QBManager

Limits, Continuity and Differentiability

80310 If \(x=2 \cos t-\cos 2 t\) and \(y=2 \sin t-\sin 2 t\), then the value of \(\frac{d^{2} y}{d x^{2}}\) at \(t=\frac{\pi}{2}\) is

1 \(\frac{3}{2}\)

2 \(\frac{-5}{2}\)

3 \(\frac{5}{2}\)

4 \(\frac{-3}{2}\)

Explanation:

(D) : Given, \(x=2 \cos t-\cos 2 t\)
\(\frac{\mathrm{dx}}{\mathrm{dt}}=-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t}\)
And, \(y=2 \sin \mathrm{t}-\sin 2 \mathrm{t}\)
\(\frac{\mathrm{dy}}{\mathrm{dt}}=2 \cos \mathrm{t}-2 \cos 2 \mathrm{t}\)
Therefore,
\(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \Rightarrow \frac{d y}{d x}=\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t}\)
\(\frac{d y}{d x}=\frac{\cos t-\cos 2 t}{-\sin t+\sin 2 t} \Rightarrow \frac{d y}{d x}=\frac{\cos t-\cos 2 t}{\sin 2 t-\sin t}\)
We know that,
\(\cos \mathrm{C}-\cos \mathrm{D}=-2 \sin \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)
\(\sin \mathrm{C}-\sin \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)
\(\frac{d y}{d x}=\frac{2 \sin \left(\frac{t+2 t}{2}\right) \cdot \sin \left(\frac{t-2 t}{2}\right)}{2 \cos \left(\frac{t+2 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}\)
\(\frac{d y}{d x}=\frac{-\sin \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}{\cos \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}\)
\(\frac{d y}{d x}=\frac{\sin \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}{\cos \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)} \Rightarrow \frac{d y}{d x}=\tan \left(\frac{3 t}{2}\right)\)
Therefore,
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}} \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\tan \left(\frac{3 \mathrm{t}}{2}\right)\right)}{(-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t})}\)
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{3}{2} \cdot \sec ^{2}\left(\frac{3 \mathrm{t}}{2}\right)}{-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t}}\)
\(\therefore \quad\left|\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right|_{\mathrm{t}=\frac{\pi}{2}}=\frac{\frac{3}{2} \cdot \sec ^{2}\left(\frac{3 \pi}{4}\right)}{-2 \sin \frac{\pi}{2}+2 \sin \pi}\)
\(=\frac{\frac{3}{2} \times(-\sqrt{2})^{2}}{-2 \times 1+0}=\frac{\frac{3}{2} \times 2}{-2}=-\frac{3}{2}\)

MHT CET-2011

Limits, Continuity and Differentiability

80311 \(y=\log \tan \frac{x}{2}+\sin ^{-1}(\cos x)\), then \(\frac{d y}{d x}\) is

1 \(\operatorname{cosec} x-1\)

2 \(\operatorname{cosec} x\)

3 \(\operatorname{cosec} x+1\)

4 \(x\)

MHT CET-2011

Limits, Continuity and Differentiability

80312 If \(\tan x=\frac{2 t}{1-t^{2}}\) and \(\sin y=\frac{2 t}{1+t^{2}}\), then the value of \(\frac{d y}{d x}\) is

1 1

2 \(\mathrm{t}\)

3 \(\frac{1}{1-\mathrm{t}}\)

4 \(\frac{1}{1+\mathrm{t}}\)

MHT CET-2011

Limits, Continuity and Differentiability

80313 If \(x y=1+\log y\) and \(k \cdot \frac{d y}{d x}+y^{2}=0\), then \(k\) is

1 \(1+x y\)

2 \(\frac{1}{x y-1}\)

3 \(x y-1\)

4 1-2xy

MHT CET-2013

Limits, Continuity and Differentiability

80314 If \(y=\sin ^{2}\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)\), then \(\frac{d y}{d x}\) is

1 \(\frac{1}{2} \mathrm{~s}\)

2 2

3 \(\frac{-1}{2}\)

4 -2

MHT CET-2013

Limits, Continuity and Differentiability

80310 If \(x=2 \cos t-\cos 2 t\) and \(y=2 \sin t-\sin 2 t\), then the value of \(\frac{d^{2} y}{d x^{2}}\) at \(t=\frac{\pi}{2}\) is

1 \(\frac{3}{2}\)

2 \(\frac{-5}{2}\)

3 \(\frac{5}{2}\)

4 \(\frac{-3}{2}\)

Explanation:

(D) : Given, \(x=2 \cos t-\cos 2 t\)
\(\frac{\mathrm{dx}}{\mathrm{dt}}=-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t}\)
And, \(y=2 \sin \mathrm{t}-\sin 2 \mathrm{t}\)
\(\frac{\mathrm{dy}}{\mathrm{dt}}=2 \cos \mathrm{t}-2 \cos 2 \mathrm{t}\)
Therefore,
\(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \Rightarrow \frac{d y}{d x}=\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t}\)
\(\frac{d y}{d x}=\frac{\cos t-\cos 2 t}{-\sin t+\sin 2 t} \Rightarrow \frac{d y}{d x}=\frac{\cos t-\cos 2 t}{\sin 2 t-\sin t}\)
We know that,
\(\cos \mathrm{C}-\cos \mathrm{D}=-2 \sin \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)
\(\sin \mathrm{C}-\sin \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)
\(\frac{d y}{d x}=\frac{2 \sin \left(\frac{t+2 t}{2}\right) \cdot \sin \left(\frac{t-2 t}{2}\right)}{2 \cos \left(\frac{t+2 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}\)
\(\frac{d y}{d x}=\frac{-\sin \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}{\cos \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}\)
\(\frac{d y}{d x}=\frac{\sin \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}{\cos \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)} \Rightarrow \frac{d y}{d x}=\tan \left(\frac{3 t}{2}\right)\)
Therefore,
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}} \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\tan \left(\frac{3 \mathrm{t}}{2}\right)\right)}{(-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t})}\)
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{3}{2} \cdot \sec ^{2}\left(\frac{3 \mathrm{t}}{2}\right)}{-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t}}\)
\(\therefore \quad\left|\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right|_{\mathrm{t}=\frac{\pi}{2}}=\frac{\frac{3}{2} \cdot \sec ^{2}\left(\frac{3 \pi}{4}\right)}{-2 \sin \frac{\pi}{2}+2 \sin \pi}\)
\(=\frac{\frac{3}{2} \times(-\sqrt{2})^{2}}{-2 \times 1+0}=\frac{\frac{3}{2} \times 2}{-2}=-\frac{3}{2}\)

MHT CET-2011

Limits, Continuity and Differentiability

80311 \(y=\log \tan \frac{x}{2}+\sin ^{-1}(\cos x)\), then \(\frac{d y}{d x}\) is

1 \(\operatorname{cosec} x-1\)

2 \(\operatorname{cosec} x\)

3 \(\operatorname{cosec} x+1\)

4 \(x\)

MHT CET-2011

Limits, Continuity and Differentiability

80312 If \(\tan x=\frac{2 t}{1-t^{2}}\) and \(\sin y=\frac{2 t}{1+t^{2}}\), then the value of \(\frac{d y}{d x}\) is

1 1

2 \(\mathrm{t}\)

3 \(\frac{1}{1-\mathrm{t}}\)

4 \(\frac{1}{1+\mathrm{t}}\)

MHT CET-2011

Limits, Continuity and Differentiability

80313 If \(x y=1+\log y\) and \(k \cdot \frac{d y}{d x}+y^{2}=0\), then \(k\) is

1 \(1+x y\)

2 \(\frac{1}{x y-1}\)

3 \(x y-1\)

4 1-2xy

MHT CET-2013

Limits, Continuity and Differentiability

80314 If \(y=\sin ^{2}\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)\), then \(\frac{d y}{d x}\) is

1 \(\frac{1}{2} \mathrm{~s}\)

2 2

3 \(\frac{-1}{2}\)

4 -2

MHT CET-2013

Limits, Continuity and Differentiability

80310 If \(x=2 \cos t-\cos 2 t\) and \(y=2 \sin t-\sin 2 t\), then the value of \(\frac{d^{2} y}{d x^{2}}\) at \(t=\frac{\pi}{2}\) is

1 \(\frac{3}{2}\)

2 \(\frac{-5}{2}\)

3 \(\frac{5}{2}\)

4 \(\frac{-3}{2}\)

Explanation:

(D) : Given, \(x=2 \cos t-\cos 2 t\)
\(\frac{\mathrm{dx}}{\mathrm{dt}}=-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t}\)
And, \(y=2 \sin \mathrm{t}-\sin 2 \mathrm{t}\)
\(\frac{\mathrm{dy}}{\mathrm{dt}}=2 \cos \mathrm{t}-2 \cos 2 \mathrm{t}\)
Therefore,
\(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \Rightarrow \frac{d y}{d x}=\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t}\)
\(\frac{d y}{d x}=\frac{\cos t-\cos 2 t}{-\sin t+\sin 2 t} \Rightarrow \frac{d y}{d x}=\frac{\cos t-\cos 2 t}{\sin 2 t-\sin t}\)
We know that,
\(\cos \mathrm{C}-\cos \mathrm{D}=-2 \sin \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)
\(\sin \mathrm{C}-\sin \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)
\(\frac{d y}{d x}=\frac{2 \sin \left(\frac{t+2 t}{2}\right) \cdot \sin \left(\frac{t-2 t}{2}\right)}{2 \cos \left(\frac{t+2 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}\)
\(\frac{d y}{d x}=\frac{-\sin \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}{\cos \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}\)
\(\frac{d y}{d x}=\frac{\sin \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}{\cos \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)} \Rightarrow \frac{d y}{d x}=\tan \left(\frac{3 t}{2}\right)\)
Therefore,
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}} \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\tan \left(\frac{3 \mathrm{t}}{2}\right)\right)}{(-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t})}\)
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{3}{2} \cdot \sec ^{2}\left(\frac{3 \mathrm{t}}{2}\right)}{-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t}}\)
\(\therefore \quad\left|\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right|_{\mathrm{t}=\frac{\pi}{2}}=\frac{\frac{3}{2} \cdot \sec ^{2}\left(\frac{3 \pi}{4}\right)}{-2 \sin \frac{\pi}{2}+2 \sin \pi}\)
\(=\frac{\frac{3}{2} \times(-\sqrt{2})^{2}}{-2 \times 1+0}=\frac{\frac{3}{2} \times 2}{-2}=-\frac{3}{2}\)

MHT CET-2011

Limits, Continuity and Differentiability

80311 \(y=\log \tan \frac{x}{2}+\sin ^{-1}(\cos x)\), then \(\frac{d y}{d x}\) is

1 \(\operatorname{cosec} x-1\)

2 \(\operatorname{cosec} x\)

3 \(\operatorname{cosec} x+1\)

4 \(x\)

MHT CET-2011

Limits, Continuity and Differentiability

80312 If \(\tan x=\frac{2 t}{1-t^{2}}\) and \(\sin y=\frac{2 t}{1+t^{2}}\), then the value of \(\frac{d y}{d x}\) is

1 1

2 \(\mathrm{t}\)

3 \(\frac{1}{1-\mathrm{t}}\)

4 \(\frac{1}{1+\mathrm{t}}\)

MHT CET-2011

Limits, Continuity and Differentiability

80313 If \(x y=1+\log y\) and \(k \cdot \frac{d y}{d x}+y^{2}=0\), then \(k\) is

1 \(1+x y\)

2 \(\frac{1}{x y-1}\)

3 \(x y-1\)

4 1-2xy

MHT CET-2013

Limits, Continuity and Differentiability

80314 If \(y=\sin ^{2}\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)\), then \(\frac{d y}{d x}\) is

1 \(\frac{1}{2} \mathrm{~s}\)

2 2

3 \(\frac{-1}{2}\)

4 -2

MHT CET-2013

Limits, Continuity and Differentiability

80310 If \(x=2 \cos t-\cos 2 t\) and \(y=2 \sin t-\sin 2 t\), then the value of \(\frac{d^{2} y}{d x^{2}}\) at \(t=\frac{\pi}{2}\) is

1 \(\frac{3}{2}\)

2 \(\frac{-5}{2}\)

3 \(\frac{5}{2}\)

4 \(\frac{-3}{2}\)

Explanation:

(D) : Given, \(x=2 \cos t-\cos 2 t\)
\(\frac{\mathrm{dx}}{\mathrm{dt}}=-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t}\)
And, \(y=2 \sin \mathrm{t}-\sin 2 \mathrm{t}\)
\(\frac{\mathrm{dy}}{\mathrm{dt}}=2 \cos \mathrm{t}-2 \cos 2 \mathrm{t}\)
Therefore,
\(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \Rightarrow \frac{d y}{d x}=\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t}\)
\(\frac{d y}{d x}=\frac{\cos t-\cos 2 t}{-\sin t+\sin 2 t} \Rightarrow \frac{d y}{d x}=\frac{\cos t-\cos 2 t}{\sin 2 t-\sin t}\)
We know that,
\(\cos \mathrm{C}-\cos \mathrm{D}=-2 \sin \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)
\(\sin \mathrm{C}-\sin \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)
\(\frac{d y}{d x}=\frac{2 \sin \left(\frac{t+2 t}{2}\right) \cdot \sin \left(\frac{t-2 t}{2}\right)}{2 \cos \left(\frac{t+2 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}\)
\(\frac{d y}{d x}=\frac{-\sin \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}{\cos \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}\)
\(\frac{d y}{d x}=\frac{\sin \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}{\cos \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)} \Rightarrow \frac{d y}{d x}=\tan \left(\frac{3 t}{2}\right)\)
Therefore,
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}} \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\tan \left(\frac{3 \mathrm{t}}{2}\right)\right)}{(-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t})}\)
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{3}{2} \cdot \sec ^{2}\left(\frac{3 \mathrm{t}}{2}\right)}{-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t}}\)
\(\therefore \quad\left|\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right|_{\mathrm{t}=\frac{\pi}{2}}=\frac{\frac{3}{2} \cdot \sec ^{2}\left(\frac{3 \pi}{4}\right)}{-2 \sin \frac{\pi}{2}+2 \sin \pi}\)
\(=\frac{\frac{3}{2} \times(-\sqrt{2})^{2}}{-2 \times 1+0}=\frac{\frac{3}{2} \times 2}{-2}=-\frac{3}{2}\)

MHT CET-2011

Limits, Continuity and Differentiability

80311 \(y=\log \tan \frac{x}{2}+\sin ^{-1}(\cos x)\), then \(\frac{d y}{d x}\) is

1 \(\operatorname{cosec} x-1\)

2 \(\operatorname{cosec} x\)

3 \(\operatorname{cosec} x+1\)

4 \(x\)

MHT CET-2011

Limits, Continuity and Differentiability

80312 If \(\tan x=\frac{2 t}{1-t^{2}}\) and \(\sin y=\frac{2 t}{1+t^{2}}\), then the value of \(\frac{d y}{d x}\) is

1 1

2 \(\mathrm{t}\)

3 \(\frac{1}{1-\mathrm{t}}\)

4 \(\frac{1}{1+\mathrm{t}}\)

MHT CET-2011

Limits, Continuity and Differentiability

80313 If \(x y=1+\log y\) and \(k \cdot \frac{d y}{d x}+y^{2}=0\), then \(k\) is

1 \(1+x y\)

2 \(\frac{1}{x y-1}\)

3 \(x y-1\)

4 1-2xy

MHT CET-2013

Limits, Continuity and Differentiability

80314 If \(y=\sin ^{2}\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)\), then \(\frac{d y}{d x}\) is

1 \(\frac{1}{2} \mathrm{~s}\)

2 2

3 \(\frac{-1}{2}\)

4 -2

MHT CET-2013

Limits, Continuity and Differentiability

80310 If \(x=2 \cos t-\cos 2 t\) and \(y=2 \sin t-\sin 2 t\), then the value of \(\frac{d^{2} y}{d x^{2}}\) at \(t=\frac{\pi}{2}\) is

1 \(\frac{3}{2}\)

2 \(\frac{-5}{2}\)

3 \(\frac{5}{2}\)

4 \(\frac{-3}{2}\)

Explanation:

(D) : Given, \(x=2 \cos t-\cos 2 t\)
\(\frac{\mathrm{dx}}{\mathrm{dt}}=-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t}\)
And, \(y=2 \sin \mathrm{t}-\sin 2 \mathrm{t}\)
\(\frac{\mathrm{dy}}{\mathrm{dt}}=2 \cos \mathrm{t}-2 \cos 2 \mathrm{t}\)
Therefore,
\(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \Rightarrow \frac{d y}{d x}=\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t}\)
\(\frac{d y}{d x}=\frac{\cos t-\cos 2 t}{-\sin t+\sin 2 t} \Rightarrow \frac{d y}{d x}=\frac{\cos t-\cos 2 t}{\sin 2 t-\sin t}\)
We know that,
\(\cos \mathrm{C}-\cos \mathrm{D}=-2 \sin \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)
\(\sin \mathrm{C}-\sin \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)
\(\frac{d y}{d x}=\frac{2 \sin \left(\frac{t+2 t}{2}\right) \cdot \sin \left(\frac{t-2 t}{2}\right)}{2 \cos \left(\frac{t+2 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}\)
\(\frac{d y}{d x}=\frac{-\sin \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}{\cos \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}\)
\(\frac{d y}{d x}=\frac{\sin \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)}{\cos \left(\frac{3 t}{2}\right) \cdot \sin \left(\frac{2 t-t}{2}\right)} \Rightarrow \frac{d y}{d x}=\tan \left(\frac{3 t}{2}\right)\)
Therefore,
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}} \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\tan \left(\frac{3 \mathrm{t}}{2}\right)\right)}{(-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t})}\)
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{3}{2} \cdot \sec ^{2}\left(\frac{3 \mathrm{t}}{2}\right)}{-2 \sin \mathrm{t}+2 \sin 2 \mathrm{t}}\)
\(\therefore \quad\left|\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right|_{\mathrm{t}=\frac{\pi}{2}}=\frac{\frac{3}{2} \cdot \sec ^{2}\left(\frac{3 \pi}{4}\right)}{-2 \sin \frac{\pi}{2}+2 \sin \pi}\)
\(=\frac{\frac{3}{2} \times(-\sqrt{2})^{2}}{-2 \times 1+0}=\frac{\frac{3}{2} \times 2}{-2}=-\frac{3}{2}\)

MHT CET-2011

Limits, Continuity and Differentiability

80311 \(y=\log \tan \frac{x}{2}+\sin ^{-1}(\cos x)\), then \(\frac{d y}{d x}\) is

1 \(\operatorname{cosec} x-1\)

2 \(\operatorname{cosec} x\)

3 \(\operatorname{cosec} x+1\)

4 \(x\)

MHT CET-2011

Limits, Continuity and Differentiability

80312 If \(\tan x=\frac{2 t}{1-t^{2}}\) and \(\sin y=\frac{2 t}{1+t^{2}}\), then the value of \(\frac{d y}{d x}\) is

1 1

2 \(\mathrm{t}\)

3 \(\frac{1}{1-\mathrm{t}}\)

4 \(\frac{1}{1+\mathrm{t}}\)

MHT CET-2011

Limits, Continuity and Differentiability

80313 If \(x y=1+\log y\) and \(k \cdot \frac{d y}{d x}+y^{2}=0\), then \(k\) is

1 \(1+x y\)

2 \(\frac{1}{x y-1}\)

3 \(x y-1\)

4 1-2xy

MHT CET-2013

Limits, Continuity and Differentiability

80314 If \(y=\sin ^{2}\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)\), then \(\frac{d y}{d x}\) is

1 \(\frac{1}{2} \mathrm{~s}\)

2 2

3 \(\frac{-1}{2}\)

4 -2

MHT CET-2013

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