Calculus Test - 1

Given:

\(y+\sin ^{-1}\left(1-x^{2}\right)=e^{x}\)

\(y=e^{x}-\sin ^{-1}\left(1-x^{2}\right)\)

Differentiating above with respect to x, we get:

\(\frac{{dy}}{{dx}}=e^{x}-\frac{1}{\sqrt{1-\left(1-{x}^{2}\right)^{2}}}(-2 {x})\)

\(\frac{{dy}}{{dx}}=e^{{x}}+\frac{2 {x}}{\sqrt{1-\left(1-2 {x}^{2}+{x}^{4}\right)}}\)

\(\frac{{dy}}{{dx}}=e^{{x}}+\frac{2 {x}}{\sqrt{2 x^{2}-{x}^{4}}}\)

\(\frac{{dy}}{{d} x}={e}^{{x}}+\frac{2}{\sqrt{2-{x}^{2}}}\)

It is given that \(f(x)\) is differentiable at \(x=c\) and every differentiable function is continuous. So, \(f(x)\) is continuous at \(x=c\).

\(\begin{array}{ll} \therefore  \lim _{x \rightarrow c^{-}} f(x)=\lim _{x \rightarrow c^{+}} f(x)=f(c) \\ \Rightarrow  \lim _{x \rightarrow c} x^2=\lim _{x \rightarrow c}(a x+b)=c^2 \\ \Rightarrow  c^2=a c+b \end{array}\) [Using def. of \(f(x)\) ]

Now, \(\quad f(x)\) is differentiable at \(x=c\cdots(i)\)

\(\begin{array}{ll} \Rightarrow  (\text { LHD at } x=c)=(\text { RHD at } x=c) \\ \Rightarrow  \lim _{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c}=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c} \\ \Rightarrow  \lim _{x \rightarrow c} \frac{x^2-c^2}{x-c}=\lim _{x \rightarrow c} \frac{(a x+b)-c^2}{x-c} \\ \Rightarrow  \lim _{x \rightarrow c} \frac{x^2-c^2}{x-c}=\lim _{x \rightarrow c} \frac{a x+b-(a c+b)}{x-c} \quad \text { [Using def. of } f(x) \text { ] } \quad \text { [Using (i)] } \\ \Rightarrow  \lim _{x \rightarrow c}(x+c)=\lim _{x \rightarrow c} \frac{a(x-c)}{x-c} a \\ \Rightarrow \lim _{x \rightarrow c}(x+c)=\lim _{x \rightarrow c} a \end{array}\)  [Using (i)]

\(\Rightarrow \quad 2 c=a\cdots(ii)\)

From (i) and (ii), we get

\(c^2=2 c^2+b \Rightarrow b=-c^2 \)

So, \(a=2 c\) and \(b=-c^2\)

It is given that: 

\(y=e^{x+e^{x+e^{x+\cdots \infty}}}\)

\(\therefore y=e^{x+\left(e^{x+e^{x+} \cdots \infty}\right)}=e^{x+y}\)

Differentiating both sides with respect to x and using the chain rule, we get:

\(\frac{d y}{d x}=\frac{d}{d x} e^{x+y}\)

\(\Rightarrow \frac{d y}{d x}=e^{x+y} \frac{d}{d x}(x+y)\)

\(\Rightarrow \frac{d y}{d x}=y\left(1+\frac{d y}{d x}\right)\)

\(\Rightarrow \frac{d y}{d x}=y+y \frac{d y}{d x}\)

\(\Rightarrow(1-y) \frac{d y}{d x}=y\)

\(\Rightarrow \frac{d y}{d x}=\frac{y}{1-y}\)

Given:

\(f(x)=x^{3}+3 x^{2}+3 x-7\)

We know that:

\(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\)

Derivative of a constant, i.e.,

\(\frac{\mathrm{d}(\text { constant })}{\mathrm{dx}}=0\)

Therefore, 

\(\frac{d f(x)}{d x}=3 x^{2}+6 x+3\)

Putting \(x=2\) in above, we get:

\(\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=3(2)^{2}+6(2)+3\)

\(=3(4)+12+3\)

\(=12+12+3\)

\(\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=27\)

The value of \(\frac{\mathrm{d} \mathrm{f}(x)}{\mathrm{dx}}\) at \(\mathrm{x}=2\) is 27.

Given:

\(y=3 t^{2}-4 t-3\)

Differentiating above with respect to t, we get:

\(\frac{\mathrm{dy}}{\mathrm{dt}}=6 \mathrm{t}-4\)

Similarly,

\(x=8 t+5\)

Differentiating above with respect to t, we get:

\(\frac{\mathrm{dx}}{\mathrm{dt}}=8\)

\(\frac{\mathrm{dx}}{\mathrm{dt}}=8\)

As we know:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}\)

Then,

\(\frac{d y}{d x}=\frac{6 t-4}{8}\)

At t = 6, we get:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(6 \times 6)-4}{8}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{32}{8}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=4\)

Given:

\(y=\frac{12 \tan x-4 \tan ^{3} x}{9-27 \tan ^{2} x}\)

\(\Rightarrow y=\frac{4\left(3 \tan x-\tan ^{3} x\right)}{9\left(1-3 \tan ^{2} x\right)}\)

We know that,

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

\(\Rightarrow y=\frac{4}{9} \tan 3 x\)

Differentiating both sides with respect to \(x\), we get:

\(\Rightarrow \frac{d y}{d x}=\frac{4}{9} \sec ^{2} 3 x \times 3\)

\(\Rightarrow \frac{d y}{d x}=\frac{4}{3} \sec ^{2} 3 x\)

Again, differentiating with respect to \(x\), we get:

\(\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{4}{3} \times 2 \times \sec 3 x \times \tan 3 x \times \sec 3 x \times 3\)

\(\Rightarrow \frac{d^{2} y}{d x^{2}}=8 \sec ^{2} 3 x \tan 3 x\)

Given that:

\(y^{2}+x^{2}+3 x+5=0\)

We know that:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}\)

Differentiating with respect to \(x\), we get:

\(2 y \frac{d y}{d x}+2 x+3(1)+0=0\)

\(2 y \frac{d y}{d x}+2 x+3=0\)

\(2 y \frac{d y}{d x}=-(2 x+3)\)

\(\frac{d y}{d x}=-\frac{2 x+3}{2 y}\)

Now at (0, -3)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2(0)+3}{2(-3)}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{3}{(-6)}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=0.5\)

We know that:

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

\(f(x)=\log x\), then \(f'(x)=\frac{1}{x}\)

\(f(x)=\) Constant, then \(f'(x)=0\)

\(\log m^{n}=n \log m\)

Given here:

\(\mathrm{f}(\mathrm{x})=\log \mathrm{x}^{3}+2 \mathrm{x}^{2}-3 \mathrm{x}\)

\(\Rightarrow \mathrm{f}(\mathrm{x})=3 \log \mathrm{x}+2 \mathrm{x}^{2}-3 \mathrm{x}\)

\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=3 \frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}} 2 \mathrm{x}^{2}-\frac{\mathrm{d}}{\mathrm{dx}} 3 \mathrm{x}\)

\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{3}{x}+4 \mathrm{x}-3\)

Putting x = 3 in above, we get:

\(\Rightarrow \mathrm{f}^{\prime}(3)=\frac{3}{3}+4(3)-3\)

\(\Rightarrow \mathrm{f}^{\prime}(3)=10\)

Since \(f(x)\) is continuous at \(x=-1\). Therefore,

\(\lim _{x \rightarrow-1} f(x)=f(-1) \)

\(\Rightarrow  \lim _{x \rightarrow-1} \frac{x^2-2 x-3}{x+1}=\lambda \)

\(\Rightarrow \lim _{x \rightarrow-1} \frac{(x-3)(x+1)}{x+1}=\lambda \Rightarrow \lim _{x \rightarrow-1}(x-3)=\lambda \Rightarrow-4=\lambda\)

So, \(f(x)\) is continuous at \(x=-1\), if \(\lambda=-4\).

For \(x \leq 1\), we have

\(f(x)=x^2+3 x+a\) (i.e. a polynomial)

For \(x>1\), we have

\(f(x)=b x+2\) (which is also a polynomial)

Since a polynomial function is everywhere differentiable.

Therefore, \(f(x)\) is differentiable for all \(x>1\) and also for all \(x<1\). Thus, we have to use the differentiability of \(f(x)\) at \(x=1\).

Now,

\(f(x)\) is differentiable at \(x=1 \)

\(\Rightarrow f(x)\) is continuous at \(x=1 \)

\(\Rightarrow \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1) \)

\(\Rightarrow  \lim _{x \rightarrow 1} x^2+3 x+a=\lim _{x \rightarrow 1} b x+2=1+3+a \)

\(\Rightarrow  1+3+a=b+2 \)

\(\Rightarrow a-b+2=0\cdots(i)\)

Again, \( f(x)\) is differentiable at \(x=1\)

\(\Rightarrow\) LHD at } x=1)=(\text { RHD at } x=1)\)

\(\Rightarrow \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\)

\(\Rightarrow \lim _{x \rightarrow 1} \frac{x^2+3 x+a-(4+a)}{x-1}=\lim _{x \rightarrow 1} \frac{(b x+2)-(4+a)}{x-1}\)

\(\Rightarrow  \lim _{x \rightarrow 1} \frac{x^2+3 x-4}{x-1}=\lim _{x \rightarrow 1} \frac{b x-2-a}{x-1}\)

\(\Rightarrow \lim _{x \rightarrow 1} \frac{(x+4)(x-1)}{x-1}=\lim _{x \rightarrow 1} \frac{b x-b}{x-1}\) [Using (i)]

\(\Rightarrow \lim _{x \rightarrow 1}(x+4)=\lim _{x \rightarrow 1} b \)

\( \Rightarrow \quad 5=b\)

Putting \(b=5\) in (i), we get \(a=3\).

So, \(a=3\) and \(b=5\).