-1, then the value of k is:2026 · asked 3×
- The value of k for which the function f(x)= casesx^2 sin 1/x, & x ≠ 0 k(x+1), & x=0 cases is a continuous…2026 · asked 3×
- If tan ((x+y)/(x-y))=k, then (dy)/(dx) is equal to2023 · asked 3×
CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.