MHT CET 2026 May 21 Shift 2 PCM Question Paper with Solution Pdf
| Updated On - May 29, 2026
The State Common Entrance Test Cell, Maharashtra conducted MHT CET 2026 May 21 Shift 2 PCM from 2 PM to 5 PM in CBT Mode. The MHT CET 2026 today’s question paper included three sections: Physics, Chemistry, and Mathematics with multiple-choice questions carrying a total of 200 marks, as per the MHT CET marking scheme, where 1 mark is awarded for every correct answer and no marks are deducted for wrong answers.
MHT CET 2026 May 21 Shift 2 PCM Question Paper with Solution Pdf is available here for download.
MHT CET 2026 May 21 Shift 2 PCM Question Paper with Solution PDF
| MHT CET 2026 May 21 Shift 2 Question Paper | Download PDF | Check Solutions |
If \( y = y(x) \) satisfies the differential equation \[ \left( \frac{2 + \sin x}{1 + y} \right) \frac{dy}{dx} = -\cos x \] and \( y(0) = 2 \), then \( y\left(\frac{\pi}{2}\right) \) is equal to:
The ratio of areas bounded by the curves \( y = \cos x \) and \( y = \cos 2x \) between \( x = 0, x = \frac{\pi}{3} \) and the \( x \)-axis is:
If \( (2 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0 \) and \( y(0) = 1 \), then find the value of \( y\left(\frac{\pi}{2}\right) \).
If \( \left( \frac{2 + \sin x}{1 + y} \right) \frac{dy}{dx} = -\cos x \) and \( y(0) = 2 \), then find the value of \( y\left(\frac{\pi}{2}\right) \).
If \( (2 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0 \) and \( y(0) = 1 \), then \( y\left(\frac{\pi}{2}\right) \) is equal to:
Solve for \( x \): \[ x + \log_{15}(5 + 3^x) = x \log_{15} 5 + \log_{15} 24 \]
For \( n \in \mathbb{N} \), if \( y = ax^{n+1} + bx^{-n} \), then \( x^2 \frac{d^2y}{dx^2} \) is equal to:
Solve for \( x \): \[ x + \log_{15}(5 + 3^x) = x \log_{15} 5 + \log_{15} 24 \]
For \( n \in \mathbb{N} \), if \( y = ax^{n+1} + bx^{-n} \), then \( x^2 \frac{d^2y}{dx^2} \) is equal to:
Let \[ f(x) = \int_1^4 \log[x] dx \] where \( [x] \) denotes the greatest integer function. Then the value of \( f(x) \) is:
MHT CET 2026 2nd Attempt Paper Analysis
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