1.Ans: Points of discontinuity: $x = \frac{1}{2}, 1, \frac{3}{2}$. Reason: $\cos(\pi x)$ takes integer values $\{-1, 0, 1\}$ in $[0, 2]$. Check intervals: for $0 < x < 1/2$, $\cos(\pi x) \in (0, 1) \Rightarrow f(x) = 0$. For $1/2 \le x < 1$, $\cos(\pi x) \in (-1, 0] \Rightarrow f(x) = -1$. Limit jumps at integer changes of the inner function.
2.Ans: $\{0, 1, 2\}$. Reason: The graph of $f(x)$ follows $x$ from $0$ to $1$, then $2-x$ from $1$ to $2$. Sharp corners exist at $x=0$, $x=1$ (intersection of $|x|$ and $|x-2|$), and $x=2$.
3.Ans: $a = 2c, b = -c^2$. Reason: For continuity at $x=c$: $c^2 = ac + b$. For differentiability: $2c = a$. Substituting $a$ back gives $b = -c^2$.
4.Ans: $k = \frac{1}{8}$. Reason: Limit $\lim_{x \to 0} \frac{2\sin^2\left(\frac{1-\cos x}{2}\right)}{x^4} = 2 \lim_{x \to 0} \left(\frac{\sin\left(\frac{x^2}{4}\right)}{x^2}\right)^2 = 2 \times \left(\frac{1}{4}\right)^2 = \frac{1}{8}$.
5.Ans: $\frac{1}{2}$. Reason: Let $x = \tan\theta$. $u = \tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right) = \frac{\theta}{2} = \frac{1}{2}\tan^{-1}x$. Let $v = \tan^{-1}x$. $du/dv = 1/2$.
6.Proof: Let $y = \sqrt{\sin x + y} \Rightarrow y^2 = \sin x + y$. Differentiating both sides implicitly: $2y\frac{dy}{dx} = \cos x + \frac{dy}{dx} \Rightarrow \frac{dy}{dx}(2y-1) = \cos x$.
7.Proof: Taking log: $y \log x = x - y \Rightarrow y(1+\log x) = x \Rightarrow y = \frac{x}{1+\log x}$. Differentiate using quotient rule: $\frac{dy}{dx} = \frac{(1+\log x)\cdot 1 - x\cdot \frac{1}{x}}{(1+\log x)^2} = \frac{\log x}{(1+\log x)^2}$.
8.Ans: $\frac{2\sqrt{2}}{a}$. Reason: $dx/dt = a(-\sin t + \frac{1}{\sin t}) = a\cos t \cot t$. $dy/dt = a\cos t$. So $dy/dx = \tan t$. $d^2y/dx^2 = \sec^2t \times \frac{dt}{dx} = \frac{\sec^2t}{a\cos t\cot t} = \frac{\sec^3 t}{a\cot t}$. At $\pi/4$, $\frac{(\sqrt{2})^3}{a(1)} = \frac{2\sqrt{2}}{a}$.
9.Ans: $1$. Reason: For $x > 1$, let $x = \tan\theta$ where $\theta \in (\pi/4, \pi/2)$. $u = \sin^{-1}(\sin 2\theta) = \pi - 2\theta$. $v = \cos^{-1}(\cos 2\theta) = 2\theta$. $\frac{du}{dx} = \frac{-2}{1+x^2}$, $\frac{dv}{dx} = \frac{-2}{1+x^2}$ (since it must match $2\theta$ for $x>0$ domain but wait, $\cos^{-1}(\frac{1-x^2}{1+x^2}) = 2\tan^{-1}x$ is valid for all $x \ge 0$). Wait, for $x>1$, $u = \pi - 2\tan^{-1}x$, $v = 2\tan^{-1}x$. $du/dv = -1$.
10.Proof: $y_1 = n(x+\sqrt{1+x^2})^{n-1} \cdot \left(1 + \frac{x}{\sqrt{1+x^2}}\right) = \frac{ny}{\sqrt{1+x^2}}$. Square and cross-multiply: $(1+x^2)y_1^2 = n^2y^2$. Differentiate again: $(1+x^2)2y_1 y_2 + 2x y_1^2 = 2n^2y y_1$. Cancel $2y_1$.
11.Proof: $x = \frac{\cos y}{\cos(a+y)}$. Diff wrt $y$: $\frac{dx}{dy} = \frac{\cos(a+y)(-\sin y) - \cos y(-\sin(a+y))}{\cos^2(a+y)} = \frac{\sin(a+y)\cos y - \cos(a+y)\sin y}{\cos^2(a+y)} = \frac{\sin a}{\cos^2(a+y)}$. Invert to get $dy/dx$.
12.Ans: $c = \frac{mb+na}{m+n}$. Reason: $f(a)=f(b)=0$. $f'(x) = m(x-a)^{m-1}(x-b)^n + n(x-b)^{n-1}(x-a)^m = 0$. Factoring out $(x-a)^{m-1}(x-b)^{n-1}$ gives $m(x-b) + n(x-a) = 0 \Rightarrow x = \frac{mb+na}{m+n}$. It divides $[a,b]$ in $n:m$ ratio internally.
13.Proof: Let $f(x) = \tan^{-1}x$ in $[a, b]$. LMVT $\Rightarrow \frac{\tan^{-1}b - \tan^{-1}a}{b-a} = \frac{1}{1+c^2}$ where $a < c < b$. Since $0 < a < c < b$, we have $\frac{1}{1+b^2} < \frac{1}{1+c^2} < \frac{1}{1+a^2}$. Multiply by $b-a$ to get the result.
14.Ans: $\text{LHD} = 0, \text{RHD} = 0$. Yes, differentiable. Reason: Around $x=2$, $|x-1| = x-1$ and $|x-3| = 3-x$. So $f(x) = x-1 + 3-x = 2$ (a constant). Derivative of constant is 0.
15.Proof: Cross multiply: $y\sqrt{1-x^2} = \sin^{-1}x$. Differentiate: $y \cdot \frac{-x}{\sqrt{1-x^2}} + \frac{dy}{dx}\sqrt{1-x^2} = \frac{1}{\sqrt{1-x^2}}$. Multiply by $\sqrt{1-x^2}$: $-xy + (1-x^2)\frac{dy}{dx} = 1$.
16.Proof: Let $x^3 = \sin\theta, y^3 = \sin\phi$. $\cos\theta + \cos\phi = a^3(\sin\theta - \sin\phi)$. Use sum-to-product: $\cot\frac{\theta-\phi}{2} = a^3$. So $\theta - \phi = 2\cot^{-1}(a^3)$. Differentiate: $\frac{3x^2}{\sqrt{1-x^6}} - \frac{3y^2}{\sqrt{1-y^6}}\frac{dy}{dx} = 0$. Isolate $dy/dx$.
17.Proof: $y_1 = \frac{a e^{a\sin^{-1}x}}{\sqrt{1-x^2}} = \frac{ay}{\sqrt{1-x^2}}$. Square and cross-multiply: $(1-x^2)y_1^2 = a^2y^2$. Differentiating yields $(1-x^2)2y_1 y_2 - 2x y_1^2 = 2a^2y y_1$. Cancel $2y_1$.
18.Ans: $\frac{2}{1+4x^2} + \frac{3}{1+9x^2}$. Reason: $y = \tan^{-1}\left(\frac{2x+3x}{1-(2x)(3x)}\right) = \tan^{-1}(2x) + \tan^{-1}(3x)$. Apply sum rule.
19.Ans: $2$ points. Reason: Sharp edges occur where $\sin x = \cos x$, which is at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ in the given interval.
20.Ans: Differentiable at $x=1$, Non-differentiable at $x=2$. Reason: The term $\sin(\pi x)$ becomes zero at integers, dampening the jump discontinuity of $[x^2]$. At $x=1$, LHD=RHD=$-\pi$. At $x=2$, jump is unrecoverable smoothly for derivative.
21.Proof: Let $f(t) = e^t$ on $[0, x]$. By LMVT, $\frac{e^x - e^0}{x - 0} = e^c$ for some $c \in (0, x)$. Since $c > 0$, $e^c > e^0 = 1$. Thus $\frac{e^x - 1}{x} > 1 \Rightarrow e^x > 1+x$.
22.Proof: $y = x^y$. Take log: $\log y = y\log x$. Differentiate implicitly: $\frac{1}{y}\frac{dy}{dx} = y \cdot \frac{1}{x} + \log x \frac{dy}{dx}$. Rearrange to isolate $\frac{dy}{dx}$.
23.Ans: $\frac{3x^2+1}{2(x^3+x)\log x} - \frac{\log(x^3+x)}{2x(\log x)^2}$. Reason: Rewrite as $y = \frac{\log(x^3+x)}{2\log x}$ using change of base formula, then apply quotient rule.
24.Proof: $y_1 = \frac{ad-bc}{(cx+d)^2}$. $y_2 = \frac{-2c(ad-bc)}{(cx+d)^3}$. $y_3 = \frac{6c^2(ad-bc)}{(cx+d)^4}$. Substitute into $2y_1y_3 = 2\left(\frac{ad-bc}{(cx+d)^2}\right)\left(\frac{6c^2(ad-bc)}{(cx+d)^4}\right) = \frac{12c^2(ad-bc)^2}{(cx+d)^6} = 3(y_2)^2$.
25.Ans: $0$. Reason: $f(x) = x^3$ for $x \ge 0$ and $-x^3$ for $x < 0$. $f'(x) = 3x^2$ for $x \ge 0$ and $-3x^2$ for $x < 0$. $f''(x) = 6x$ for $x \ge 0$ and $-6x$ for $x < 0$. At $x=0$, LHD of $f'$ is 0 and RHD of $f'$ is 0, so $f''(0) = 0$.
26.Proof: Let $t = x+\sqrt{x^2-1}$. Notice $x-\sqrt{x^2-1} = 1/t$. $y = t^m + t^{-m}$. Find $y_1$, then square and cross-multiply to form a diff equation involving $(x^2-1)y_1^2$. Diff again to get result.
27.Ans: $a=0$. Reason: $y = x^{1/2} + a x^{-1/2}$. $y_1 = \frac{1}{2}x^{-1/2} - \frac{a}{2}x^{-3/2}$. $y_2 = -\frac{1}{4}x^{-3/2} + \frac{3a}{4}x^{-5/2}$. Substitute into eq: $2x(-\frac{1}{4}x^{-3/2} + \frac{3a}{4}x^{-5/2}) + 3(\frac{1}{2}x^{-1/2} - \frac{a}{2}x^{-3/2}) = x^{-1/2} + 0 \cdot x^{-3/2}$. We need this to be 0 for all x, which requires $a=0$ and the structure dictates it works when $a=0$ wait, actually only $a$ matters for cancellation.
28.Ans: Continuous at $x=1$. Reason: Limit $\lim_{x \to 1} |x-1|$ is 0, and $([x] - [-x])$ is a bounded function around $x=1$. $0 \times (\text{bounded}) = 0$. Function value is also 0.
29.Ans: $\frac{1}{e}$. Reason: $y = t^t$ where $t=\log_e x$. $dy/dx = (dy/dt) \cdot (dt/dx) = t^t(1+\log_e t) \cdot \frac{1}{x}$. At $x=e$, $t=1$. $y'(e) = 1^1(1+0) \cdot \frac{1}{e} = \frac{1}{e}$.
30.Ans: $\frac{1}{2\sqrt{1-x^2}}$. Reason: Substitute $x = \cos 2\theta$. $y = \tan^{-1}\left(\frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4}-\theta\right)\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x$. Diff gives result.
