Mathematics Test - 2

Three- digit numbers are divisible by 9 are: 

108, 117, 126 . . . . 999

Series of AP:

108,117, 126 . . . . 999

T_n =999

a = 108

d = 117 - 108 = 9

As we know that, 

T_n = a + (n - 1) d

⇒ 999 = 108 + (n - 1) 9

⇒ 891 = (n - 1) 9

⇒ 99 = n - 1

⇒ n = 100

Given,

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

Here, \(1-x^{2} \geq 0\)

\(\Rightarrow x^{2}-1 \leq 0\)

\(\Rightarrow(x-1)(x+1) \leq 0\)

when, \(x-1=0 \Rightarrow x=1\)

when, \(x+1=0 \Rightarrow x=-1\)

thus, domain of \(x=[-1,1]\)

Given that:

\(\underset{{{x \rightarrow 0}}}{\lim} \frac{\left(e^{4 x^{2}}-1\right)}{x \sin x}\)

Let,

\(L=\underset{{{x \rightarrow 0}}}{\lim} \frac{\left(e^{4 x^{2}}-1\right)}{x \sin x}\) 

\(L=\underset{{{x \rightarrow 0}}}{\lim}\frac{\left(e^{4 x^{2}}-1\right)}{x \sin x} \times \frac{4 x}{4 x}\)

\(L=\underset{{{{x} \rightarrow 0}}}{\lim}\frac{\left({e}^{4 x^{2}}-1\right)}{4 {x}^{2}} \times\left(\frac{{x}}{\sin {x}}\right) \times 4\)

We know that:

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

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

\(\therefore \underset{{{{x} \rightarrow 0}}}{\lim} \frac{\left(e^{4 x^{2}}-1\right)}{4 x^{2}}=1\)

and, \( \underset{{{{x} \rightarrow 0}}}{\lim} \left(\frac{{x}}{\sin {x}}\right)=1\)

and, 

Now, 

\(L=1 \times 1 \times 4=4\)

Statement- 2 is true. 

Now, $\frac{1}{5}(1+5p),\frac{1}{3}(1+2p)$

$\frac{1}{3}(1-p),\frac{1}{5}(1-3p)\ge 0$

$\Rightarrow p\ge -1/5,p\ge -1/2,p\le 1,p\le 1/3$

$\Rightarrow -1/5\le p\le 1/3$ ..(1)

and $\frac{1}{5}(1+5p)+\frac{1}{3}(1+2p)+\frac{1}{3}(1-p)+\frac{1}{5}(1-3p)\le 1$

$\begin{array}{ll}\Rightarrow & \frac{1}{5}(2+2p)+\frac{1}{3}(2+p)\le 1\\ \Rightarrow & 6+6p+10+5p\le 15\end{array}$

$\Rightarrow 11p\le -1 \Rightarrow p\le -\frac{1}{11}$ ..(2)

From (1) and (2) we get $-1/5\le p\le -\frac{1}{11}$

$\Rightarrow$ there are infinite values of $p$.

\(\begin{aligned} & a=\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{2 n}{n^2+k^2} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{2}{1+\left(\frac{k}{n}\right)^2} \\ & a=\int_0^1 \frac{2}{1+x^2} d x=2 \tan ^{-1} x \int_0^1=\frac{\pi}{2} \\ & f(x)=\sqrt{\frac{1-\cos x}{1+\cos x}}, x \in(0,1) \\ & f(x)=\frac{1-\cos x}{\sin x}=\operatorname{cosec} x-\cot x \\ & f^{\prime}(x)=\operatorname{cosec}^2 x-\operatorname{cosec} x \cot x \\ & \left.\begin{array}{l}f\left(\frac{a}{2}\right)=f\left(\frac{\pi}{4}\right)=\sqrt{2}-1 \\ f^{\prime}\left(\frac{a}{2}\right)=f^{\prime}\left(\frac{\pi}{4}\right)=2-\sqrt{2}\end{array}\right\} f^{\prime}\left(\frac{a}{2}\right)=\sqrt{2} \cdot f\left(\frac{a}{2}\right) \\ & \end{aligned}\)

Given:

\(\mathrm{P}(\mathrm{n}): \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+5)\) is a multiple of 3.

For \(\mathrm{n}=1\)

\( \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+5)=1.2 \cdot 6=12=3.4\)

\(\mathrm{P}(\mathrm{n})\) is true for \(\mathrm{n}=1\)

Suppose \(\mathrm{p}(\mathrm{k})\) is true for \(\mathrm{n}=\mathrm{k}\)

\(k(k+1)(k+5)=3 m\) (let) or \(k^{3}+6 k^{2}+5 k=3 m\).......(i)

Replacing \(\mathrm{k}\) by \(\mathrm{k}+1\), we get

\((k+1)(k+2)(k+6)=k\left(k^{2}+8 k+12\right)+\left(k^{2}+8 k+12\right) \)

\(k^{3}+9 k^{2}+20 k+12=\left(k^{3}+6 k^{2}+5 k\right)+\left(3 k^{2}+15 k+12\right)\)

\(=3 \mathrm{~m}+3 \mathrm{k}^{2}+15 \mathrm{k}+12 \quad\) [.......from (i) \(]\)

\(=3\left(\mathrm{~m}+\mathrm{k}^{2}+5 \mathrm{k}+4\right)\)

\((k+1)(k+2)(k+6)\) is a multiple of 3 i.e., \(P(k+1)\) is multiple of 3 , if \(P(k)\) is a multiple of 3 i.e., \(\mathrm{P}(\mathrm{k}+1)\) is true whenever \(\mathrm{P}(\mathrm{k})\) is true.

So, \(\mathrm{P}(\mathrm{n})\) is true for all \(\mathrm{n} \in \mathrm{N}\).

Let \(X Y\) plane divides the line joining the points \(A(2,3,-5)\) and \(B(-1,-2,-3)\) in the ratio \(k: 1\).

When the line segment is divided internally in the ratio \(m: n\), we use the formula: 

\(\Leftrightarrow({x}, {y})=\left(\frac{{mx}_{2}+{nx}_{1}}{{~m}+{n}}, \frac{{my}_{2}+{ny}_{1}}{{~m}+{n}}\right)\)

Using the section formula, the coordinate of the point of intersection is given by:

\(\left(\frac{-k+2}{k+1}, \frac{-2 k+3}{k+1}, \frac{-3 k-5}{k+1}\right)\)

As we know, on the XY plane Z-coordinate is zero.

Therefore, \(\frac{-3 {k}-5}{{k}+1}=0\)

\(\Rightarrow-3 k-5=0\)

\(\Rightarrow-3 k=5\)

\(\Rightarrow \frac{k}{1}=\frac{-5}{3}\)

Therefore, the ratio is \(5: 3\) externally.

Given that: 

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

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

\(\Rightarrow \frac{2 {x}}{{a}^{2}}+\frac{2 {y}}{{b}^{2}} \frac{{dy}}{{dx}}=0\)

\(\Rightarrow \frac{2 {y}}{{b}^{2}} \frac{{dy}}{{dx}}=-\frac{2 {x}}{a^{2}}\)

\(\therefore \frac{{dy}}{{dx}}=-\frac{{b}^{2} {x}}{{a}^{2} {y}}\)

Given: Equation of curve is \(y=\sqrt{5 x-3}-2\) and the tangent to the curve \(y=\sqrt{5 x-3}-2\) is parallel to the line \(4 x-2 y+3=0\)

The given line \(4 x -2 y +3=0\) can be re-written as:

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

Now by comparing the above equation of line with \(y = m x + c\) we get,

\(m =2\) and \(c =\frac{3}{2}\)

\(\because\) The line \(4 x-2 y+3=0\) is parallel to the tangent to the curve \(y=\sqrt{5 x-3}-2\)

As we know that if two lines are parallel then their slope is same.

So, the slope of the tangent to the curve \(y=\sqrt{5 x-3}-2\) is \(m =2\)

Let, the point of contact be \(\left( x _{1}, y _{1}\right)\)

As we know that slope of the tangent at any point say \(\left(x_{1}, y_{1}\right)\) to a curve is given by:

\(m=\left[\frac{d y}{d x}\right]_{\left(x_{1}, y_{1}\right)}\)

\(\Rightarrow \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\sqrt{5 x-3}} \cdot 5-0=\frac{5}{2 \sqrt{5 x-3}}\)

\(\Rightarrow\left[\frac{d y}{d x}\right]_{\left(x_{1}, y_{1}\right)}=\frac{5}{2 \sqrt{5 x_{1}-3}}\)

\(\because\) Slope of tangent to the curve \(y=\sqrt{5 x-3}-2\) is \(m =2\)

\(\Rightarrow 2=\frac{5}{2 \sqrt{5 x_{1}-3}}\)

By squaring both the sides of the above equation we get:

\(4=\frac{25}{4 \cdot\left(5 x_{1}-3\right)}\)

\(\Rightarrow x_{1}=\frac{73}{80}\)

\(\because\left( x _{1}, y _{1}\right)\) is point of conctact i.e., \(\left( x _{1}, y _{1}\right)\) will satisfy the equation of curve:

\(y=\sqrt{5 x-3}-2\)

\(\Rightarrow y_{1}=\sqrt{5 x_{1}-3}-2\)

By substituting \(x_{1}=\frac{73}{80}\) in the above equation we get:

\(y _{1}=-\frac{3}{4}\)

So, the point of contact is: \(\left(\frac{73}{80},-\frac{3}{4}\right)\)

As we know that equation of tangent at any point say \(\left(x_{1}, y_{1}\right)\) is given by:

\(y-y_{1}=\left[\frac{d y}{d x}\right]_{\left(x_{1}, y_{1}\right)} \cdot\left(x-x_{1}\right)\)

\(\Rightarrow y+\frac{3}{4}=2 \cdot\left(x-\frac{73}{80}\right)\)

\(\Rightarrow 80 x-40 y-103=0\)

So, the equation of tangent to the given curve at the point \(\left(\frac{73}{80},-\frac{3}{4}\right)\) is \(80 x -40 y-103=0\)

\(y d x+\left(x+x^{2} y\right) d y=0\)

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

\(y d x+x d y=-x^{2} y d y\) \(\left(\because \frac{d}{d x}(x y)=y d x+x d y\right)\)

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

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

\(\frac{\frac{d}{d x}(x y)}{(x y)^{2}}=\frac{-d y}{y}\).....(i)

integrating eq. (1) both side, we get \(\int \frac{\frac{d}{d x}(x y)}{(x y)^{2}}=-\int \frac{d y}{y}\)

\(\frac{1}{(x y)}=-\log y+C\) (where C is integral constant)

\(-\frac{1}{x y}+\log y=C\)