Show that the lines
Question: Show that the lines 3x 4y + 5 = 0, 7x 8y + 5 = 0 and 4x + 5y = 45 are concurrent. Also find their point of intersection. Solution: Given: $3 x-4 y+5=0$ $7 x-8 y+5=0$ and $4 x+5 y=45$ or $4 x+5 y-45=0$ To show: Given lines are concurrent The lines $a_{1} x+b_{1} y+c_{1}=0, a_{1} x+b_{1} y+c_{1}=0$ and $a_{1} x+b_{1} y+c_{1}=0$ are concurrent if We know that, We have, $a_{1}=3, b_{1}=-4, c_{1}=5$ $a_{2}=7, b_{2}=-8, c_{2}=5$ $a_{3}=4, b_{2}=5, c_{3}=-45$ Now, expanding along first row, w...
Read More →If tanx = b/a then find the value of
Question: If tanx = b/a then find the value of $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$ Solution: According to the question, tan x = b/a Let, $y=\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$ $\therefore y=\sqrt{\frac{a\left(1+\frac{b}{a}\right)}{a\left(1-\frac{b}{a}\right)}}+\sqrt{\frac{a\left(1-\frac{b}{a}\right)}{a\left(1+\frac{b}{a}\right)}}$ $=\sqrt{\frac{(1+\tan x)}{(1-\tan x)}}+\sqrt{\frac{(1-\tan x)}{(1+\tan x)}}$ $=\frac{\sqrt{1+\tan x}}{\sqrt{1-\tan x}}+\frac{\sqrt{1-\tan x}}{\s...
Read More →Show that f(x)=tanx is an increasing function
Question: Show that $f(x)=\tan x$ is an increasing function on $(-\pi / 2, \pi / 2)$. Solution: Given:- Function $f(x)=\tan x$ Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$. (i) If $f^{\prime}(x)0$ for $a l l_{x} \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- (i) Obtain the function and put it equal to $f(x)$ (ii) Find $f^{\prime}(x)$ (iii) Put $f^{\prim...
Read More →Prove the following
Question: If $\cos (\alpha+\beta)=\frac{4}{5}$ and $\sin (\alpha-\beta)=\frac{5}{13}$,where lie between 0 and /4, find value of tan 2 [Hint: Express tan 2 as tan ( + + ] Solution: According to the question, cos( + ) = 4/5 (i) We know that, sin x = (1 cos2x) Therefore, sin ( + ) = (1 cos2( + )) ⇒sin ( + ) = (1 (4/5)2) = 3/5(ii) Also, sin( ) = 5/13 {given} (iii) we know that, cos x = (1 sin2x) Therefore, cos ( ) = (1 sin2( )) ⇒cos ( ) = (1 (5/13)2) = 12/13(iv) Therefore, tan 2 = tan ( + + ) We kno...
Read More →Show that f(x)=cos x is a decreasing function
Question: Show that $f(x)=\cos x$ is a decreasing function on $(0, \pi)$, increasing in $(-\pi, 0)$ and neither increasing nor decreasing in $(-\pi, \pi)$. Solution: Given:- Function $f(x)=\cos x$ Theorem:- Let $\mathrm{f}$ be a differentiable real function defined on an open interval $(\mathrm{a}, \mathrm{b})$. (i) If $f^{\prime}(x)0$ for $a l l_{X} \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:-...
Read More →Show that the lines
Question: Show that the lines x + 7y = 23 and 5x + 2y = a 16 intersect at the point (2, 3). Solution: Suppose the given two lines intersect at a point P(2, 3). Then, (2, 3) satisfies each of the given equations. So, taking equation $x+7 y=23$ Substituting $x=2$ and $y=3$ Lhs = x + 7y $=2+7(3)$ $=2+21$ $=23$ $=\mathrm{RHS}$ Now, taking equation $5 x+2 y=16$ Substituting $x=2$ and $y=3$ LHS = 5x + 2y $=5(2)+2(3)$ $=10+6$ $=16$ $=\mathrm{RHS}$ In both the equations pair (2, 3) for (x, y) satisfies ...
Read More →If m sin θ = n sin (θ + 2α),
Question: If m sin = n sin ( + 2), then prove that tan ( + ) cot = (m + n)/(m n) [Hints: Express sin( + 2) / sin = m/n and apply componendo and dividend] Solution: According to the question, m sin = n sin ( + 2) To prove: tan ( + )cot =(m + n)/(m n) Proof: m sin = n sin ( + 2) ⇒sin( + 2) / sin = m/n Applying componendo-dividendo rule, we have, $\Rightarrow \frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}=\frac{m+n}{m-n}$ By transformation formula of T-ratios, We know...
Read More →If [2sinα / (1+cosα+sinα)] = y, then prove that [(1– cosα+sinα) / (1+sinα)]
Question: If [2sin / (1+cos+sin)] = y, then prove that [(1 cos+sin) / (1+sin)] is also equal to y. $\left[\right.$ Hint : Express $\left.\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}=\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}\right]$ Solution: According to the question, y =2sin /(1+cos+sin) Multiplying numerator and denominator by (1 cos + sin ), We get, $\Rightarrow y=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha} \times ...
Read More →Find the points of intersection of the lines
Question: Find the points of intersection of the lines $4 x+3 y=5$ and $x=2 y-7$ Solution: Suppose the given two lines intersect at a point $P\left(x_{1}, y_{1}\right) .$ Then, $\left(x_{1}, y_{1}\right)$ satisfies each of the given equations. $\therefore 4 x+3 y=5$ or $4 x+3 y-5=0 \ldots$ (i) and $x=2 y-7$ or $x-2 y+7=0$...(ii) Now, we find the point of intersection of eq. (i) and (ii) Multiply the eq. (ii) by 4, we get $4 x-8 y+28=0$ ...............(iii) On subtracting eq. (iii) from (i), we g...
Read More →Prove that
Question: Prove that $\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$ Solution: According to the question, $\mathrm{LHS}=\frac{\tan \mathrm{A}+\sec \mathrm{A}-1}{\tan \mathrm{A}-\sec \mathrm{A}+1}$ $=\frac{\frac{\sin A}{\cos A}+\frac{1}{\cos A}-1}{\frac{\sin A}{\cos A}-\frac{1}{\cos A}+1}$ $=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}$ $=\frac{\sin A+(1-\cos A)}{\sin A-(1-\cos A)}$ Using the identity, sin2A + cos2A = 1, we get, sin A + (1 cos A). $\therefore \mathrm{LHS}=\frac{\sin ...
Read More →Show that
Question: Show that $f(x)=\sin x$ is an increasing function on $(-\pi / 2, \pi / 2)$. Solution: Given:- Function $f(x)=\sin x$ Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$. (i) If $f^{\prime}(x)0$ for $a l l_{x} \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- (i) Obtain the function and put it equal to $f(x)$ (ii) Find $f^{\prime}(x)$ (iii) Put $f^{\prim...
Read More →The perpendicular distance of a line from the origin is 5 units,
Question: The perpendicular distance of a line from the origin is 5 units, and its slope is -1. Find the equation of the line. Solution: Given: perpendicular distance from orgin is 5 units, and the slope is -1 To find : the equation of the line Formula used : We know that the perpendicular distance from a point $\left(x_{0}, y_{0}\right)$ to the line $a x+b y+c=0$ is given by $D=\frac{|a x+b y+c|}{\sqrt{a^{2}+b^{2}}}$ The equation of a straight line is given by $y=m x+c$ where $m$ denotes the sl...
Read More →Prove that the line
Question: Prove that the line 12x 5y 3 = 0 is mid-parallel to the lines 12x 5y + 7 = 0 and 12x 5y 13 = 0 Solution: Given: parallel lines $12 x-5 y-3=0,12 x-5 y+7=0,12 x-5 y-13=0$ To Prove : line $12 x-5 y-3=0$ is mid-parallel to the lines $12 x-5 y+7=0$ and $12 x-5 y$ $-13=0$ Formula used: The distance between the parallel lines $a x+b y+c=0$ and $a x+b y+d$ $=0$ is, $D=\frac{|d-c|}{\sqrt{a^{2}+b^{2}}}$ The equation of line 1 is $12 x-5 y+7=0$ The equation of line $m$ is $12 x-5 y-3=0$ The equat...
Read More →Find the distance between the parallel lines p
Question: Find the distance between the parallel lines p(x + y) = q = 0 and p(x + y) r =0 Solution: Given: parallel lines $p(x+y)=q=0$ and $p(x+y)-r=0$ To find : distance between the parallel lines $p(x+y)-q=0$ and $p(x+y)-r=0$ Formula used: The distance between the parallel lines $a x+b y+c=0$ and $a x+b y+d$ $=0$ is, $D=\frac{|d-c|}{\sqrt{a^{2}+b^{2}}}$ The parallel lines are $p(x+y)-q=0$ and $p(x+y)-r=0$ The parallel lines are $p x+p y-q=0$ and $p x+p y-r=0$ Here $a=p, b=p, c=-q, d=-r$ $D=\fr...
Read More →Find the distance between the parallel lines
Question: Find the distance between the parallel lines $y=m x+c$ and $y=m x+d$ Solution: Given: parallel lines y = mx + c and y = mx + d To find : distance between the parallel lines Formula used: The distance between the parallel lines $a x+b y+c=0$ and $a x+b y+d$ $=0$ is, $D=\frac{|d-c|}{\sqrt{a^{2}+b^{2}}}$ The parallel lines are $m x-y+c=0$ and $m x-y+d=0$ Here $a=m, b=-1, c=c, d=d$ $D=\frac{|d-c|}{\sqrt{m^{2}+1^{2}}}=\frac{|d-c|}{\sqrt{m^{2}+1}}$ The distance between the parallel lines $y=...
Read More →Find the distance between the parallel lines
Question: Find the distance between the parallel lines 8x + 15y 36 = 0 and 8x + 15y + 32 = 0. Solution: Given: parallel lines 8x + 15y 36 = 0 and 8x + 15y + 32 = 0. To find : distance between the parallel lines Formula used: The distance between the parallel lines ax $+b y+c=0$ and $a x+b y+d$ $=0$ is, $D=\frac{|d-c|}{\sqrt{a^{2}+b^{2}}}$ Here $a=8, b=15, c=-36, d=32$ $D=\frac{|32-(-36)|}{\sqrt{8^{2}+15^{2}}}=\frac{|32+36|}{\sqrt{64+225}}=\frac{68}{\sqrt{289}}=\frac{68}{17}=4$ The distance betwe...
Read More →Find the distance between the parallel lines
Question: Find the distance between the parallel lines 4x 3y + 5 = 0 and 4x 3y + 7 = 0 Solution: Given: parallel lines $4 x-3 y+5=0$ and $4 x-3 y+7=0$ To find : distance between the parallel lines Formula used: The distance between the parallel lines $a x+b y+c=0$ and $a x+b y+d$ $=0$ is, $D=\frac{|d-c|}{\sqrt{a^{2}+b^{2}}}$ Here $a=4, b=-3, c=5, d=7$ $D=\frac{|7-5|}{\sqrt{4^{2}+(-3)^{2}}}=\frac{|2|}{\sqrt{16+9}}=\frac{2}{\sqrt{25}}=\frac{2}{5}$ The distance between the parallel lines $4 x-3 y+5...
Read More →Show that
Question: Show that $f(x)=\cos ^{2} x$ is a decreasing function on $(0, \pi / 2)$. Solution: Given:- Function $f(x)=\cos ^{2} x$ Theorem:- Let $\mathrm{f}$ be a differentiable real function defined on an open interval $(\mathrm{a}, \mathrm{b})$. (i) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- (i) Obtain the function and put it equal to $f(x)$ (ii) Find $f^{\prime...
Read More →A vertex of a square is at the origin and its one side lies along the line
Question: A vertex of a square is at the origin and its one side lies along the line 3x 4y 10 = 0. Find the area of the square. Solution: Given: $A B C D$ is a square and equation of $B C$ is $3 x-4 y-10=0$ To find : Area of the square Formula used: We know that the length of perpendicular from $(m, n)$ to the line $a x+b y+c=0$ is given by, $D=\frac{|a m+b n+c|}{\sqrt{a^{2}+b^{2}}}$ The given equation of the line is $3 x-4 y-10=0$ Here $m=0$ and $n=0, a=3, b=-4, c=-10$ The given equation of the...
Read More →Let f(x) = √x and g (x) = x be two functions
Question: Let f(x) = xand g (x) = x be two functions defined in the domain R+{0}. Find (i) (f + g) (x) (ii) (f g) (x) (iii) (fg) (x) (iv) (f/g) (x) Solution: (i) (f + g)(x) ⇒(f + g)(x) = f(x) + g(x) ⇒f(x) + g(x) = x + x (ii) (f g)(x) ⇒(f g)(x) = f(x) g(x) ⇒f(x) g(x) = xx (iii) (fg)(x) ⇒(fg)(x) = f(x) g(x) ⇒(fg)(x) = (x)(x) ⇒f(x)g(x)= xx (iv) (f/q)(x) = f(x)/g(x) $\Rightarrow\left(\frac{f}{g}\right)(x)=\frac{\sqrt{x}}{x}$ Multiplying and dividing by $\sqrt{x}$, We get $=\frac{\sqrt{x}}{x} \times ...
Read More →Show that
Question: Show that $f(x)=x^{3}-15 x^{2}+75 x-50$ is an increasing function for all $x \in R$. Solution: Given:- Function $f(x)=x^{3}-15 x^{2}+75 x-50$ Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$. (i) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- (i) Obtain the function and put it equal to $f(x)$ (ii) Find $f^{\prime}(x)...
Read More →Find all the points on the line x + y = 4 that lie at a unit distance from the
Question: Find all the points on the line x + y = 4 that lie at a unit distance from the line 4x+3y=10. Solution: Given: points lie on the line x + y = 4 , perpendicular distance = 1 units To find : points on the line x + y = 4 Formula used: We know that the distance between a point $(m, n)$ and a line $a x+b y+c$ $=0$ is given by, $D=\frac{|a m+b n+c|}{\sqrt{a^{2}+b^{2}}}$ The equation of the line is $4 x+3 y-10=0$ and $D=1$ units Here $m=x$ and $n=4-x$ (from the equation $x+y=4$ ), $a=4, b=3, ...
Read More →Prove the following
Question: $f(x)=\frac{x-1}{x+1}$, then show that: (i) $f\left(\frac{1}{x}\right)=-f(x)$ (ii) $f\left(-\frac{1}{x}\right)=\frac{-1}{f(x)}$ Solution: (i) $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-1}{\mathrm{x}+1}$ Substituting $\mathrm{x}$ by $1 / \mathrm{x}$, we get $f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}$ $=\frac{\frac{1-x}{x}}{\frac{1+x}{x}}$ $=\frac{1-x}{1+x}$ $=\frac{-(x-1)}{1+x}$ $=-\frac{x-1}{x+1}$ Therefore, We get, $f\left(\frac{1}{x}\right)=-f(x)$ Hence proved (ii) ...
Read More →Show that
Question: Show that $f(x)=x-\sin x$ is increasing for all $x \in R$. Solution: Given:- Function $f(x)=x-\sin x$ Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$. (i) If $f^{\prime}(x)0$ for $a l l x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- (i) Obtain the function and put it equal to $f(x)$ (ii) Find $f^{\prime}(x)$ (iii) Put $f^{\prime}(x)0$ and solve...
Read More →What are the points on the x-axis whose perpendicular distance from the
Question: What are the points on the x-axis whose perpendicular distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is 4 units? Solution: Given: perpendicular distance is 4 units and line $\frac{x}{3}+\frac{y}{4}=1$ To find : points on the x-axis Formula used: We know that the length of the perpendicular from (m,n) to the line ax + by + c = 0 is given by, $D=\frac{|a m+b n+c|}{\sqrt{a^{2}+b^{2}}}$ The equation of the line is $4 x+3 y-12=0$ Any point on the $x$-axis is given by $(x, 0)$ Here $m=x$...
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