Differentiability JEE Main Mathematics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: c

Explanation: ​_a_ = –2, _b_ = –3​ ​Here, $f^{'} \left(x\right) = \begin{cases} 2 b x + a , & x \geq - 1 \\ 2 a x , & x < - 1 \end{cases}$ Given, _f_ '(_x_) is continuous everywhere, ​$\therefore \underset{x \rightarrow - 1^{-}}{lim } f^{'} \left(x\right) = \underset{x \rightarrow - 1^{+}}{lim } f^{'} \left(x\right)$ ​$\Rightarrow - 2 a = - 2 b + a \Rightarrow 3 a = 2 b$ ​$\Rightarrow a = 2 , b = 3$ or _a_ = –2, _b_ = –3

Correct Answer: 3

Explanation: $3$ $f \left(x\right) = \left|x - 2\right| - \left|x\right| + \left|2 x + 4\right| \Rightarrow \textrm{ } f \left(x\right) = \left\{\begin{matrix} \begin{matrix}- 2 x - 2 & ; & x < - 2 \\ 2 x + 6 & ; & - 2 \leq x < 0 \\ 6 & ; & 0 \leq x < 2\end{matrix} \\ \textrm{ }\textrm{ }\textrm{ } 2 x + 2                 ;                           \textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ } x \geq 2 \end{matrix}\right$ $\Rightarrow \textrm{ } f ' \left(x\right) = \left\{\begin{matrix} - 2 & ; & x < - 2 & \\ 2 & ; & - 2 \leq x < 0 & \\ 0 & ; & 0 \leq x < 2 & \\ 2 & ; & x \geq 2 & \end{matrix}\right$ $\Rightarrow \textrm{ } f \left(x\right) \textrm{ }\text{is}\textrm{ }\text{non}-\text{differentiable}\textrm{ }\text{at}\textrm{ }\text{3}\textrm{ }\text{points}.$

Correct Answer: d

Explanation: $a = \frac{1}{2}$ $\begin{matrix}\begin{matrix}f \left(x\right) = \left\{\right. \begin{matrix}x + a , & x 1 \\ a x^{2} + 1 , & x < 1\end{matrix}\end{matrix} \\ \begin{matrix}f \left(x\right) i s d e r i v a b l e a t x = 1\end{matrix} \\ \begin{matrix}f^{'} \left(1^{+}\right) = f^{'} \left(1^{}\right)\end{matrix} \\ \begin{matrix}\begin{matrix}1 = 2 a x \\ a t x = 1\end{matrix}\end{matrix} \\ \begin{matrix}2 a = 1 \\ a = \frac{1}{2}\end{matrix}\end{matrix}$

Correct Answer: b

Explanation: ​continuous at _x_ = 0   _f_(_x_) = sin<$^{–1}$ (cos _x_) ​LHL: ​ ​$\underset{x \rightarrow 0^{-}}{lim } \left(sin \right)^{- 1} \left(cos x\right) = \underset{h \rightarrow 0}{lim } \left(sin \right)^{- 1} \left(cos \left(0 - h\right)\right)$ $\underset{h \rightarrow 0}{lim } \left(sin \right)^{- 1} \left(cos \left(- h\right)\right) = \underset{h \rightarrow 0}{lim } \left(sin \right)^{- 1} \left(cos 0\right) = \frac{\pi}{2}$ RHL: $\underset{x \rightarrow 0^{+}}{lim } \left(sin \right)^{- 1} \left(cos x\right)$ ​$= \underset{h \rightarrow 0}{lim } \left(sin \right)^{- 1} cos \left(0 + h\right)$ = sin<$^{–1}$ cos 0 ​$= \frac{\pi}{2}$ Thus, LHL = RHL $= f \left(0\right) = \frac{\pi}{2}$ RHD: ​$\underset{h \rightarrow 0}{lim } \frac{f \left(x + h\right) - f \left(x\right)}{h} = \frac{\left(sin \right)^{- 1} \left(cos h\right) - 1}{h}$ ​$\underset{h \rightarrow 0}{lim } \frac{\left(sin \right)^{- 1} \left(cos h\right) - 1}{h} = \frac{1 - sin h}{\sqrt{1 - \left(cos \right)^{2} h}} = \frac{- sin h}{sin h} = - 1$​​ LHD: $\underset{h \rightarrow 0}{lim } \frac{f \left(x - h\right) - f \left(x\right)}{- h} = \underset{h \rightarrow 0}{lim } \frac{\left(sin \right)^{- 1} \left(cos \left(- h\right)\right) - 1}{- h}$ ​$\underset{h \rightarrow 0}{lim } \frac{\left(sin \right)^{- 1} \left(cos h\right) - 1}{- h} = \frac{- sin h}{- sin h} = 1$​​ ​LHD ≠ RHD Thus, function is not differentiable at _x_ = 0.

Correct Answer: c

Explanation: continuous at   $x=\frac{\pi }{2}$ but not differentiable at  $x=\frac{\pi }{2}$  $f \left(x\right) = \begin{cases} - 1 + x ; & - \infty < x < 0 \\ - 1 + sin x ; & 0 \leq x < \frac{\pi}{2} \\ cos x ; & \frac{\pi}{2} \leq x < \infty \end{cases}$ To check the differentiability at $x   =   0$ $L H D = \underset{h \rightarrow 0}{lim} \frac{f \left(0 - h\right) - f \left(0\right)}{- h} = \underset{h \rightarrow 0}{lim} \frac{1 + 0 - h - \left(- 1\right)}{- h} = 1$ $R H D = \underset{h \rightarrow 0}{lim} \frac{f \left(0 + h\right) - f \left(0\right)}{- h} = \underset{h \rightarrow 0}{lim} \frac{- 1 + sin \textrm{ } h + 1}{h} = 1$ $\because \textrm{ }\textrm{ }\textrm{ }\textrm{ } L H D = R H D$ $\therefore$Differentiable at $x   =   0$ $\Rightarrow$   Continuous at x = 0. To check the continuity at  $x = \frac{\pi}{2}$ $L H L \textrm{ }\textrm{ }\textrm{ }\textrm{ } \underset{x \rightarrow \left(\frac{\pi}{2}\right)^{-}}{lim} f \left(x\right) = \underset{x \rightarrow \left(\frac{\pi}{2}\right)^{-}}{lim} \left(- 1 + sin x\right) = 0$ $R H L \textrm{ }\textrm{ }\textrm{ }\textrm{ } \underset{x \rightarrow \left(\frac{\pi}{2}\right)^{+}}{lim} f \left(x\right) = \underset{x \rightarrow \left(\frac{\pi}{2}\right)^{+}}{lim} cos x = 0$ Continuous at  $x = \frac{\pi}{2}$. To check the differentiability at  $x = \frac{\pi}{2}$ $L H D \textrm{ }\textrm{ }\textrm{ }\textrm{ } = \underset{h = 0}{lim} \frac{f \left(\frac{\pi}{2} - h\right) - f \left(\frac{\pi}{2}\right)}{- h} = \underset{h \rightarrow 0}{lim} \frac{- 1 + cos \textrm{ } h - 0}{- h} = 0$ $R H D \textrm{ }\textrm{ }\textrm{ }\textrm{ } = \underset{h = 0}{lim} \frac{f \left(\frac{\pi}{2} + h\right) - f \left(\frac{\pi}{2}\right)}{h} = \underset{h \rightarrow 0}{lim} \frac{- sin \textrm{ } h - 0}{h} = - 1$ $L H D \neq R H D$ Not differentiable at  $x = \frac{\pi}{2}$.