Maths M.C.Q (1 Marks)

4. $\text{f}(\text{e}^\text{x})$

Solution:

Let $\text{y}=\text{f}(\text{e}^\text{x}),$ then

$\text{y}_1=\text{f}'(\text{e}^\text{x})\text{e}^\text{x}$

$\text{y}_2=\text{f}''(\text{e}^\text{x})\text{e}^\text{x}\text{e}^\text{x}+\text{f}'(\text{e}^\text{x})\text{e}^\text{x}$

$=\text{e}^{2\text{x}}\text{f}''(\text{e}^\text{x})+\text{f}'(\text{e}^\text{x})\text{e}^\text{x}$

3. -y

Solution:

$\text{y}=\text{a}\cos(\log_\text{e}\text{x})+\text{b}\sin(\log_\text{e}\text{x})$

$\Rightarrow\text{y}_1=-\text{a}\sin(\log_\text{e}\text{x})\frac{1}{\text{x}}+\text{b}\cos(\log_\text{e}\text{x})\frac{1}{\text{x}}$

$\Rightarrow\text{y}_2=\frac{-\text{a}\sin(\log_\text{e}\text{x})+\text{b}\cos(\log_\text{e}\text{x})}{\text{x}}$

$\Rightarrow\text{y}_2=\frac{-\text{a}(\log_\text{e}\text{x})-\text{b}\sin(\log_\text{e}\text{x})-\{-\text{a}\sin(\log_\text{e}\text{x})+\text{b}\cos(\log_\text{e}\text{x})\}}{\text{x}^2}$

$\Rightarrow\text{x}^2\text{y}_2=-\{\text{a}\cos(\log_\text{e}\text{x})+\text{b}\sin(\log_\text{e}\text{x})\}\\-\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log_\text{e}\text{x})\}$

$\Rightarrow\text{x}^2\text{y}_2=-\text{y}-\text{xy}_1$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=-\text{y}$

2. Continuous on [-1, 1] and differentiable on $(-1,0)\cup(0,1)$

Solution:

We have,

$\text{f(x)}=\sqrt{1-\sqrt{1-\text{x}^2}}$

Here, function will be defined for those values of x for which

$1-\text{x}^2\geq0$

$\Rightarrow1\geq\text{x}^2$

$\Rightarrow\text{x}^2\leq1$

$\Rightarrow|\text{x}\leq1|$

$\Rightarrow-1\leq\text{x}\leq1$

Therefore, function is continuous in -1, 1

Now, we need to check the differentiability of $\text{f(x)}=\sqrt{1-\sqrt{1-\text{x}^2}}$ in the interval -1, 1

Now,

We will check the differentiable at x = 0

LHL at x = 0

$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}-0}{\text{x}}$

$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}}{\text{x}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-(0-\text{h})^2}}}{0-\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-\text{h}^2}}}{-\text{h}}$

$=-\infty$

RHL at x = 0

$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}-0}{\text{x}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-(0+\text{h})^2}}}{0+\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-\text{h}^2}}}{\text{h}}$

$=\infty$

So, the function is not differentiable at x = 0.

4. $\frac{\text{g(x)}}{\text{f(x)}}$

Solution:

We have f(x) = 2x and $\text{g(x)}=\frac{\text{x}^2}{2}+1,$ which are continuous functions,

So, $\text{f(x)}+\text{g(x)},\text{f(x)}-\text{g(x)}$ and $\text{f(x)}\times\text{g(x)}$ are also continuous.

But $\frac{\text{g(x)}}{\text{f(x)}}$ is discontinuous where f(x) = 0, i.e., x = 0

3. 3

Solution:

$\text{xy}-\log_\text{e}\text{y}=1$

$\Rightarrow\text{xy}_1+\text{y}-\frac{\text{y}_1}{\text{y}}=0$

$\Rightarrow\text{xyy}_1+\text{y}^2-\text{y}_1=0$

$\Rightarrow\text{yy}_1+\text{xy}_1\text{y}_1+\text{xyy}_2+2\text{yy}_1-\text{y}_2=0$

$\Rightarrow\text{x}(\text{y}_1^2+\text{yy}_2)-\text{y}_2+3\text{yy}_2=0$

$\therefore\lambda=3$

4. $\frac{1}{2\text{e}}$

Solution:

We have, $\text{f(x)}=\log_{\text{x}^2}(\log\text{x})$

$\Rightarrow\text{f(x)}=\frac{\log(\log\text{x})}{\log\text{x}^2}$

$\Rightarrow\text{f(x)}=\frac{\log(\log\text{x})}{2\log\text{x}}$

$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\frac{\text{d}}{\text{dx}}\bigg\{\frac{\log(\log\text{x})}{\log\text{x}}\bigg\}$

$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\log\text{x}}\times\frac{1}{\text{x}}\times\log\text{x}-\frac{\log(\log)\text{x}}{\text{x}}}{(\log\text{x})^2}\Bigg\}$

$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\text{x}}-\frac{\log(\log)\text{x}}{\text{x}}}{(\log\text{x})^2}\Bigg\}$

$\Rightarrow\text{f}'\text{(e)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\text{e}}-\frac{\log(\log)\text{e}}{\text{e}}}{(\log\text{e})^2}\Bigg\}$

[Putting x = e]

$\Rightarrow\text{f}'\text{(e)}=\frac{1}{2}\times\bigg\{\frac{\frac{1}{\text{e}}}{1}\bigg\}$

$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2\text{e}}$

2. f is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}.$

Solution:

Let $\text{u(x)}=\sin\text{x}$ and $\text{v(x)}=|\text{x}|$

$\therefore\ \text{f(x)}=\text{vou(x)}=\text{v}[\text{u(x)}]$

Since, u(x)and v(x) both are continuous functions.

Hence, f(x) = vou(x) is also a continuous function but v(x) is not differentiable at x = 0.

So, f(x) is not differentiable where $\sin\text{x}=0$

$\Rightarrow\ \text{x}=\text{n}\pi,\text{n}\in\text{Z}$

Hence, f(x) is continuous everywhere but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}.$