Mathematics Test - 33

\(\cos \theta=\frac{2 \times(-1)+5 \times 8+(-3) \times 4}{\sqrt{2^{2}+5^{2}+(-3)^{2}} \sqrt{(-1)^{2}+8^{2}+4^{2}}}\)
\(\Rightarrow \cos \theta=\frac{-2+40-12}{9 \sqrt{38}}=\left(\frac{26}{9 \sqrt{38}}\right)\)
\(\Rightarrow \theta=\cos ^{-1}\left(\frac{26}{9 \sqrt{38}}\right)\)
Angle between two lines in \(3 D\) Let \(\theta\) be the angle between two straight lines whose direction cosines are \(l_{1}, \mathrm{m}_{1}, n_{1}\) and \(l_{2}, \mathrm{m}_{2}, n_{2}\) respectively, is given by \(\cos \theta=l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\)
If direction ratios of two lines \(a_{1}, b_{1}, c_{1}\) and \(a_{2}, b_{2}, c_{2}\) are given, then angle between these two lines is
\(\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\)

since the given function is continuous at \(x=-\frac{\pi}{2}\)
\(\therefore-4 \sin \left(-\frac{\pi}{2}\right)+\cos \left(-\frac{\pi}{2}\right)=a \sin \left(-\frac{\pi}{2}\right)+b\)
\(\Rightarrow 4=b-a\)
again since the given function is continuous at \(x=\frac{\pi}{2}\)
\(\therefore a \sin \left(\frac{\pi}{2}\right)+b=\cos \left(\frac{\pi}{2}\right)+2\)
\(\Rightarrow a+b=2\)
Solving (i) and (ii)
\(\Rightarrow a=-1, b=3\)

Continuity at a Point Calculus
A function \(f(x)\) 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).
Continuity in Interval
A function is said to be continuous on the interval \((a, b)\) if it is continuous at each point in the interval.
A function \(f(x)\) is continuous at the closed interval \([a, b]\) if :
A function \(f(x)\) is continuous at the closed interval \([a, b]\) if
(1) \(f(x)\) is continuous at \(x\) for all values of \(x\) belonging to the open interval (a,b).
(2) \(f(a)=\lim _{x \rightarrow a^{+}} f(x)\) continuous

Let the third vertex be \((x, y)\)
\(\therefore \frac{5-2+x}{3}=5\) and \(\frac{4+4+y}{3}=6\)
\(\mathrm{x}=12\) and \(\mathrm{y}=10\)
Hence third vertex is (12,10)

Centroid of a Triangle in Coordinate Geometry
Centroid of a triangle: The centroid of a triangle is the point of intersection of its medians. The centroid divides the medians in the ratio 2: 1 (vertex : base)
If \(A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\) and \(C\left(x_{3}, y_{3}\right)\) are the vertices of a triangle. If \(G\) be the centroid upon one of the median (say) \(A D,\) then \(A G: G D=2: 1\)
\(\Rightarrow \quad\) Co-ordinate of \(G\) are \(\left(\frac{x_{1}+x_{2}+x_{1}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)

\[
\because \quad f(x)=a-(x-3)^{8 / 9}
\]
\(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})=0-\frac{8}{9}(\mathrm{x}-3)^{-1 / 9}\)
At \(x=3, f^{\prime}(x)\) is not defined.
hence sign scheme of \(f^{\prime}(x)\) is
\(\therefore f(x)\) is maximum at \(x=3\)
Hence maximum value of \(f(x)\) is eaual to "a"

Maxima and Minima in Calculus

A function f (x) is said to attain a maximum at x = a if there exists a neighbourhood (a−δ,a+δ)a-δa+δ such that f(x)≤f(a)fx≤fa ∀xε(a−δ,a+δ)∀xεa-δa+δ

A function f (x) is said to attain a minimum at x = a if there exists a neighbourhood (a−δ,a+δ)a-δa+δ such that f(x)≥f(a)fx≥fa ∀xε(a−δ,a+δ)

Maxima
For a twice differentiable function if an interior point is a point of maxima then \(\frac{\mathrm{dy}}{\mathrm{dx}}=0, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}<0\) at that point.
Minima
For a twice differentiable function if an interior point is a point of minima then \(\mathrm{dy} / \mathrm{dx}=0, \mathrm{d}^{2} \mathrm{y} / \mathrm{dx}^{2}>0\) at that point.
Second Derivative Test
If the function \(f\) is twice differentiable at a critical point \(x\) (i.e.f' \((x)=0\) ), then i.e.f'x \(=0\), then :
If \(f^{\prime \prime}(x)<0\) thenfhas a local maximum atx" If \(f^{\prime \prime} x<0\) thenfhas a local maximum at
If \(f^{\prime \prime}(x)>0\) then \(f\) has a local minimum atx* If \(f^{\prime \prime} x>0\) then \(f\) has a local minimum atx
If \(f^{\prime \prime}(x)=0,\) the test is inconclusive

\(\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\)
\(=\left[\frac{a x^{6}}{6}+\frac{b x^{4}}{4}+\frac{c x^{2}}{2}+k x\right]_{-3}^{3}\)
\(=\frac{a .3^{6}}{6}+\frac{b .3^{4}}{4}+\frac{c .3^{2}}{2}+k(3)-\frac{a .3^{6}}{6}-\frac{b .3^{4}}{4}-\frac{c .3^{2}}{2}+k(3)\)
\(=6 k\)
ie, Integral depends upon \(k\)
Alternate Solution
since \(x^{5}, x^{3}\) and \(x\) are odd functions, their integrals from -a to a will be zero.
Hence the given integral only depends on \(\mathrm{k}\)

if \(x^{2}+y^{2}=t+\frac{1}{t}\)
and \(x^{4}+y^{4}=t^{2}+\frac{1}{z^{2}}\)
squaring both sides,the equation given below
\(x^{2}+y^{2}=t+\frac{1}{t}\)
we get
\(\left(x^{2}+y^{2}\right)^{2}=\left(t-\frac{1}{t}\right)^{2}\)
\(x^{4}+y^{4}+2 x^{2} y^{2}=t^{2}+\frac{1}{t^{2}}-2 t\left(\frac{1}{t}\right)\)
from another equation put the value of \(x^{4}+y^{4}\) in the eq
\(t^{2}+\frac{1}{t^{2}}+2 x^{2} y^{2}=t^{2}+\frac{1}{t^{2}}-2\)
\(2 x^{2} y^{2}=-2\)
\(x^{2} y^{2}=-1 \ldots e q 1\)
or
\(y=\frac{-1}{x^{2} y} \ldots \ldots e q 2\)
take derivative of eq1 with respect to \(x\)
\(2 x y^{2}+2 x^{2} y \frac{d y}{d x}=0\)
\(2 x^{2} y \frac{d y}{d x}=2 x y^{2}\)

\(\frac{d y}{d x}=\frac{u}{x}\)
put value of \(y\) from eq 2
\(\frac{d y}{d x}=\left(\frac{1}{x}\right)\left(\frac{-1}{x^{2} y}\right)\)
\(\frac{d y}{d x}=\frac{-1}{x^{3} y}\)

We have \(R=\{(a, b): a \leq b\}\). Let \(a, b \in R\)
Reflexive: for any \(a \in R\) we have \(a \leq a .\) So, \(R\) is reflexive.
Symmetric: we observe that (2,3)\(\in \mathrm{R}\) but (3,2)\(\in \mathrm{R}\). So, \(\mathrm{R}\) is not symmetric.
Transitivity: \((a, b) \in R\) and \((b, c) \in R \Rightarrow a \leq b\) and \(b \leq c \Rightarrow a \leq c \Rightarrow(a, c) \in R\)
So, \(R\) is transitive. Hence, \(R\) is reflexive and transitive but not symmetric

Reflexive Relation
(1) Reflexive relation : A relation R on a set A is said to be reflexive if every element of A is related to itself.
Thus, R is reflexive ⇔ (a, a) ∈ R for all a ∈ A.
Example : Let A {1, 2, 3} and R = {(1, 1); (1, 3)}
Then R is not reflexive since 3 ∈ A but (3, 3) ∉ R
A reflexive relation on A is not necessarily the identity relation on A. The universal relation on a non - void set A is reflexive.
Symmetric Relation
Symmetric relation: A relation R on a set A is said to be a symmetric relation iff(a, b) ∈R⇒ (b, a) ∈ R for all a, b ∈ A i.e., aRb ⇒ bRa for all a, b ∈ A. It should be noted that R is symmetric iff R-1 = R.
The identity and the universal relations on a non - void set are symmetric relations. A reflexive relation on a set A is not necessarily symmetric.

\(\int_{0}^{1} \frac{1-x}{\sqrt{1-x^{2}}} d x=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x+\frac{1}{2} \int_{0}^{1} \frac{-2 x}{\sqrt{1-x^{2}}} d x\)
\(=\left(\sin ^{-1} x\right)_{0}^{1}+\frac{1}{2} \cdot 2(\sqrt{1-x^{2}})_{0}^{1}=\frac{\pi}{2}-1\)

Properties of Definite Integrals
(1) \(\int_{a}^{b} f(x) \mathrm{d} \mathrm{x}=\int_{a}^{b} f(\mathrm{t}) \mathrm{dt}\) i.e., The value of a definite integral remains unchanged if its variable is replaced by any other symbol.
(2) \(\int_{a}^{b} f(x) \mathrm{d} \mathrm{x}=-\int_{b}^{a} f(x) \mathrm{d} \mathrm{x}\) i.e., by the interchange in the limits of definite integral, the sign of the integral is changed.
(3) \(\left.\int_{a}^{b} f(x) \mathrm{d} \mathrm{x}=\int_{a}^{c} f(x) \mathrm{d} \mathrm{x}+\int_{c}^{b} f(x) \mathrm{dx}, \text { (where } aor \(\int_{a}^{b} f(x) \mathrm{d} \mathrm{x}=\int_{a}^{c_{1}} f(x) \mathrm{d} \mathrm{x}+\int_{c_{1}}^{c_{2}} f(x) \mathrm{d} \mathrm{x}+\ldots \ldots+\int_{c_{\mathrm{n}}}^{b} f(x) \mathrm{d} \mathrm{x}\)
; (where generally \(\left.aGenerally this property is used when the integrand has two or more rules in the integration interval.
This is useful when \(f(x)\) is not continuous in \([a, b]\) because we can break up the integral into several integrals at the points of discontinuity so that the function is continuous in the sub-intervals.
(4) \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x:\) In particular \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\). It is generally
used for those complicated integrals whose denominators are unchanged when \(x\) is replaced by \((a-\)x)

(5) \(\int_{-a}^{a} f(x) d x=\int_{0}^{a}[f(x)+f(-x)] d x=\left[\begin{array}{cl}2 \int_{0}^{a} f(x) d x & \text { if } f(x) \text { is an even function } \\ 0 & \text { if } f(x) \text { is an odd function }\end{array}\right.\)
(6) \(\int_{0}^{2 a} f(x) d x=\int_{0}^{a}[f(x)+f(2 a-x)] d x\)
Definition of Definite Integrals Consider a continuous function \(f(x)\) in \([a, b]\)

Given, \(x=a(\cos \theta+\theta \sin \theta)\)
and \(y=a(\sin \theta-\theta \cos \theta)\)
\(\therefore \quad \frac{d x}{d \theta}=a(-\sin \theta+\theta \cos \theta+\sin \theta)=a \theta \cos \theta\)
and \(\frac{d y}{d \theta}=a(\cos \theta+\theta \sin \theta-\cos \theta)=a \theta \sin \theta\)
\(\therefore \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\tan \theta\)

Differentiation of a Function w.r.t Another Function
Let \(u=f(x)\) and \(v=g(x)\) be two functions of \(x\). Then to find the derivative of \(f(x) w . r \cdot t g(x)\), i.e., to find
\(\frac{d u}{d v}\) we use the following formula \(\frac{d u}{d v}=\frac{\frac{d u}{d y}}{\frac{d v}{d x}}\)
Thus, to find the derivative of \(f(x)\) w.r.t \(g(x),\) we first differentiate both w.r.t. \(x\). and then divide the derivative of \(f(x)\) wrt \(x\) by the derivative of \(g(x)\) w.rt. \(x\)