Vardaan Learning Institute

Ans: $k = \frac{5}{3}$. Reason: $\lim_{x \to 0} \frac{\sin 5x}{3x} = \frac{5}{3} \lim_{x \to 0} \frac{\sin 5x}{5x} = \frac{5}{3}$. For continuity, limit must equal $f(0)=k$.

Ans: Continuous at both $x=1$ and $x=2$. Reason: Modulus functions are continuous everywhere. The sum of two continuous functions is also continuous.

Ans: $3a - 3b = 2$. Reason: $\text{LHL} = f(3) = 3a+1$. $\text{RHL} = 3b+3$. Equating them gives $3a+1 = 3b+3$.

Ans: Yes, continuous at $x=0$. Reason: As $x \to 0$, $x \to 0$ and $\sin(1/x)$ oscillates between $[-1, 1]$. Hence limit is $0 \times (\text{bounded value}) = 0$, which equals $f(0)$.

Ans: $k = 6$. Reason: Put $x = \frac{\pi}{2} - h$. Limit becomes $\lim_{h \to 0} \frac{k\sin h}{2h} = \frac{k}{2}$. So, $k/2 = 3 \Rightarrow k=6$.

Ans: Continuous at $x=2.5$ but discontinuous at $x=3$. Reason: $[x]$ is discontinuous at all integral points.

Proof: Let $g(x) = \cos x$ and $h(x) = x^2$. Both are continuous everywhere. Their composition $(g \circ h)(x) = \cos(x^2)$ is also continuous everywhere.

Ans: Discontinuous at $x = 2$. Reason: The denominator becomes zero, making the function undefined at that point.

Proof: It's a modulus function (continuous). $\text{LHD at } x=3 = \lim_{h \to 0} \frac{-(3-h-3) - 0}{-h} = -1$. $\text{RHD} = \lim_{h \to 0} \frac{(3+h-3) - 0}{h} = 1$. Since $\text{LHD} \neq \text{RHD}$, it is not differentiable.

Ans: $\text{LHD} = -1$ and $\text{RHD} = 1$.

Ans: It is differentiable. Reason: $f'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h) = 0$. Both LHD and RHD are $0$.

Ans: Yes. Reason: For $x \le 1$, $f'(x) = 2 \Rightarrow \text{LHD} = 2$. For $x > 1$, $f'(x) = 2x \Rightarrow \text{RHD} = 2$. Also, $\text{LHL}=\text{RHL}=f(1)=5$. Hence, continuous and differentiable.

Ans: $\frac{dy}{dx} = \frac{1}{x \log_e x \cdot \log_e 7}$.

Ans: $\frac{dy}{dx} = -\frac{\sin(\sqrt{x})}{2\sqrt{x}}$.

Ans: $6\sin^2(2x+1)\cos(2x+1)$.

Ans: $\frac{\sec^2\sqrt{x}}{4\sqrt{x\tan\sqrt{x}}}$.

Ans: $\frac{dy}{dx} = \frac{2}{\cos y - 3}$.

Ans: $\frac{dy}{dx} = -\frac{2x+y}{x+2y}$.

Ans: $\frac{dy}{dx} = \frac{y\sin(xy)}{\sin(2y) - x\sin(xy)}$.

Ans: $\frac{e^x(\sin x - \cos x)}{\sin^2 x}$.

Ans: $\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}$.

Ans: $\frac{dy}{dx} = -\frac{3x^2+2xy+y^2}{x^2+2xy+3y^2}$.

Ans: $\frac{2}{1+x^2}$. Reason: Put $x = \tan\theta$. The function simplifies to $y = 2\tan^{-1}x$.

Ans: $\frac{3}{1+x^2}$. Reason: Put $x = \tan\theta$. The function simplifies to $y = 3\tan^{-1}x$.

Ans: $\frac{2}{1+x^2}$. Reason: Put $x = \tan\theta$. Function simplifies to $y = 2\tan^{-1}x$.

Ans: $-\frac{2}{\sqrt{1-x^2}}$. Reason: Put $x = \cos\theta$. $y = \sec^{-1}(\sec 2\theta) = 2\cos^{-1}x$.

Ans: $\frac{1}{2}$. Reason: $y = \tan^{-1}\left(\frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)}\right) = \tan^{-1}(\tan(x/2)) = \frac{x}{2}$.

Ans: $\frac{1}{2}$. Reason: $y = \tan^{-1}\left(\sqrt{\tan^2(x/2)}\right) = \frac{x}{2}$.

Ans: $-1$. Reason: $y = \cos^{-1}(\cos(\frac{\pi}{2}-x)) = \frac{\pi}{2} - x$.

Ans: $-\frac{1}{2(1+x^2)}$. Reason: Put $x=\tan\theta$. $y = \cot^{-1}(\tan(\theta/2)) = \frac{\pi}{2} - \frac{\theta}{2} = \frac{\pi}{2} - \frac{1}{2}\tan^{-1}x$.

Ans: $x^{\sin x} \left(\cos x \log x + \frac{\sin x}{x}\right)$.

Ans: $(\sin x)^{\cos x} \left(\cos x \cot x - \sin x \log(\sin x)\right)$.

Ans: $\frac{dy}{dx} = \frac{y(y-x\log y)}{x(x-y\log x)}$.

Ans: $(\log x)^x \left[\log(\log x) + \frac{1}{\log x}\right] + x^{\log x} \left[\frac{2\log x}{x}\right]$.

Ans: $\frac{y}{2} \left[\frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4}\right]$.

Ans: $\frac{dy}{dx} = \frac{\log y}{(1-\log y)^2}$. Reason: Taking log gives $x\log y = y - x \Rightarrow x = \frac{y}{1+\log y}$. Differentiating implicitly gives the result.

Ans: $\frac{1}{x \log_e a}$.

Ans: $\frac{dy}{dx} = -\frac{y(y+x\log y)}{x(x+y\log x)}$.

Ans: $-\cot \theta$.

Ans: $\frac{1}{t}$.

Ans: $\cot\left(\frac{\theta}{2}\right)$. Reason: $\frac{dy}{dx} = \frac{a\sin\theta}{a(1-\cos\theta)} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)}$.

Ans: $\tan t$. Reason: $\frac{dx}{dt} = a t \cos t$ and $\frac{dy}{dt} = a t \sin t$.

Proof: Multiply $x$ and $y$: $xy = \sqrt{a^{\sin^{-1}t} \cdot a^{\cos^{-1}t}} = \sqrt{a^{\pi/2}}$ (which is constant). Diff both sides: $x\frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.

Ans: $\frac{1}{\sqrt{2}}$. Reason: $\frac{dy}{dx} = \frac{3a\tan^2\theta\sec^2\theta}{3a\sec^2\theta\sec\theta\tan\theta} = \sin\theta$. At $\theta = \pi/4$, $\sin(\pi/4) = 1/\sqrt{2}$.

Ans: $2$.

Ans: $-2\sin x - x\cos x$.

Proof: $y_1 = A\cos x - B\sin x$. $y_2 = -A\sin x - B\cos x = -(A\sin x + B\cos x) = -y$. Hence $y_2 + y = 0$.

Proof: $y_1 = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x\cos x$. $y_2 = 2e^x(\cos x - \sin x)$. Substituting $y, y_1, y_2$ into the equation yields $0$.

Proof: $y_1 = \frac{2\tan^{-1}x}{1+x^2} \Rightarrow y_1(1+x^2) = 2\tan^{-1}x$. Diff again using product rule: $(1+x^2)y_2 + y_1(2x) = \frac{2}{1+x^2}$. Multiply by $(1+x^2)$ to get the result.

Ans: $-\frac{b}{a^2 \sin^3 t}$. Reason: $\frac{dy}{dx} = -\frac{b}{a}\cot t$. $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(-\frac{b}{a}\cot t\right) \times \frac{dt}{dx} = \frac{b}{a}\text{cosec}^2t \times \left(-\frac{1}{a\sin t}\right)$.

Ans: Verified with $c = 0$. Reason: $f(-2)=f(2)=6$. Continuous and differentiable. $f'(c) = 2c = 0 \Rightarrow c=0 \in (-2, 2)$.

Ans: Verified with $c = \frac{\pi}{2}$. Reason: $f(0)=f(\pi)=0$. Continuous and differentiable. $f'(c) = \cos c = 0 \Rightarrow c=\pi/2 \in (0, \pi)$.

Ans: Verified with $c = 2.5$. Reason: $f(1)=-6, f(4)=-3$. $f'(c) = 2c-4 = \frac{-3 - (-6)}{4-1} = 1 \Rightarrow 2c = 5 \Rightarrow c=2.5 \in (1, 4)$.

Ans: $\left(\frac{7}{2}, \frac{1}{4}\right)$. Reason: By LMVT, $f'(c) = \text{slope of chord} = \frac{1-0}{4-3} = 1$. Thus $2(c-3) = 1 \Rightarrow c=3.5$.

Ans: Rolle's Theorem is NOT applicable. Reason: Although $f(-1) = f(1)$, $f(x) = |x|$ is not differentiable at $x = 0$ which lies in $(-1, 1)$.