List of top Questions asked in MHT CET- 2026

The rate of reduction is proportional to the square root of a persons existing assets at that moment. If his assets at the beginning are 10000 and they dwindle down to 5625 in 2 years, then the person will be bankrupt in

The solution of the differential equation \((x+2y^3)\frac{dy}{dx}-y = 0\) is

Let \(\overset{̄}{a} = \hat{i}+\hat{j}\), \(\overset{̄}{c} = \hat{i}-\hat{j}\) and a vector \(\overset{̄}{b}\) be such that \(\overset{̄}{a}\times \overset{̄}{b} = \overset{̄}{c}\) and \(\overset{̄}{a}\cdot \overset{̄}{b} = 3\) then \(|\overset{̄}{b}| =\)

If a vector \(3\hat{i}+4\hat{j}-5\hat{k}\) is rotated through a certain angle about the origin in the anti-clockwise direction, then the components of the new vector are \(a+1,-3,5\) . The possible values of \(a\) is

Let A, B, C, D be the points in the plane with position vectors \(-2\hat{i}-\hat{j}\), \(4\hat{i}\), \(3\hat{i}+3\hat{j}\) and \(-3\hat{i}+2\hat{j}\) respectively, then \(◻\)ABCD is

\(\int \frac{(x+1)(x+logx)^2}{x}\,dx =\)

\(\int \frac{cos^3x}{sin^2x+sinx}\,dx =\)

\(\int _{π/6}^{π/3}\frac{sinx-cosx}{1+sinxcosx}\,dx =\)

\(\int cot^4x\,dx\) is equal to

The integrating factor of the differential equation \((1+t^2)+(x-e^{tan^{-1}t})\frac{dt}{dx} = 0\) is

\(\int _0^{log5}\frac{e^x\sqrt{e^x-1}}{e^x+3}\,dx =\)

The derivative of \(log_8(log_5x)\) w. r. t. \(x\) is

The minimum value of \(\frac{logx}{x}\) in the interval \((2,\infty )\) is

The function \(f(x) = tan^{-1}(sinx+cosx)\) is an increasing function in the interval.....

A triangle has two fixed vertices A \((a,0)\) and B\((0,b)\) . Let its third vertex C is moving along the line \(x = y\). If \(s\) is the area of triangle ABC, then \(\frac{ds}{dx} =\)

If \(y^{\frac{1}{m}}+y^{\frac{-1}{m}} = 2x\) , then \((x^2-1)y_1^2 =\)

If \(A = [\begin{array}{cc}5a & -b \\ 3 & 2\end{array}]\) and \(A\,(\text{adj }A) = AA^T\), Then \(5a+b =\)

If \(y = tan^{-1}[\frac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}]\) , then \(\frac{dy}{dx} =\)

\(sin^{-1}(\frac{12}{13})+cos^{-1}(\frac{4}{5})+tan^{-1}(\frac{63}{16}) =\)

Inverse of \([\begin{array}{cc}3 & -2 \\ 1 & 4\end{array}]\) is

\(cos(cos^{-1}(-\frac{1}{2})+\frac{π}{3}) =\)

If \(f(x) = \frac{4x+3}{6x-4}\), \(x\neq \frac{2}{3}\) and \((\text{fof})(x) = g(x)\) where \(g:R-\{\frac{2}{3}\}\rightarrow R-\{\frac{2}{3}\}\), then \((g\,o\,g\,o\,g\,o\,g\,o\,g)\,(3) =\)

\(\underset{x\rightarrow 0}{lim}\frac{(5^x-1)^4\,\text{cosec}\,(xlog5)}{tan(xlog5)\cdot log(1+x^2log25)} = \ldots \ldots\)

With usual notations in \(△\)ABC, if \(bcos^2\frac{C}{2}+ccos^2\frac{B}{2} = \frac{3a}{2}\) then

Simplest form of the following switching circuit is