Mathematics Test - 31

We can solve this problem using sum of n terms of an A.P.
We know that
The sum of n terms of an AP with first term 'a' and common difference 'd' is given by
\(\mathrm{Sn}=\mathrm{n} / 2(2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d})\)
By using this formula we can find the answer.
We have
The numbers which are divisible by 11 between 100 and 500 are \(110,121, \ldots . ., 495\)
This is an A.P sequence.
First find number of terms using the formula
\(\mathrm{n}=\frac{\text { last term-first term }}{\text { common difference }}+1\)
Here,
last term \(=495,\) first term \(=110\)
Common difference, \(d=121-110=11\)

\(\mathrm{n}=\frac{495-110}{11}+1\)
\(=\frac{385}{11}+1\)
\(=35+1\)
\(=36\)
\(\mathrm{S}_{36}=\frac{36}{2}[2 \times 110+(36-1) 11]\)
\(\left[\because \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d})\right]\)
here, \(\mathrm{a}=110, \mathrm{d}=11\)
\(=18[220+35 \times 11]\)
\(=18[220+385]\)
\(=18 \times 605\)
\(=10890\)

Therefore, the sum of all integers between 100 and 500 which are divisible by 11 is 10890 .

\(\lim _{x \rightarrow \frac{\pi}{4}} f(x)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x}\)
\(=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{4}=\frac{1}{4}[\text { by } L \text { 'Hospital's rule }]\)
\(\operatorname{since}, f(x)\) is continuous at \(x=\frac{\pi}{4}\)
\(\therefore \lim _{x \rightarrow \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right) \quad \Rightarrow \quad \frac{1}{4}=a\)

KEY CONCEPT
Continuity at a Point Calculus
A function \(f(x) f x\) is said to be continuous at a point \(x=a\) of its domain if and only if it satisfies the following three conditions:
(1) \(f(a)\) exists. ('a' lies in the domain of \(f\) )
(2) \(\lim _{x \rightarrow a^{-}} f(x)\) exist i.e. \(\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x)\) or \(R .\) H.L. at \(x=a=L . H . L .\) at \(x=a\)
(3) \(\lim _{x \rightarrow a} f(x)=f(a)\) (limit equals the value of function)

R. of line is 7,-3,-1
since given line is perpendicular to the plane
\(\therefore\) Required plane is, \(7(x+10)-3(y-5)-(z-4)=0\) or \(7 x-3 y-z+89=0\)
Equation of a Plane through one Point and one perpendicular Vector Equation of plane passing through a given point:
(1) Equation of plane passing through the point \(\left(x_{1}, y_{1}, z_{1}\right)\) is \(A\left(x-x_{1}\right)+B\left(y-y_{1}\right)+C\left(z-z_{1}\right)=0,\) where \(A, B\) and \(C\) are \(d . r . '\) 's of normal to the plane.
(2) Equation of plane passing through point \(A(\vec{a})\) and having \(\vec{n}\) as it's normal vector is \(\vec{r} . \vec{n}=\vec{a} . \vec{n}\)
Plane Passing through a point and Perpendicular to a line \((3 D)\) Vector equation of a plane through the point \(A(a)\) and perpendicular to the vector \(n\) is \((\mathrm{r}-\mathrm{a}) \cdot n=0\) or \(\mathrm{r} \cdot n=\mathrm{a} \cdot n\) or \(\mathrm{r} \cdot n=\mathrm{d},\) where \(\mathrm{d}=\mathrm{a} \cdot n .\) This is known as the scalar product form of a plane.

Given \(y=\int_{0}^{x} \frac{d t}{1+t^{3}}\)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{1+x^{3}}\)
\(\Rightarrow \mathrm{m}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{x=1}=\frac{1}{1+1}=\frac{1}{2}\)
KEY CONCEPTS
Newton Leibniz Theorem
(1) If \(f(x)\) is continuous and \(u(x), v(x),\) are differentiable functions in the interval \([a, b],\) then,
\(\frac{\mathrm{d}}{\mathrm{dx}}\left(\int_{u(x)}^{v(x)} f(t) \mathrm{dt}\right)=f\{v(x)\} \frac{\mathrm{d}}{\mathrm{dx}}\{v(x)\}-f\{u(x)\} \frac{\mathrm{d}}{\mathrm{dx}}\{u(x)\}\)
(2) If the function \(\phi(x)\) and \(\psi(x)\) are defined on \([a, b]\) and differentiable at a point \(x e(a, b),\) and \(f(x, \mathrm{t})\) is continuous, then,
\(\frac{d}{d x}\left[\int_{\phi(x)}^{\psi(x)} f(x, t) d t\right]=\int_{\phi(x)}^{\psi(x)} \frac{d}{d x} f(x, t) d t+\left\{\frac{d \psi(x)}{d x}\right\} f(x, \psi(x))\)
\(-\left\{\frac{\mathrm{d} \phi(\mathbf{x})}{\mathrm{d} x}\right\} \mathrm{f}(x, \phi(x))\)

\(A-2 I=\left[\begin{array}{cc}4 & 2 i \\ i & 1\end{array}\right]-2\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\)
\(=\left[\begin{array}{cc}2 & 2 i \\ i & -1\end{array}\right]\)
\(A-3 I=\left[\begin{array}{cc}4 & 2 i \\ i & 1\end{array}\right]-3\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\)
\(\therefore(A-2 I)(A-3 I)=\left[\begin{array}{cc}2 & 2 \mathrm{i} \\ \mathrm{i} & -1\end{array}\right]\left[\begin{array}{cc}1 & 2 \mathrm{i} \\ \mathrm{i} & -2\end{array}\right]\)
\(=\left[\begin{array}{cc}1 & 2 i \\ i & -2\end{array}\right]\)
\(\therefore\left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right]\)

KEY CONCEPTS

Subtraction is performed in analogous way.
Example:
\[
\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]-\left[\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right]=\left[\begin{array}{lll}
1-5 & 2-6 \\
3-7 & 4-8
\end{array}\right]=\left[\begin{array}{ll}
-4 & -4 \\
-4 & -4
\end{array}\right]
\]

Multiplication of matrix

When the number of elements in row vector is the same as the number of rows in the second matrix then this matrix multiplication can be performed.
Example:
\[
\underbrace{\left[\begin{array}{lll}
1 & 2 & 3
\end{array}\right] \cdot\left[\begin{array}{lll}
2 & 1 & 3 \\
3 & 3 & 2 \\
4 & 1 & 2
\end{array}\right]}_{1 \times 3}=\left[\begin{array}{l}
1 \cdot 2+2 \cdot 3+3 \cdot 4 \\
1 \cdot 1+2 \cdot 3+3 \cdot 1 \\
1 \cdot 3+2 \cdot 2+3 \cdot 2
\end{array}\right]=\underbrace{\left[\begin{array}{l}
20 \\
10 \\
13
\end{array}\right]}_{1 \times 3}
\]

\(\alpha, \beta\) are roots of \(\mathrm{x}^{2}-3 \mathrm{x}+1=0\)
\(\frac{1}{\alpha-2}\) or \(\frac{1}{\beta-2}=\mathbf{x}\)
\(\Rightarrow(\alpha-2)\) or \((\beta-2)=\frac{1}{x}\)
\(\Rightarrow \alpha\) or \(\beta=\frac{1}{x}+2 \Rightarrow x \rightarrow \frac{1}{x}+2\)
required quadratic equation
\(=\left(\frac{1}{x}+2\right)^{2}-3\left(\frac{1}{x}+2\right)+1=0\)
\(\Rightarrow 4 x^{2}+4 x+1-3 x-6 x^{2}+x^{2}=0\)
\(\Rightarrow x^{2}-x-1=0\)

Equation in terms of Roots of another Equation Equation in terms of the roots of another equation : If \(\alpha, \beta\) are roots of the equation \(a x^{2}+b x+c=0,\) then the equation whose roots are
(i) \(-\alpha,-\beta \Rightarrow a x^{2}-\mathrm{bx}+c=0 \quad\) (Replace \(\mathrm{x}\) by \(-\mathrm{x}\) )
(ii) \(\frac{1}{a}, \frac{1}{\beta} \Rightarrow c x^{2}+b x+a=0\)
(Replace \(\left.x \text { by } \frac{1}{x}\right)\)
(iii) \(\alpha^{n}, \beta^{n} ; n \in N \Rightarrow a\left(x^{\frac{1}{n}}\right)^{2}+b\left(x^{\frac{1}{n}}\right)+c=0 \quad\left(\text { Replace } x \text { by } x^{\frac{1}{n}}\right)\)
(iv) \(k \alpha, k \beta \Rightarrow a x^{2}+k b x+k^{2} c=0\)
(v) \(\left.k+\alpha, k+\beta \Rightarrow a(x-k)^{2}+b(x-k)+c=0 \quad \text { (Replace } \mathbf{x} \text { by }(x-k)\right)\)
(vi) \(\frac{\alpha}{k}, \frac{\beta}{k} \Rightarrow k^{2} a x^{2}+k b x+c=0\)
\((\text {Replace } x \text { by } k x)\)
(vii) \(\alpha^{\frac{1}{n}}, \beta^{\frac{1}{n}} ; n \in \mathrm{N} \Rightarrow \alpha\left(x^{2}\right)^{2}+b\left(x^{n}\right)+c=0 \quad\left(\text { Replace } \mathrm{x} \text { by } x^{n}\right)\)

\begin{array}{l}
\because f(x) \cos x=\frac{1}{2} \cdot 2 f(x) f^{\prime}(x) \\
\text { Then, } f^{\prime}(x)=\cos x \\
\therefore f(x)=\sin x+c
\end{array}

\(\frac{d}{d x}(c)=0\)
\(\frac{d}{d x}(x)=1\)
\(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\)

\(\int \sin x \mathrm{d} \mathrm{x}=-\cos x+c\)
\(\int \cos x \mathrm{d} \mathrm{x}=\sin x+c\)
\(\int \sec ^{2} x \mathrm{d} \mathrm{x}=\tan x+c\)
\(\int \operatorname{cosec}^{2} x \mathrm{dx}=-\cot x+c\)
\(\int \sec x \tan x \mathrm{dx}=\sec x+c\)
\(\int \operatorname{cosec} x \cot x \mathrm{dx}=-\operatorname{cosec} x+c\)

\(\mathrm{M}\) is the foot of the perpendicular
Slope of \(\mathrm{OM}=\frac{2-0}{-1-0}=\frac{2}{-1}=-2\)
Slope of \(A B=m\)
\(\therefore \mathrm{OM} \perp \mathrm{AB}\)
\(\Rightarrow m \times(-2)=-1\)
\(\Rightarrow m=\frac{1}{2}\)
\(\mathrm{M}(-1,2)\) lies on \(\mathrm{AB}\) whose equation is
\(y=m x+c\) or \(y=\frac{1}{2} x+c\)
\(2=\frac{1}{2} \times(-1)+c \Rightarrow c=2+\frac{1}{2}=\frac{5}{2}\)
\(\therefore \quad m=\frac{1}{2}\) and \(c=\frac{5}{2}\)

KEY CONCEPTS

Condition for Parallel & Perpendicular Lines
Equation of parallel and perpendicular lines to a given line
(1) Equation of a line which is parallel to ax+by+c=0 is ax+by+λ=0
(2) Equation of a line which is perpendicular to ax+by+c=0 is bx-ay+λ=0. The value of λ in both cases is obtained with the help of additional information given in the problem.
Point Slope form of Line
Equation of a line throught the point (x_1, y₁) and having slope m is y - y₁= m(x - x₁).

\(x^{2}+\frac{1}{x^{2}}=\left(x+\frac{1}{x}\right)^{2}-2=4 \cos ^{2} \theta-2\)
\(\Rightarrow x^{2}+\frac{1}{x^{2}}=2 \cos 2 \theta\)
Again \(x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right)=8 \cos ^{3} \theta\)
\(x^{3}+\frac{1}{x^{3}}=8 \cos ^{3} \theta-6 \cos \theta=2 \cos 3 \theta\)
Similarly, \(x^{n}+\frac{1}{x^{2}}=2 \cos n \theta\)
Alternate solution
Given \(x+\frac{1}{x}=2 \cos \theta\)
\(x^{2}-2 x \cos \theta+1=0\)
\(x=\cos \theta \pm i \sin \theta\)
so \(x=e^{i \theta}\)
\(x^{n}+x^{-n}=e^{i n \theta}+e^{-i n \theta}\)
\(=2 \cos n \theta\)

Expansion of cos2A Trigonometry \(\cos 2 \mathrm{A}=2 \cos ^{2} \mathrm{A}-1=1-2 \sin ^{2} \mathrm{A}=\cos ^{2} \mathrm{A}-\sin ^{2} \mathrm{A}=\frac{1-\tan ^{2} \mathrm{A}}{1+\tan ^{2} \mathrm{A}}\)
\(\left(\text {where } A \neq(2 n+1) \frac{\pi}{2}\right)\)
Expansion of \(\cos 3 \mathrm{A}\)
\(\cos 3 \mathrm{A}=4 \cos ^{3} \mathrm{A}-3 \cos \mathrm{A}=4 \cos \left(60^{\circ}-\mathrm{A}\right) \cdot \cos \mathrm{A} \cdot \cos \left(60^{\circ}+\mathrm{A}\right)\)