Mathematics Test - 18

Observe the graph of the function \(f(x)=|x|\).

At \(x=1\) the curve is smooth and thus the function is differentiable at \(x=1\).

On the other hand the function is not differentiable at \(x=0\) as it has a corner at \(x=0\).

Observe the graph of the function \(f(x)=e^{x}\).

At \(x=0\) the curve is smooth so it is differentiable at \(x=0\).

Thus, only the \(2\)nd statement is true.

\(x-2 y-2 z=\lambda x\)

\( (1-\lambda) x-2 y-2 z=0\) .....(i)

\(x+2 y+z=\lambda y\)

\(x+(2-\lambda) y+z=0\)

\(-x-y=\lambda z\)

\(x+y+\lambda z=0\)

Since the system of given linear equations has a non-trivial solution.

Therefore,

\(\left|\begin{array}{ccc}1-\lambda & -2 & -2 \\ 1 & 2-\lambda & 1 \\ 1 & 1 & \lambda\end{array}\right|=0\)

\(((1-\lambda)(\lambda(2-\lambda)-1)-2(\lambda-1)-2(1-(2-\lambda=0\)

\((1-\lambda)^{3}=0\)

\(\Rightarrow \lambda=1\)

Thus, the set of values of \(\lambda\) contains exactly one element.

Given,

N = {1, 2, 3, …, 100}

B = {y | y = x + 2, x ∈ N}

Hence, for 1 ∈ N

y = 1 + 2 = 3

For 2 ∈ N

y = 2 + 2 = 4 etc

Therefore, B = {3, 4, 5, 6, … , 100}

We have \((x+y)^{n}={ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1} . y+{ }^{n} C_{2} x^{n-2}. y^{2}+\ldots+{ }^{n} C_{n} y^{n}\)

General term: General term in the expansion of \((x+y)^{n}\) is given by:

\(\mathrm{T}_{(\mathrm{r}+1)}={ }^{n} C_{\mathrm{r}} \times x^{\mathrm{n}-\mathrm{r}} \times y^{\mathrm{r}}\)

We have to find term independent of \(x\) in \(\left(x^{2}-\frac{1}{x^{3}}\right)^{10}\).

We know that,

\(\mathrm{T}_{(\mathrm{r}+1)}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \times {x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}\)

\(\Rightarrow \mathrm{T}_{(\mathrm{r}+1)}={ }^{10} \mathrm{C}_{\mathrm{r}} \times\left({x}^{2}\right)^{10-\mathrm{r}} \times\left(\frac{-1}{{x}^{3}}\right)^{\mathrm{r}}\)

\(=(-1)^{\mathrm{r}} \times{ }^{10} \mathrm{C}_{\mathrm{r}} \times({x})^{20-2 \mathrm{r}} \times\left({x}^{3}\right)^{-\mathrm{r}}\)

\(=(-1)^{\mathrm{r}} \times{ }^{10} \mathrm{C}_{\mathrm{r}} \times({x})^{20-2 \mathrm{r}} \times({x})^{-3 \mathrm{r}}\)

\(=(-1)^{\mathrm{r}} \times{ }^{10} \mathrm{C}_{\mathrm{r}} \times({x})^{20-5 \mathrm{r}}\)

For the term independent of \(x\), power of \(x\) should be zero

Therefore, \(20-5 r=0\)

\(\Rightarrow \mathrm{r}=4\)

\(\mathrm{T}_{(4+1)}=(-1)^{4} \times{ }^{10} \mathrm{C}_{4}={ }^{10} \mathrm{C}_{4}\)

I implies II means ≡ I → II

\(X^{-1}=\frac{\operatorname{Adj}(\mathrm{X})}{|X|}\)

If \(|X| \neq 0\) then \(X^{-1}\)

\(|X|=\frac{\operatorname{Adj}(X)}{X^{-1}}\)

If \(X^{-1}\) then \(|X| \neq 0\) also \(\mid\) Adj \(\left.X|=| X\right|^{n-1}\) then \(\mid\) Adj \(X \mid \neq 0\)

If \(X^{-1}\) then \(|X| \neq 0\)

I implies II and II implies I

∴ Both I and II are equivalent.

Note:

\(X^{-1}\) means \(X\) is invertible

|X| determinant of \(X\)

As given,

\([(A-B) \cup B]^{C}\)

\((A-B) \cup B=\left(A \cap B^{C}\right) \cup B\)

By De Morgan's Theorem, if A and B are any two sets, then

\(A-B=A \cap B^{C}\)

\(=(A \cup B) \cap\left(B^{C} \cup B\right)\)

\(=(A \cup B) \cap(U)\)

\(=A \cup B\)

So \([(A-B) \cup B]^{C}=(A \cup B)^{C}=A^{C} \cap B^{C}\)

Given,

\(\frac{x}{4}<\frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}\)

\(=\frac{x}{4}<\frac{5(5 x-2)-3(7 x-3)}{15}\)

On simplifying we get

\(=\frac{x}{4}<\frac{25 x-10-21 x+9}{15} \)

\(=\frac{x}{4}<\frac{4 x-1}{15} \)

\(=15 x<4(4 x-1) \)

\(=15 x<16 x-4 \)

\(=4

All the real numbers of \(x\) which are greater than \(4\) are the solutions of the given inequality

Hence, \((4, \infty)\) will be the solution for the given inequality.

Given that,

\(y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}\)

\(\Rightarrow y=\sqrt{x+y}\)

Squarring on both sides the above, we get:

\( y^{2}=x+y\)

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

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

\(\Rightarrow(2 y-1) \frac{d y}{d x}=1\)

\(\Rightarrow \frac{d y}{d x}=\frac{1}{2 y-1}\)

Therefore, if \({y}=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\ldots \infty}}}\),

Then,

\(\frac{{dy}}{{dx}}=\frac{1}{2 {y}-1}\)

Distance between parallel lines:

The distance between the lines \(y=m x+c_{1}\) and \(y=m x+c_{2}\) is \(\frac{\left|c_{1}-c_{2}\right|}{\sqrt{1+m^{2}}}\)

The distance between the lines \(a x+b y+c_{1}=0\) and \(a x+b y+c_{2}=0\) is \(\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a^{2}+b^{2}}}\)

Calculation:

Given lines are 6x + 8y + 15 = 0 and 3x + 4y + 9 = 0

⇒ 6x + 8y + 15 = 0

Take 2 common from above equation, we get

\(\Rightarrow 3 x+4 y+\frac{15 }{ 2}=0\) .....(i)

And 3x + 4y + 9 = 0 .....(ii)

Equation (i) and (ii) are parallel to each other.

\(\therefore\) The distance between the lines \(=\frac{\left|\frac{15}{2}-9\right|}{\sqrt{3^{2}+4^{2}}}=\frac{\left(\frac{3}{2}\right)}{5}=\frac{3}{10}\)