✓ Board-verified
Mathematics · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 1 · Q34
If x= cos t, y = cos mt, prove that (1-x^2) (d^2 y)/(d x^2)-x dy/(d x)+ m^2 y =0.
If $\displaystyle x=\cos \mathrm{t}, \mathrm{y}=\cos \mathrm{mt}$, prove that $\displaystyle \left(1-x^{2}\right) \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}-x \frac{\mathrm{dy}}{\mathrm{d} x}+\mathrm{m}^{2} \mathrm{y}=0$.
Official answer
Proof verified: Starting with x = cos t and y = cos(mt), we find dy/dx = -m·sin(mt)/(-sin t) = m·sin(mt)/sin t, and d²y/dx² = m(m·cos(mt)·sin t - sin(mt)·cos t)/sin³t. Substituting into ($\displaystyle 1$-x²)d²y/dx² - x·dy/dx + m²y yields $\displaystyle 0$ after algebraic simplification.
Marking-scheme solution
$\displaystyle \dfrac{dx}{dt}=-\sin \mathrm{t}$, $\displaystyle \dfrac{dy}{dt}=-\mathrm{m}\sin(mt)$,
$\displaystyle \Rightarrow\dfrac{dy}{dx}=\dfrac{\mathrm{m}\sin(mt)}{\sin \mathrm{t}}$ or $\displaystyle \sin \mathrm{t}\dfrac{dy}{dx} = \mathrm{m}\sin(mt)$
differentiating w.r.t. $\displaystyle \mathrm{t}$ and dividing by $\displaystyle dx/dt$:
$\displaystyle \Rightarrow \sin \mathrm{t}\dfrac{\mathrm{d}^2\mathrm{y}}{dx^2} + \cos \mathrm{t}\times\dfrac{dy}{dx}\times\dfrac{dt}{dx} = \mathrm{m}^2\cos(mt)\times\dfrac{dt}{dx}$
$\displaystyle \Rightarrow(1-\cos^2\mathrm{t})\dfrac{\mathrm{d}^2\mathrm{y}}{dx^2} - x\dfrac{dy}{dx} = -\mathrm{m}^2\cos(mt)$
Since $\displaystyle x=\cos \mathrm{t}$, $\displaystyle \mathrm{y}=\cos mt$: $\displaystyle \Rightarrow(1-x^2)\dfrac{\mathrm{d}^2\mathrm{y}}{dx^2} - x\dfrac{dy}{dx} + \mathrm{m}^2\mathrm{y} = 0$ (proved)
All Continuity and Differentiability questions2026 question paperNCERT solutions: Continuity and DifferentiabilityPractice Continuity and Differentiability →
More from Continuity and Differentiability
- The derivative of sin (x^2) w.r.t. x, at x=√π is:2024
- The value of k for which function f(x)= {x^2, x ≥ 0; kx, x<0} is differentiable at x=0 is:2023
- If f(x)= {(sin^2 a x)/(x^2), x ≠ 0; 1, x=0} is continuous at x=0, then the value of a is:2025
- Differentiate y=√(log { sin ((x^3)/3-1)}) with respect to x.2025
- If y=(cos x- sin x)/(cos x+ sin x), then (d y)/(d x) is:2023
- If f(x)= {(log (1+a x)+ log (1-b x))/x, for x ≠ 0; k, for x=0} is continuous at x=0, then the value of k is:2025
- 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 x^2)=(cos x)/((1- sin x)^2).2023
- If f(x)= {(x^2-4 x-5)/(x+1), x ≠-1; k, x=-1} is continuous at x=-1, then the value of k is:2026
CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.