Jee Main Math Class 12 Limits Continuity And Differentiability JEE Main Physics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: 62

Explanation: $f(g(x))=\left[2g^2(x)\right]+1$ $=\begin{cases} \left[2(2x-3)^2\right]+1 & x<0 \\ \left[2(2x+3)^2\right]+1 & x\ge 0 \end{cases}.$ $\therefore$ fog is discontinuous whenever $2(2x-3)^2$ or $2(2x+3)^2$ belongs to integer except $x=0$ . $\therefore 62$ points of discontinuity.

Correct Answer: (d)

Explanation: $f(x)=\;\frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3}$ is continuous at $x=0$ $\lim\limits_{x\to 0}=\;\frac{3x-\frac{(3x)^3}{x^3}+\dots+\alpha\left(x-\frac{x^3}{x^3}\dots\right)-\beta\left(1-\frac{(3x)^2}{x^2}\dots\right)}{x^3}=f(0)$ $\lim\limits_{x\to 0}=\;\frac{-\beta+x(3+\alpha)+\frac{9\beta x^2}{x^2}+\left(\frac{-27}{3!}-\frac{\alpha}{x^3}\right)x^3\dots}{x^3}=f(0)$ $\;\text{ for exist }$ $\beta=0,\;3+\alpha=0,\;-\frac{27}{3!}-\frac{\alpha}{x^3}=f(0)$ $\alpha=-3,\;-\frac{27}{6}-\frac{(-3)}{6}=f(0)$ $f(0)=\;\frac{-27+3}{6}=-4$

Correct Answer: 8

Explanation: Probable values of $x$ where $\left[ \frac{x^2}{2} \right]$ may be discontinuous on $x \in [0,4]$ are $= 1,2,3,4,5,6,7,8$ $x = \sqrt{2}, 2, \sqrt{6}, 2\sqrt{2}, \sqrt{10}, 2\sqrt{3}, \sqrt{14}, 4$ And for $\left[ \sqrt{x} \right]$ corresponding values are $x=1,2$ On checking for continuity at these points we get the $f(x)$ is discontinuous at $x=1, \sqrt{2}, 2, \sqrt{6}, 2\sqrt{2}, \sqrt{10}, 2\sqrt{3}, \sqrt{14}$ Hence, $f(x)$ is discontinuous for 8 values of $x \in [0,4]$

Correct Answer: (a)

Explanation: $\lim\limits_{x \rightarrow 1}\;\frac{f(x)}{x-1}$ $f(x)+f'(x)+f''(x)=x^5+64$ Let $f(x)=x^5+a x^4+b x^3+c x^2+d x+e$ $f'(x)=5 x^4+4 a x^3+3 b x^2+2 c x+d$ $f''(x)=20 x^3+12 a x^2+6 b x+2 c$ $x^5+(a+5)x^4+(b+4a+20)x^3+(c+3b+12a)x^2+(d+2c+6b)x+e+d+2c=x^5+64$ $\Rightarrow a+5=0$ $b+4a+20=0$ $c+3b+12a=0$ $d+2c+6b=0$ $e+d+2c=64$ $\therefore a=-5, b=0, c=60, d=-120, e=64$ $\therefore f(x)=x^5-5x^4+60x^2-120x+64$ Now, $\lim\limits_{x \rightarrow 1}\;\frac{x^5-5x^4+60x^2-120x+64}{x-1}$ is ( $\frac{0}{0}$ form) By L' Hospital rule $\lim\limits_{x \rightarrow 1}\;\frac{5x^4-20x^3+120x-120}{1}$ $=-15$

Correct Answer: (c)

Explanation: $f:(0,\infty) \longrightarrow R$ $f(x)=e^{-\left|\log_e x\right|}$ $f(x)=\;\frac{1}{e^{|\ln x|}}=\begin{cases} \frac{1}{e^{-\ln x}} & 0 < x<1 \\ \frac{1}{e^{\ln x}} & x \ge 1 \end{cases}$ $\begin{cases} \frac{1}{\frac{1}{x}} = x & 0 < x<1 \\ \frac{1}{x}, & x \ge 1 \end{cases}$  $m=0$ (No point at which function is not continuous) $n=1$ (Not differentiable) $\therefore m+n=1$