Calculus Test - 2

Since \(\mathrm{f}(\mathrm{x})\) is given to be continuous at \(\mathrm{x}=0\),

\(\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\)

Also, \(\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x)\) because \(f(x)\) is same for \(x>0\) and \(x<0\).

\(\therefore \lim _{x \rightarrow 0} f(x)=f(0)\)

We know that:

\(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\)

\(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\)

Therefore,

\(\Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x}{e^{2 x}-1}=k-2\)

Multiplying and dividing by 3x in numerator and by 2x in denominator, we get:

\(\Rightarrow \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{3 x} \times 3 x}{\frac{e^{2 x}-1}{2 x} \times 2 x}=k-2\)

\(\Rightarrow \frac{3}{2}=k-2\)

\(\Rightarrow k=\frac{7}{2}\)

Since \(f(x)\) is continuous at \(x=3\). Therefore,

\( \lim _{x \rightarrow 3} f(x)=f(3) \)

\(\Rightarrow \lim _{x \rightarrow 3} f(x)=k\)

\(\Rightarrow  \lim _{x \rightarrow 3} \frac{x^2-9}{x-3}=k\)

\(\Rightarrow  \lim _{x \rightarrow 3} \frac{(x-3)(x+3)}{x-3}=k\)

\(\Rightarrow  \lim _{x \rightarrow 3}(x+3)=k \)

\(\Rightarrow 6=k\)

Thus, \(f(x)\) is continuous at \(x=3\), if \(k=6\).

\(f(x)\) being a polynomial function, is everywhere differentiable. 

The derivative of \(f\) at \(x\) is given by

\( f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \)

\(\Rightarrow  f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\left|(x+h)^2-6(x+h)+8\right|-\left|x^2-6 x+8\right|}{h} \)

\(\Rightarrow  f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{2 h x-6 h+h^2}{h} \)

\(\Rightarrow  f^{\prime}(x)=\lim _{h \rightarrow 0}(2 x-6+h) \)

\(\Rightarrow  f^{\prime}(x)=2 x-6 \)

\(\therefore  f^{\prime}(5)-3 f^{\prime}(2)=(2 \times 5-6)-3(2 \times 2-6)=4+6=10 \)

and checking from the options,

\(f^{\prime}(8)=2 \times 8-6=10 \)

So, \(f^{\prime}(5)-3 f^{\prime}(2)=f^{\prime}(8) \)

Given: 

\(\sqrt{y}=\sin x+\cos x\)  

Squaring both sides, we get:

\(\Rightarrow y=(\sin x+\cos x)^{2}\)

\(\Rightarrow y=\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\)

\(\Rightarrow y=1+\sin 2 x \quad(\because 2 \sin x \cos x=\sin 2 x)\)

Differentiating above with respect to x, we get:

\(\frac{{dy}}{{dx}}=0+2 \times \cos 2 {x}\)

\(=2 \cos 2 {x}\)

Given:

\(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}+\frac{1}{\mathrm{x}}\)

We know that:

If \(f(x)=x^{n}\), then:

\(f'(x)=n x^{n-1}\)

Differentiating with respect to x, we get:

\(f'(x)=6 x^{2}-\frac{1}{x^{2}}\)

Putting \(x=1\) in above, we get:

\(f'(1)=6 \times 1^{2}-\frac{1}{1^{2}}\)

\(f'(1)=6-1=5\)

\(\therefore\) The value of \(f^{\prime}(1)\) is 5.

When \(x \leq 2\), we have

\(f(x)=k x^2\), which being a polynomial function is continuous at each \(x<2\).

When \(x>2\), we have

\(f(x)=3\), which being a constant function is continuous at each \(x>2\)

So, consider the point \(x=2\).

We have,

\((\) LHL at \(x=2)=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2} k x^2=4 k\), \(\quad \left[\because f(x)=k a^2\right.\) for \(\left.x \leq 2\right]\)

\((\) RHL at \(x=2)=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2} 3=3\), \(\quad [\because f(x)=3\) for \(x>2]\)

and, \(\quad f(2)=k(2)^2=4 k\).

As \(f(x)\) is continuous in its domain.

Therefore, it is also continuous at \(x=2\).

Consequently,

\(\lim _{x \rightarrow 2^{-}} f(x)=f(2)=\lim _{x \rightarrow 2^{+}} f(x) \Rightarrow 4 k=3\) 

\(\Rightarrow k=\frac{3}{4}\)

Let, \(y=\frac{1}{3 x^{2}}\) ....(1)

We know that:

\(\frac{d x^{n}}{dx}=nx^{n-1}\)

Differentiating (1) with respect of x, we get:

\(\frac{dy}{dx} =\frac{d}{dx}\left(\frac{1}{3 x^{2}}\right)\)

\(=\frac{1}{3} \frac{d\left(x^{-2}\right)}{dx}\)

\(=\frac{1}{3} \times-2 \times x^{-2-1}\)

\(=\frac{-2}{3 x^{3}}\)

Given:

\(\lim _{x \rightarrow 2} \frac{\sin \left(e^{x-2}-1\right)}{\log (x-1)}\) .....(1)

We know that:

\(\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=l=\lim _{x \rightarrow a} f(x)\)

On substituting, \(\mathrm{h}=\mathrm{x}-2\) in (1), we get:

\(=\lim _{h \rightarrow 0} \frac{\sin \left(e^{h}-1\right)}{\log (1+h)}\)

This can be rearranged as:

\(=\lim _{h \rightarrow 0} \frac{\sin \left(e^{h}-1\right)}{e^{h}-1} \cdot \frac{e^{h}-1}{h} \cdot \frac{h}{\log (1+h)}\)

\(=1 \cdot 1 \cdot 1\)

\(=1\)

And, \(f(2)=k\)

∴ For the function to be continuous the value of the function f(x) at x = 2 must equal the limiting value of 1, i.e., k = 1

Let, \(v=\cos ^{-1}\left(\frac{1}{2 x^{2}-1}\right)\) and \(u=\sqrt{1-x^{2}}\)

Substituting \(x=\cos \theta\), 

\(\Rightarrow \theta=\cos ^{-1} x\),  we get:

\(v=\cos^{-1}\left(\frac{1}{2 \cos ^{2} \theta-1}\right)\)

\(=\cos ^{-1}\left(\frac{1}{\cos ^{2} \theta-\sin ^{2} \theta}\right)\)

\(=\cos ^{-1}\left(\frac{1}{\cos 2 \theta}\right)\)

\(=\cos ^{-1}(\sec 2 \theta)\)

\(=\cos^{-1}\left(\cos \left(90^{\circ}-2 \theta\right)\right)\)

\(=90^{\circ}-\) \(2 \theta\).

And, \(u=\sqrt{1-x^{2}}\)

\(=\sqrt{1-\cos ^{2} \theta}\)

\(=\sqrt{\sin ^{2} \theta}=\sin \theta\)

Now, \(\frac{d v}{d \theta}=-2\) 

And, \(\frac{d u}{d \theta}=\cos \theta\)

\(\therefore \frac{\mathrm{dv}}{\mathrm{du}} =\frac{\mathrm{dv}}{\mathrm{d} \theta} \times \frac{\mathrm{d} \theta}{\mathrm{d} u}\)

\(=\frac{-2}{\cos \theta}=\frac{-2}{x}\)

Given that:

\(\mathrm{y}=\tan ^{-1}\left[\frac{8 \mathrm{x}}{1-15 \mathrm{x}^{2}}\right]\)

As we know that,

\(\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\tan ^{-1}\left[\frac{x+\mathrm{y}}{1-\mathrm{xy}}\right]\)

So,

\(y=\tan ^{-1}\left[\frac{5 x+3 x}{1-5 x \cdot 3 x}\right]\)

\(=\tan ^{-1} 5 x+\tan ^{-1} x\)

Also, we know:

\(\frac{\mathrm{d}\left(\tan ^{-1} \mathrm{x}\right)}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}^{2}}\)

Differentiating y with respect to x, we get:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\tan ^{-1} 5 \mathrm{x}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\tan ^{-1} 3 \mathrm{x}\right)}{\mathrm{dx}}\)

\(=\frac{5}{1+(5 x)^{2}}+\frac{3}{1+(3 \mathrm{x})^{2}}\)

\(=\frac{5}{1+25 x^{2}}+\frac{3}{1+9 x^{2}}\)