List of top Mathematics Questions on Second Order Derivative asked in MHT CET

Let \(y^{2}=a\cos ^{2}x+b\sin ^{2}x\) be such that \(a+b=2024\) and \(a>0,b>0\) follows \(y^{4}+y^{3}y^{''}=k\). If \(\lambda\) is the maximum possible value of \(k\), then \(\lambda =\)
If \(y^{1/n}+y^{-1/n}=2x\), then \((x^{2}-1)\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}=\ldots\)
If \(y=\cos ^{-1}x\), then \(\frac{d^{2}y}{dx^{2}}=\)
If \(y=x-e^{x}\), then \(\frac{d^{2}x}{dy^{2}}=\) ...
If \(f(x)=\begin{vmatrix} x^{2} & \sin x & e^{x} \\ 2 & 1 & -1 \\ p & p^{2} & 1 \end{vmatrix}\) then \(f^{''}(0)=\)
If \(x=at^{2}\) and \(y=2at,t\neq 0\) then ...
If \(x = asin^3t\) and \(y = acos^3t\), then the value of \(\frac{d^2y}{dx^2}\) at \(t = \frac{π}{3}\) is equal to
If \(x = 4t^3+3,y = 3t^4+4\) and \(\frac{\frac{d^2x}{dy^2}}{(\frac{dx}{dy})^n}\) is constant then the value of \(n\) is
If \(x^m+y^m = k,(m\neq 1)\) and \(y^{''} = \frac{ax^b}{y^c}\) such that \(a+b+c = 0\), then the value of k is...
Let \(x = at^2-1\), where \(a > 0\) and \(y = t^3+1\). If at \(t = 1\), \(\frac{d^2y}{dx^2} = \frac{3}{16}\), then the value of \(a\) is...
If \(f(x) = cot^{-1}(\frac{3x-x^3}{1-3x^2})\) and \(g(x) = cos^{-1}(\frac{1-x^2}{1+x^2})\), then \(\underset{x\rightarrow a}{lim}\frac{f(x)-f(a)}{g(x)-g(a)}\), \((0 < a < \frac{1}{2})\) is
If \(y^{\frac{1}{m}}+y^{\frac{-1}{m}} = 2x\) , then \((x^2-1)y_1^2 =\)
If \(e^y+xy = e\), then the ordered pair \((\frac{\text{d}y}{\text{d}x},\frac{\text{d}^2y}{\text{d}x^2})\) at \(x = 0\) is equal to
If \(y = x+e^x\), then \(\frac{d^2x}{dy^2}\) is equal to
If \(x^3+y^3 = 6\) and \(\frac{d^2y}{dx^2}\cdot \frac{d^2x}{dy^2} = \frac{m}{(xy)^n}\), where \(m,n\in R\) then \(\frac{m}{n} =\)
If \(y = \sqrt{x^2+8x+3}\), then the value of \(\frac{d^2y}{dx^2}\) at \(x = 3\) is
If \( y = \cos^{-1} x \), find \( \frac{d^2y}{dx^2} \) in terms of \( y \).
For \(n\in\mathbb{N}\), if \[ y=ax^{n+1}+bx^{-n}, \] then \[ x^2\frac{d^2y}{dx^2} \] is equal to:
For \(n\in\mathbb{N}\), if \[ y=ax^{n+1}+bx^{-n}, \] then \[ x^2\frac{d^2y}{dx^2} \] is equal to:
If $x = a \sin t - b \cos t$ and $y = a \cos t + b \sin t$, and it is given that $\frac{d^2y}{dx^2} = 0$, then:

If \(y=\mathrm{a}^x \cdot \mathrm{~b}^{2 x-1}\), then \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) is equal to

If \(\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1\), then \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) is

If \((\mathrm{a}+\mathrm{b} x) \mathrm{e}^{\frac{y}{x}}=x\), then \(x^3 \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\) is equal to

If \[ x = \sqrt{e^{\sin^{-1} t}} \quad \text{and} \quad y = \sqrt{e^{\cos^{-1} t}}, \] then find \[ \frac{d^2 y}{dx^2}. \]
If $y = \tan^{-1}\left[\frac{\log\left(\frac{e}{x^2}\right)}{\log(ex^2)}\right] + \tan^{-1}\left[\frac{3 + 2\log x}{1 - 6\log x}\right]$, then $\frac{d^2y}{dx^2} = $