Maths
Find the shortest distance between the given lines.
Find the shortest distance between the given lines.
Find the shortest distance between the given lines.
Find the shortest distance between the given lines.
Find the shortest distance between the given lines.
Find the shortest distance between the given lines.
Find the shortest distance between the given lines.
Find the shortest distance between the given lines. $$ \begin{array}{l} \square \vec{r}=(\hat{i}+\hat{j})+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\ \vec{r}=(2 \hat{i}+\hat{j}-\hat{k})+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k}) \end{array} $$
Answer Given equations: r¯=(ı^+ȷ^)+λ(2ı^-ȷ^+k^) \overline{\mathrm{r}}=(\hat{\imath}+\hat{\jmath})+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})...
Find the shortest distance between the given lines. $$ \begin{array}{l} \square \vec{r}=(\hat{i}+\hat{j})+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\ \vec{r}=(2 \hat{i}+\hat{j}-\hat{k})+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k}) \end{array} $$
Answer Given equations: r¯=(ı^+ȷ^)+λ(2ı^-ȷ^+k^) \overline{\mathrm{r}}=(\hat{\imath}+\hat{\jmath})+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})...
If A(1, 2, 3), B(4, 5, 7), C(-4, 3, -6) and D(2, 9, 2) are four given points then find the angle between the lines AB and CD.
Answer Given - A = (1,2,3) B = (4,5,7) C = (-4,3,-6) D = (2,9,2) Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by ((a’-a),(b’-b),(c’-c))...
Show that the lines x = – y = 2z and x + 2 = 2y – 1 = – z + 1 are perpendicular to each other.
To prove – The lines are perpendicular to each other Direction ratios of L1 = (2,-2,1) Direction ratios of L2 = (2,1,-2) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be...
prove that
To find – The value of λ Direction ratios of L1 = (-3,2λ,2) Direction ratios of L2 = (3λ,1,-5) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then...
To prove – The lines are perpendicular to each other Direction ratios of L1 = (2,-3,4) Direction ratios of L2 = (2,4,2) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be...
Find the angle between each of the following pairs of lines:
To find – Angle between the two pair of lines Direction ratios of L1 = (-3,-2,0) Direction ratios of L2 = (1,-3,2) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that...
Find the angle between each of the following pairs of lines:
To find – Angle between the two pair of lines Direction ratios of L1 = (1,0,-1) Direction ratios of L2 = (3,4,5) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of...
Find the angle between each of the following pairs of lines:
To find – Angle between the two pair of lines Direction ratios of L1 = (2,1,-3) Direction ratios of L2 = (3,2,-1) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of...
Find the angle between each of the following pairs of lines:
To find – Angle between the two pair of lines Direction ratios of L1 = (4,3,5) Direction ratios of L2 = (1,-1,1) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of...
Find the angle between each of the following pairs of lines
To find – Angle between the two pair of lines Direction ratios of L1 = (1,1,2) Direction ratios of L2 = (3,5,4) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of...
Find the angle between each of the following pairs of lines:
To find – Angle between the two pair of lines Direction ratios of L1 = (2,-2,1) Direction ratios of L2 = (1,2,-2) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of...
Find the angle between each of the following pairs of lines
Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by
The position vectors of three points A, B and C are
It can thus be written as: A = (-4,2,-3) B = (1,3,-2) C = (-9,1,-4) To prove – A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line...
Find the values of λ and μ so that the points A(-1, 4, -2), B(λ, μ 1) and C(0, 2, -1) are collinear.
Answer Given - A = (-1,4,-2) B = (λ,μ,1) C = (0,2,-1) To find – The value of λ and μ so that A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios...
Find the values of λ and μ so that the points A(3, 2, -4), B(9, 8, -10) and C(λ, μ -6) are collinear.
Answer Given - A = (3,2,-4) B = (9,8,-10) C = (λ,μ,-6) To find – The value of λ and μ so that A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction...
Find the value of λ for which the points A(2, 5, 1), B(1, 2, -1) and C(3, λ, 3) are collinear.
Answer Given - A = (2,5,1) B = (1,2,-1) C = (3,λ,3) To find – The value of λ so that A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the...
Show that the points A(2, 3, -4), B(1, -2, 3) and C(3, 8, -11) are collinear.
Answer Given - A = (2,3,-4) B = (1,-2,3) C = (3,8,-11) To prove – A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by...
Show that the points A(2, 1, 3), B(5, 0, 5) and C(-4, 3, -1) are collinear.
Answer Given - A = (2,1,3) B = (5,0,5) C = (-4,3,-1) To prove – A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by...
Find the image of the point (2, -1, 5) in the line
Find the image of the point $(2,-1,5)$ in the line r→=(11i^-2j^-8k^)+λ(10i^-4j^-11k^) \overrightarrow{\mathrm{r}}=(11 \hat{i}-2 \hat{j}-8 \hat{\mathrm{k}})+\lambda(10 \hat{i}-4...
Find the image of the point (0, 2, 3) in the line
Find the image of the point $(0,2,3)$ in the line $\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}$. Answer Given: Equation of line is...
Find the coordinates of the foot of the perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1).
Find the coordinates of the foot of the perpendicular drawn from the point $A(1,8,4)$ to the line joining the points $B(0,-1,3)$ and $C(2,-3,-1)$ Answer Given: perpendicular drawn from point...
Find the coordinates of the foot of the perpendicular drawn from the point A(1, 2, 1) to the line joining the points B(1, 4, 6) and C(5, 4, 4).
Find the coordinates of the foot of the perpendicular drawn from the point $A(1,2,1)$ to the line joining the points $B(1,4,6)$ and $C(5,4,4)$. Answer Given: perpendicular drawn from point...
Find the vector equation of a line passing through the point having the position vector
Find the vector equation of a line passing through the point having the position vector $(\hat{i}+2 \hat{j}-3 \hat{k})$ and parallel to the line joining the points with position vectors...
Find the vector equation of a line passing through the point A(3, -2, 1) and parallel to the line joining the points B(-2, 4, 2) and C(2, 3, 3). Also, find the Cartesian equations of the line.
Find the vector equation of a line passing through the point $A(3,-2,1)$ and parallel to the line joining the points $\mathrm{B}(-2,4,2)$ and $\mathrm{C}(2,3,3) .$ Also, find the Cartesian equations...
Find the vector and Cartesian equations of the line joining the points whose position vectors are
Find the vector and Cartesian equations of the line joining the points whose position vectors are $(\hat{i}-2 \hat{j}+\hat{k})$ and $(\hat{i}+3 \hat{j}-2 \hat{k})$ Answer Given: line passes through...
Find the vector and Cartesian equations of the line passing through the points A(2, -3, 0) and B(-2, 4, 3).
Find the vector and Cartesian equations of the line passing through the points $A(2,-3,0)$ and $B(-2,4,$, 3). Answer Given: line passes through the points $(2,-3,0)$ and $(-2,4,3)$ To find: equation...
Find the vector and Cartesian equations of the line passing through the points A(3, 4, -6) and B(5, -2, 7).
Find the equations of the line passing through the point $(1,-2,3)$ and parallel to the line $\frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-2}{-4}=\frac{\mathrm{Z}+7}{5}$. Also find the vector form of...
Find the coordinates of the foot of the perpendicular drawn from the point (1, 2, 3) to the line
Find the coordinates of the foot of the perpendicular drawn from the point $(1,2,3)$ to the line $\frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-7}{2}=\frac{\mathrm{z}-7}{-2}$. Also, find the length of the...
Show that the lines and do not intersect each other.
Show that the lines $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{3}=z$ and $\frac{\mathrm{x}+1}{5}=\frac{\mathrm{y}-2}{1}, Z=2$ do not intersect each other. Answer Given: The equations of the two...
Show that the lines and intersect each other. Also, find the point of their intersection.
Show that the lines $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{Z}-3}{4}$ and $\frac{\mathrm{x}-4}{5}=\frac{\mathrm{y}-1}{2}=\mathrm{Z}$ intersect each other. Also, find the point...
Prove that the lines
Prove that the lines $\frac{\mathrm{x}-4}{1}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}+1}{7}$ and $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{-3}=\frac{\mathrm{z}+10}{8}$ intersect each other and find...
Find the Cartesian and vector equations of the line passing through the point (1, 2, -4) and
Find the Cartesian and vector equations of the line passing through the point $(1,2,-4)$ and perpendicular to each of the lines...
Find the equations of the line passing through the point (-1, 3, -2) and perpendicular to each of the lines and
Find the equations of the line passing through the point $(-1,3,-2)$ and perpendicular to each of the lines $\frac{\mathrm{X}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{Z}}{3}$ and...
Find the Cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line
Find the Cartesian and vector equations of a line which passes through the point $(1,2,3)$ and is parallel to the line $\frac{-\mathrm{X}-2}{1}=\frac{\mathrm{y}+3}{7}=\frac{2 Z-6}{3}$ Answer Given:...
Find the equations of the line passing through the point (1, -2, 3) and parallel to the line Also find the vector form of this equation so obtained.
Find the equations of the line passing through the point $(1,-2,3)$ and parallel to the line $\frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-2}{-4}=\frac{\mathrm{Z}+7}{5}$. Also find the vector form of...
Find the Cartesian equations of the line which passes through the point (1, 3, -2) and is parallel to the line Also, find the vector form of the equations so obtained.
Find the Cartesian equations of the line which passes through the point $(1,3,-2)$ and is parallel to the line given by $\frac{\mathrm{x}+1}{3}=\frac{\mathrm{y}-4}{5}=\frac{\mathrm{z}+3}{-6}$. Also,...
The Cartesian equations of a line are 3x + 1 = 6y – 2 = 1 – z. Find the fixed point through which it passes, its direction ratios and also its vector equation.
The Cartesian equations of a line are $3 x+1=6 y-2=1-z$. Find the fixed point through which it passes, its direction ratios and also its vector equation. Answer Given: Cartesian equation of line are...
The Cartesian equations of a line Find the vector equation of the line.
The Cartesian equations of a line are $\frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}+2}{-5}=\frac{\mathrm{Z}-6}{4}$. Find the vector equation of the line. Answer Given: Cartesian equation of line...
A line is drawn in the direction and it passes through a point with position vector Find the equations of the line in the vector as well as Cartesian forms.
A line is drawn in the direction of $(\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})$ and it passes through a point with position vector $(2 \hat{i}-\hat{j}-4 \hat{k}) .$ Find the equations...
Find the vector equation of the line passing through the point with position vector and parallel to the vector Deduce the Cartesian equations of the line.
Find the vector equation of the line passing through the point with position vector $(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-5 \hat{\mathrm{k}})$ and parallel to the vector $(\hat{i}+3...
A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.
A line passes through the point $(2,1,-3)$ and is parallel to the vector $(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$. Find the equations of the line in vector and Cartesian forms....
A line passes through the point (3, 4, 5) and is parallel to the vector Find the equations of the line in the vector as well as Cartesian forms.
Answer Given: line passes through point $(3,4,5)$ and is parallel to $2 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}$ To find: equation of line in vector and Cartesian forms Formula Used: Equation of a...
Let $I=\int _{\pi /6}^{\pi /2}\frac{\cos ecx\cot x}{1+\cos e{{c}^{2}}x}dx=\int _{\pi /6}^{\pi /2}\frac{\cos x}{1+{{\sin }^{2}}x}dx$ Let $\sin x=t$ $\Rightarrow \cos xdx=dt$ Also, when...
Let $I=\int _{1}^{2}\frac{1}{x{{(1+{{\log }_{e}}x)}^{2}}}dx$ Let $1+{{\log }_{e}}x=t$ $\Rightarrow \frac{1}{x}dx=dt$ Also, when $x=1,t=1$and when $x=2$ $t=1+{{\log }_{e}}2$ Hence, $I=\int...
Let $I=\int _{0}^{{{\left( \frac{\pi }{2} \right)}^{\frac{1}{3}}}}{{x}^{2}}\sin \left( {{x}^{3}} \right)dx$ Let ${{x}^{3}}=t$ $\Rightarrow 3{{x}^{2}}=dt$ Also, when $x=0,t=0$and when $t=\frac{\pi...
Let $I=\int _{\pi /4}^{\pi /2}\frac{\cos x}{{{\left( \cos \left( \frac{x}{2} \right)+\sin \left( \frac{x}{2} \right) \right)}^{3}}}dx$ $\cos x={{\cos }^{2}}\left( \frac{x}{2} \right)-{{\sin...
Let $I=\int _{2}^{3}\frac{2-x}{\sqrt{5x-6-{{x}^{2}}}}dx$ Let $2-x=\frac{ad}{dx}\left( 5x-6-{{x}^{2}} \right)+b$ $=-2ax+5a+b$ Hence $-2a=-1$and $5a+b=2$ Solving these equations, We get and...
Let $I=\int _{0}^{\pi /2}\left( \sqrt{\tan x}+\sqrt{\cos tx} \right)dx=\int _{0}^{\pi /2}\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}dx$ Let $\sin x-\cos x=t$ $\Rightarrow \left( \cos x+\sin x...
Let $I=\int _{0}^{1}\frac{1-{{x}^{2}}}{{{(1+{{x}^{2}})}^{2}}}dx$ Let $x=\tan t$ $\Rightarrow dx={{\sec }^{2}}tdt$ Also, when $x=0,t=0$and when $x=1,t=\frac{\pi }{4}$ Hence, $I'=\int _{0}^{\pi...
Let $I=\int _{0}^{1}{{x}^{3}}\sqrt{1+3{{x}^{4}}}dx$ Let $1+3{{x}^{4}}=t$ $\Rightarrow 12{{x}^{3}}dx=dt$ Also, when $x=0,t=1$and when $x=1,t=4$ $I=\frac{1}{12}\int _{1}^{4}\sqrt{t}dt$...
Let $I=\int _{0}^{9}\frac{1}{1+\sqrt{x}}$ Let $\sqrt{x}=u$ $\Rightarrow \frac{1}{2\sqrt{x}}=dx=du$ Or $dx=2udu$ $=\frac{1}{2u}dx$ Also, when $x=0,u=0$and $x=9,u=3$ Hence, $I=\int...
Let $I=\int _{0}^{a}{{\sin }^{-1}}\sqrt{\frac{x}{a+x}}dx$ Let $x=a{{\tan }^{2}}y$ $\Rightarrow dx=2a\tan y{{\sec }^{2}}ydy$ Also, when $x=0,y=0$ And when $x=a$ $y=\frac{\pi }{4}$ Hence, $I=\int...
Let $I=\int _{0}^{1}{{\sin }^{-1}}\sqrt{x}dx$ Let $\sqrt{x}=1$ $\Rightarrow \frac{1}{2\sqrt{x}}=dx=dt$ Or $dx=2tdt$ When $x=0,t=0$ And when $x=1,t=1$ Hence, $I=2\int _{0}^{1}t{{\sin }^{-1}}tdt$...
Let $I=\int _{0}^{1}x{{(a{{\tan }^{-1}}x)}^{2}}dx$ Using integration by parts, we get $I=\frac{{{\left( {{\tan }^{-1}}x \right)}^{2}}{{x}^{2}}}{2}|_{0}^{1}-\int _{0}^{1}\frac{2{{\tan...
Let $I=\int _{0}^{1}{{\left( {{\cos }^{-1}}x \right)}^{2}}dx$ Let $x=\cos t\Rightarrow dx=-\sin tdt$ Also, when $x=0,t=\frac{\pi }{2}$ And when $x=1,t=0$ Hence, $I=-\int _{\pi /2}^{0}{{t}^{2}}\sin...
Let \[I=\int _{\pi /3}^{\pi /2}\frac{\sqrt{1+\cos x}}{(1-\cos x)\frac{5}{2}}dx\] Using $1+\cos x=2\cos {{x}^{2}}\left( \frac{x}{2} \right)$ $1-\cos x=2{{\sin }^{2}}\left( \frac{x}{2} \right)$ we get...
Let $I=\int _{0}^{\pi /2}\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx$ Using $\sin 2x=2\sin x\cos x$, we get $=2\int _{0}^{\pi /2}\frac{\tan x{{\sec }^{2}}x}{({{\tan }^{4}}x+1)}dx$ $I=\int...
Let $I=\int _{0}^{\pi /2}\frac{\sin x\cos x}{{{\cos }^{2}}x+3\cos x+2}dx$ Let $\cos x=t$ $\Rightarrow -\sin xdx=dt$ Also, when $x=0,t=1$ And when $t=0$ $x=\frac{\pi }{2}$ Hence, $I=-\int...
Let $I=\int _{0}^{\pi /4}\frac{{{\tan }^{3}}x}{1+\cos 2x}dx$ Using $1+\cos 2x=2{{\cos }^{2}}x$, we get $I=\frac{1}{2}\int _{0}^{\pi /4}{{\tan }^{3}}x{{\sec }^{2}}xdx$ Let $\tan x=t$ $\Rightarrow...
Let $I=\int _{0}^{\pi }\frac{1}{3+\cos x+2\sin x}dx$ Using $\sin x=\frac{2\tan \left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)}$ $\cos x=\frac{1-{{\tan }^{2}}\left( \frac{x}{2}...
$f(x)$is continuous in $\left[ a,b \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=\left( b-a \right)/n$ $\int...
$f\left( x \right)$is continuous in $\left[ 1,3 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=2/n$ $\int _{1}^{3}\left(...
$f\left( x \right)$is continuous in $\left[ 0,2 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=2/n$ $\int _{0}^{2}\left(...
Since it is modulus function so we need to break the function and then solve it $f\left( x \right)=\int _{0}^{1/3}\left( 1-3x \right)dx+\int _{1/3}^{1}(3x-1)dx$it is continuous in $\left[ 0,1...
$f\left( x \right)$is continuous in $\left[ 0,3 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=3/n$ $\int _{0}^{3}\left(...
$f\left( x \right)$is continuous in $\left[ 0,2 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=2/n$ $\int _{0}^{2}\left(...
$f(x)$is continuous in $\left[ 2,4 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=3/n$ \[\int...
$I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{\cos x+2\sin x}dx$
Let $I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{\cos x+2\sin x}dx$ Using $\sin x=\frac{2\tan \left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)}$ And $\cos x=\frac{1-{{\tan...
$I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{\cos x+2\sin x}dx$
Let $I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{\cos x+2\sin x}dx$ Using $\sin x=\frac{2\tan \left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)}$ And $\cos x=\frac{1-{{\tan...
$I=\int{_{0}^{\pi }}\frac{1}{5+4\cos x}dx$
Let $I=\int{_{0}^{\pi }}\frac{1}{5+4\cos x}dx$ Using, we get $\cos x=\frac{1-{{\tan }^{2}}\left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)}$ $I=\int{_{0}^{\pi...
$I=\int{_{0}^{\pi }}\frac{1}{6-\cos x}dx$
$I=\int{_{0}^{\pi }}\frac{1}{6-\cos x}dx$ Using, we get $\cos x=\frac{1-{{\tan }^{2}}\left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)}$ $I=\int{_{0}^{\pi }}\frac{1-{{\tan...
$I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{5+4\sin x}dx$
Let $I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{5+4\sin x}dx$ $$$\sin x=\frac{2\tan \left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)}$...
$I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{4+9{{\cos }^{2}}x}dx$
Let $I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{4+9{{\cos }^{2}}x}dx$ Dividing by ${{\cos }^{2}}x$ in numerator and denominator, we get $$$I=\int{_{0}^{\frac{\pi }{2}}}\frac{{{\sec }^{2}}x}{2{{\sec...
$I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{1+{{\cos }^{2}}x}dx$
Let $I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{1+{{\cos }^{2}}x}dx$ Dividing by ${{\cos }^{2}}x$in numerator and denominator, we get $I=\int{_{0}^{\frac{\pi }{2}}}\frac{{{\sec...
$I=\int{_{0}^{\frac{\pi }{2}}\frac{1}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx}$
Let $I=\int{_{0}^{\frac{\pi }{2}}\frac{1}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx}$ Dividing by ${{\cos }^{2}}x$in numerator and denominator, we get $I=\int{_{0}^{\frac{\pi...
$I=\int{_{0}^{\frac{\pi }{2}}}\sqrt{1+\sin x}dx$
Let $I=\int{_{0}^{\frac{\pi }{2}}}\sqrt{1+\sin x}dx$ Using and ${{\sin }^{2}}\frac{x}{2}+\cos \frac{x}{2}=1$ $\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$ $I=\int{_{0}^{\frac{\pi }{2}}}\sqrt{{{\left(...
$I=\int{_{0}^{\frac{\pi }{2}}}\sqrt{1+\cos x}dx$
Let $I=\int{_{0}^{\frac{\pi }{2}}}\sqrt{1+\cos x}dx$ Using, we get $1+\cos x=2{{\cos }^{2}}\frac{x}{2}$ $I=\sqrt{2\sin }\left( \frac{x}{2} \right)\left| _{0}^{\frac{\pi }{2}} \right.=2$
$I=\int{_{0}^{1}}{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)dx$
Let $I=\int{_{0}^{1}}{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)dx$ Let $\int{\left( x \right)}={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)$ Let $x=\tan \theta $ $\Rightarrow \theta...
$I=\int{_{0}^{2}x\sqrt{2-xdx}}$
Let $I=\int{_{0}^{2}x\sqrt{2-xdx}}$ Using the property that, we get $\int{_{a}^{b}f(x)dx=\int{_{a}^{b}f(a+b-x)dx}}$ $I=\int{_{0}^{2}(2-x)\sqrt{xdx}}$...
$I=\int{_{0}^{1}\frac{x}{\sqrt{{{a}^{2}}+{{x}^{2}}}}dx}$
Let $I=\int{_{0}^{1}\frac{x}{\sqrt{{{a}^{2}}+{{x}^{2}}}}dx}$ Let ${{a}^{2}}+{{x}^{2}}={{t}^{2}}$ $\Rightarrow xdx=tdt.$ Also when $x=0,t=a$ And when x=a, $t=\sqrt{2a}$ Hence ,...
$I=\int{_{0}^{a}\frac{{{x}^{4}}}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}$
Let $I=\int{_{0}^{a}\frac{{{x}^{4}}}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}$ Let x=a sin t a cos t dt=dx also, when $x=0,t=0$ and when x=a , $t=\frac{\pi }{2}$ Hence, $I=\int{_{0}^{\frac{\pi...
$I=\int{_{0}^{\sqrt{2}}\sqrt{2-{{x}^{2}}}dx}$
Let $I=\int{_{0}^{\sqrt{2}}\sqrt{2-{{x}^{2}}}dx}$ Consider, $I=\int{_{0}^{\sqrt{2}}\sqrt{2-{{x}^{2}}}dx}$ Let x=a sin t $\Rightarrow a\cos ttdt=dx$ Also, when $x=0,t=0$ and when x=a, $t=\frac{\pi...
$I=\int{_{0}^{a}\sqrt{{{a}^{2}}-{{x}^{2}}}dx}$
Let $I=\int{_{0}^{a}\sqrt{{{a}^{2}}-{{x}^{2}}}dx}$ Let x=a sin t $\Rightarrow a\cos tdt=dx$ Also, $x=0,t=0$ And When $x=a,t=\frac{\pi }{2}$ Hence, $I=\int{_{0}^{\frac{\pi...
$I=\int{_{0}^{\frac{\pi }{2}}\frac{\sin x\cos x}{1+{{\sin }^{4}}x}dx}$
$I=\int{_{0}^{\frac{\pi }{2}}\frac{\sin x\cos x}{1+{{\sin }^{4}}x}dx}$ Let ${{\sin }^{2}}x=t$ $\Rightarrow 2\sin x\cos x=dt.$ Also, When $x=0,t=0$ And When $x=\frac{\pi }{2},t=1.$ Hence,...
$I=\int{_{0}^{\frac{\pi }{2}}\sqrt{\sin x{{\cos }^{5}}xdx}}$
Let $I=\int{_{0}^{\frac{\pi }{2}}\sqrt{\sin x{{\cos }^{5}}xdx}}$ Let $\sin x=t$ $\Rightarrow \cos xdx=dt.$ Also, When $x=0,t=0$ And When $x=\frac{\pi }{2},t=1.$ Consider ${{\cos }^{5}}x={{\cos...
$I=\int{_{0}^{\frac{\pi }{2}}\frac{\sin x}{\sqrt{1+\cos x}}dx}$
Let $I=\int{_{0}^{\frac{\pi }{2}}\frac{\sin x}{\sqrt{1+\cos x}}dx}$ Let $1+cosx=t$ $\Rightarrow -\sin xdx=dt.$ Also, when $x=0,t=2$ And When $x=\frac{\pi }{2},t=1$ Hence,...
$I=\int{_{0}^{1}}\sqrt{\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}dx}$
Let $I=\int{_{0}^{1}}\sqrt{\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}dx}$ Let ${{\tan }^{-1}}x=t$ $\Rightarrow \frac{1}{1+{{x}^{2}}}dx=dt$ When $x=1$, $t=\frac{\pi }{4}$ Hence, $I=\int{_{0}^{\frac{\pi...
$f\left( x \right)$is continuous in $\left[ 0,2 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=2/n$ $\int _{0}^{2}\left(...
$f\left( x \right)$is continuous in $\left[ 1,3 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ $\int _{1}^{3}\left(...
$f\left( x \right)$is continuous in $\left[ 1,3 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=3/n$ \[\int...
$f\left( x \right)$is continuous in $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=3/n$ \[\int...
$f\left( x \right)$is continuous in $\left[ 2,5 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=\left( b-a \right)/n$ Here $h=3/n$ $\int...
$f(x)$is continuous in $\left[ 2,5 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+r-h)$, where $h=(b-a)/n$ Here $h=3/n$ $\int...
$f(x)$is continuous in $\left[ 0,3 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=b-a/n$ Here $h=3/n$ \[\int...
$f\left( x \right)$is continuous in $\left[ 1,3 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=\left( b-a \right)/n$ Here $h=2/n$ $\int...
$f(x)$is continuous in $\left[ 1,2 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=1/n$ $\int...
$f(x)$is continuous in $\left[ 0,2 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=2n$ $\int _{0}^{2}\left( x+4...
$y=\int _{0}^{4}\left| x \right|+\left| x-2 \right|+\left| x-4 \right|dx$ $y=\int _{0}^{2}\left| x \right|+\left| x-2 \right|+\left| x-4 \right|dx+\int _{2}^{4}\left| x \right|+\left| x-2...
$y=\int _{0}^{4}f(x)dx$ $y=\int _{0}^{2}f(x)dx+\int _{2}^{4}f(x)dx$ $y=\int _{0}^{2}3{{x}^{2}}+4dx+\int _{2}^{4}9x-2dx$ $y=\left( {{x}^{3}}+4x \right)_{0}^{2}+\left( \frac{9{{x}^{2}}}{2}-2x...
$y=\int _{1}^{3}f(x)dx$ $y=\int _{1}^{2}f(x)dx+\int _{2}^{3}f(x)dx$ $y=\int _{1}^{2}2x+1dx+\int _{2}^{3}{{x}^{2}}+1dx$ $y=\left( {{x}^{2}}+x \right)_{1}^{2}+\left( \frac{{{x}^{3}}}{3}+x...
We know that $\left| \sin x \right|=-\sin x$in $\left[ -\pi /4,0 \right)$ $\left| \sin x \right|=\sin x$in $\left[ 0,\pi /4 \right]$ $y=\int _{-\pi /4}^{0}\left| \sin x \right|dx+\int _{0}^{\pi...
We know that $\left| \cos x \right|=\cos x$in $\left[ 0,\pi /2 \right)$ $\left| \cos x \right|=-\cos x$in $\left[ \pi /2,3\pi /2 \right)$ $\left| \cos x \right|=\cos x$in $\left[ 3\pi /2,2\pi ...
We know that $\left| x-5 \right|=-(x-5)$in $\left[ 0,5 \right)$ $\left| x-5 \right|=(x-5)$in $\left[ 5,8 \right]$ $y=\int _{0}^{5}\left| x-5 \right|dx+\int _{5}^{8}\left| x-5 \right|dx$ $y=\int...
From the differential equation of the family of circles having centers on y-axis and radius $2$ units.
Equation of the family of circles having centers on $y$-axis and radius $2$ units can be represented by $(x)^{2}+(y-a)^{2}=4$, where $a$ is an arbitrary constant. $(y-a)^{2}+x^{2}=4(1)$...
We know that $\left| x+1 \right|=-(x+1)$in $\left[ -2,-1 \right)$ $\left| x+1 \right|=(x+1)$in $\left[ -1,2 \right]$ $y=\int _{-2}^{-1}\left| x+1 \right|dx+\int _{-1}^{2}\left| x+1 \right|dx$...
We know that $\left| x \right|=-x$in $\left[ -1,0 \right)$ $\left| x \right|=x$in $\left[ 0,1 \right]$ $y=\int _{-1}^{0}{{e}^{\left| x \right|}}dx+\int _{0}^{1}{{e}^{\left| x \right|}}dx$ $y=\int...
$y=\int _{-\pi }^{\pi }{{x}^{12}}{{\sin }^{9}}xdx......(1)$ Use king theorem of definite integral $\int _{a}^{b}f(t)dt=\int _{a}^{b}f(a+b-t)dt$ $y=\int _{-\pi }^{\pi }{{\left( \pi -\pi -x...
$y=\int _{-\pi }^{\pi }{{\sin }^{75}}x+{{x}^{125}}dx.....(1)$ Use king theorem of definite integral $\int _{a}^{b}f(t)dt=\int _{a}^{b}f(a+b-t)dt$ $y=\int _{-\pi }^{\pi }{{\sin }^{75}}\left( \pi -\pi...
$y=\int _{-a}^{a}{{x}^{3}}\sqrt{{{a}^{2}}-{{x}^{2}}dx}.....(1)$ Use king theorem of definite integral $\int _{a}^{b}f(t)dt=\int _{a}^{b}f(a+b-t)dt$ $y=\int...
Form the differential equation of the family of all circles touching the $y-axis$ at the origin.
Equation of the family of all circles touching the $y$-axis at the origin can be represented by $(x-a)^{2}+y^{2}=a^{2}$, where $a$ is an arbitrary constants. $(x-a)^{2}+y^{2}=a^{2}(1)$...
Form the differential equation of the family of hyperbolas having foci on the $x-axis$ and centers at the origin.
Equation of the family of an ellipse having foci on the $y$-axis and centers at the origin can be represented by $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $......(1)$ Differentiating the above...
Let $x=\tan t$ $\Rightarrow dx={{\sec }^{2}}tdt$ At $x=0,t=0$ At $x=1,t=\pi /4$ \[y=\int _{0}^{\pi /4}\frac{\log \left( 1+\tan t \right)}{1+{{\tan }^{2}}t}{{\sec }^{2}}tdt\] \[y=\int _{0}^{\pi...
Use integration by parts $\int I\times IIdt=I\int IIdt-\int \frac{d}{dt}I\left( \int IIdt \right)dt$ $y=\log x\int \frac{\sqrt{1}}{\sqrt{1-{{x}^{2}}}}dx-\int \frac{d}{dx}\log x\left( \int...
Form the differential equation of the family of curves given by $x^{2}+y^{2}-\;2ay\;=\;a^{2},$ where a is an arbitrary constant.
Equation of the family of curves, $x^{2}+y^{2}-2 a y=a^{2}$, where $a$ is an arbitrary constant. $\begin{array}{l} x^{2}-2 a x+a^{2}+2 y^{2}=a^{2} \\ x^{2}-2 a x+2 y^{2}=0 \end{array}$...
Let, $x=\sin t$ $\Rightarrow dx=\cos tdt$ At $x=0,t=0$ At $x=1,t=\pi /2$ $y=\int _{0}^{\pi /2}\frac{{{\sin }^{-1}}\sin t}{\sin t}\cos tdt$ $y=\int _{0}^{\pi /2}\frac{t\cos t}{\sin t}dt$ $y=\int...
Form the differential equation of the family of an ellipse having foci on the $y-axis$ and centers at the origin.
Equation of the family of an ellipse having foci on the $y$-axis and centers at the origin can be represented by...
Use integration by parts $\int I\times IIdx=I\int IIdx-\int \frac{d}{dx}I\left( \int IIdx \right)dx$ $y=x\int \cot xdx-\int \frac{d}{dx}x\left( \int \cot xdx \right)dx$ $y=\left( x\log \sin x...
$y=\int _{1}^{4}\frac{\sqrt{x}}{\sqrt{5-x}+\sqrt{x}}dx$
Use king theorem of definite integral $\int _{a}^{b}f(x)=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{1}^{4}\frac{\sqrt{4+1-x}}{\sqrt{4+1-x}+\sqrt{x}}dx$ $y=\int...
$y=\int _{a/4}^{3a/4}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx…..(1)$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{a/4}^{3a/4}\frac{\sqrt{\frac{3a}{4}+\frac{a}{4}-x}}{\sqrt{\frac{3a}{4}+\frac{a}{4}-x}+\sqrt{x}}dx$...
Form the differential equation of the family of curves given by $(x\;-\;a)^{2}+\;2y^{2}=a^{2},$ where $a$ is an arbitrary constant.
Equation of the family of curves, $(x-a)^{2}+2 y^{2}=a^{2}$, where $a$ is an arbitrary constant. $\begin{array}{l} x^{2}-2 a x+a^{2}+2 y^{2}=a^{2} \\ x^{2}-2 a x+2 y^{2}=0 \end{array}$...
Form the differential equation of the family of parabolas having a vertex at the origin and axis along positive $y-axis.$
Equation of the family of parabolas having a vertex at the origin and axis along positive $y$-axis can be represented by $(x)^{2}=4 a y$, where $a$ is an arbitrary constants. $x^{2}=4 a y(1)$...
$y=\int _{\pi /4}^{3\pi /4}\frac{x}{1+\sin x}dx….(1)$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{\pi /4}^{3\pi /4}\frac{\left( \frac{3\pi }{4}+\frac{\pi }{4}-x \right)}{1+\sin \left( \frac{3\pi...
Mark the tick against the correct answer in the following: The vertices of a a $A B C$ are $A(-2,4), B(2,-6)$ and $C(5,4) .$ The area of a $A B C$ is
A. $17.5 \mathrm{sq}$ units
B. 35 sq units
C. 32 sq units
D. $28 \mathrm{sq}$ units
Solution: Option(B) To find: Area of $A B C$ Given: $A(-2,4), B(2,-6)$ and $C(5,4)$ Formula used: $\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\...
$y=\int _{\pi /4}^{3\pi /4}\frac{1}{2{{\cos }^{2}}\frac{x}{2}}dx$
$y=\frac{1}{2}\int _{\pi /4}^{3\pi /4}{{\sec }^{2}}\frac{x}{2}dx$ $y=\frac{1}{2}\left( \frac{\frac{\tan x}{2}}{\frac{1}{2}} \right)_{\pi /4}^{3\pi /4}$ $y=\tan \frac{3\pi }{8}-\tan \frac{\pi }{8}$...
$y=\int _{\pi /6}^{\pi /3}\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{\pi /6}^{\pi /3}\frac{\sqrt{\cos \left( \frac{\pi }{3}+\frac{\pi }{6}-x \right)}}{\left( \sqrt{\sin...
Form the differential equation of the family of circles passing through the fixed point $(a,0)$ and $(-a,0),$ where $a$ is the parameter.
Now, it is not necessary that the centre of the circle will lie on origin in this case. Hence let us assume the coordinates of the centre of the circle be $(0, \mathrm{~h})$. Here, $\mathrm{h}$ is...
Form the differential equation of the family of curves $y^{2}=\;m(a^{2}\;-\;x^{2}),$ where $a$ and $m$ are parameters.
Equation of the family of curves, $y^{2}=m\left(a^{2}-x^{2}\right)$, where $a$ and $m$ are parameters. Differentiating the above equation with respect to $x$ on both sides, we have,...
Mark the tick against the correct answer in the following: The solution set of the equation $\left|\begin{array}{ccc}\mathrm{x} & 3 & 7 \\ 2 & \mathrm{x} & 2 \\ 7 & 6 & \mathrm{x}\end{array}\right|=0$ is
A. $\{2,-3.7\}$
B. $\{2,7 .-9\}$
C. $[-2,3,-7\}$
D. none of these
Solution: Option(C) To find: Value of $x$ We have, $\left|\begin{array}{lll}\mathrm{x} & 3 & 7 \\ 2 & \mathrm{X} & 2 \\ 7 & 6 & \mathrm{x}\end{array}\right|=0$ Applying...
$y=\int _{\frac{\pi }{8}}^{\frac{3\pi }{8}}\frac{\cos x}{\cos x+\sin x}dx…(1)$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{\pi /8}^{3\pi /8}\frac{\cos \left( \frac{3\pi }{8}+\frac{\pi }{8}-x \right)}{\sin \left( \frac{3\pi...
Form the differential equation of the family of circles having centers on the $x-axis$ and radius unity.
Equation of the family of circles having centers on the $x$-axis and radius unity can be represented by $(x-a)^{2}+(y)^{2}=1$, where $a$ is an arbitrary constants. $(x-a)^{2}+y^{2}=1(1)$...
Mark the tick against the correct answer in the following: $\left|\begin{array}{ccc} x+1 & x+2 & x+4 \\ x+3 & x+5 & x+8 \\ x+7 & x+10 & x+14 \end{array}\right|=?$
A. $-2$
B. 2
C. $x^{2}-2$
D. $x^{2}+2$
Solution: Option(A) To find: Value of $\left|\begin{array}{ccc}\mathrm{x}+1 & \mathrm{x}+2 & \mathrm{x}+4 \\ \mathrm{x}+3 & \mathrm{x}+5 & \mathrm{x}+8 \\ \mathrm{x}+7 &...
Form the differential equation of the family of curves $y\;=\;ae^{bx},$ where $a$ and $b$ are arbitrary constants.
Equation of the family of curves, $y=a e^{b x}$, where $a$ and $b$ are arbitrary constants. Differentiating the above equation with respect to $x$ on both sides, we have, $\begin{array}{l} y=a e^{b...
\[y=\int _{0}^{\pi /2}\log \left( \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} \right)dx\]
\[y=\int _{0}^{\pi /2}\log \frac{1}{\sin x\cos x}dx\] \[y=-\left( \int _{0}^{\pi /2}\log \sin xdx+\int _{0}^{\pi /2}\log \cos xdx \right)\] Let, $I=\int _{0}^{\pi /2}\log \sin xdx......(1)$ Use king...
Mark the tick against the correct answer in the following: $\left|\begin{array}{ccc} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{array}\right|=?$
A. $4 \mathrm{abc}$
B. $2(a+b+c)$
C. $(a b+b c+c a)$
D. none of these
Solution: Option(A) To find: Value of $\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|$ We have, $\left|\begin{array}{ccc}b+c & a...
Mark the tick against the correct answer in the following: $\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\mathrm{x} & 1 \\ 1 & 1 & 1+\mathrm{y} \end{array}\right|=?$
A. $(x+y)$
B. $(x-y)$
C. $x y$
D. none of these
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|$ We have, $\left|\begin{array}{ccc}1 & 1 & 1...
Form the differential equation of the family of curves $x\;=\;A \;cos \;nt\;+\;B\;sin\;nt,$ where $A$ and $B$ are arbitrary constants.
Equation of the family of curves, $x=A \cos n t+B \sin n t$, where $A$ and $B$ are arbitrary constants. Differentiating the above equation with respect to $t$ on both sides, we have,...
Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.
Equation of the family of circles in the second quadrant and touching the coordinate axes can be represented by $(x-(-a))^{2}+(y-a)^{2}=a^{2}$, where $a$ is an arbitrary constants....
Mark the tick against the correct answer in the following: $\left|\begin{array}{ccc} x+y & x & x \\ 5 x+4 y & 4 x & 2 x \\ 10 x+8 y & 8 x & 3 x \end{array}\right|=?$
A. 0
B. $x^{3}$
C. $y^{3}$
D. none of these
Solution: Option( B) To find: Value of $\left|\begin{array}{ccc}x+y & x & x \\ 5 x+4 y & 4 x & 2 x \\ 10 x+8 y & 8 x & 3 x\end{array}\right|$ We have,...
Mark the tick against the correct answer in the following: $\left|\begin{array}{ccc} \sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta) \end{array}\right|=?$
A. 0
B. 1
C. $\sin (\alpha+\delta)+\sin (\beta+\delta)+\sin (\gamma+\delta)$
D. none of these
Solution: Option(A) To find: Value of $\left|\begin{array}{lll}\operatorname{sind} & \cos \alpha & \sin (\alpha+\bar{\delta}) \\ \sin \beta & \cos \beta & \sin (\beta+\bar{\delta})...
Form the differential equation of the family of curves, $y\;=\;a\;sin\;(bx+c),$ Where $a$ and $c$ are parameters.
Equation of the family of curves, $y=a \sin (b x+c)$, Where $a$ and $c$ are parameters. Differentiating the above equation with respect to $x$ on both sides, we have, $\begin{array}{l} y=a \sin (b...
Mark the tick against the correct answer in the following: $\left|\begin{array}{lll} 1 ! & 2 ! & 3 ! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 ! \end{array}\right|=?$
A. 2
B. 6
C. 24
D. 120
Solution: Option(24) To find: Value of $\left|\begin{array}{lll}1 ! & 2 ! & 3 ! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 !\end{array}\right|$ We have,...
Mark the tick against the correct answer in the following: If $\omega$ is a complex cube root of unity then the value of $\left|\begin{array}{ccc}1 & \omega & 1+\omega \\ 1+\omega & 1 & \omega \\ \omega & 1+\omega & 1\end{array}\right|$ is
A. 2
B. 4
C. 0
D. $-3$
Solution: Option(0) To find: Value of $\left|\begin{array}{ccc}1 & \omega & 1+\omega \\ 1+\omega & 1 & \omega \\ \omega & 1+\omega & 1\end{array}\right|$ Formula used: (i)...
Form the differential equation of the family of concentric circles x^{2} + y^{2}=a^{2}, where a>0 and a is a parameter.
X^2+y^2=a^2 Differentiating w.r.t x, 2x+2ydx/dy=0. Now as a is eliminated so this is the differential equation of the family of concentric circles. ⇒dx/dy=−y/x
Form the differential equation of the family of straight lines $y\;=\;mx\;+\;c,$ where $m$ and $c$ are arbitrary constants.
The equation of a straight line is represented as, $Y=m x+c$ Differentiating the above equation with respect to $x$, $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}$ Differentiating the above equation...
$\text { Verify that } y=e^{-3 x} \text { is a solution of the differential equation } \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-6 y=0$
As per the given question,
$\text { Verify that } y=\log \left(x+\sqrt{x^{2}+a^{2}}\right) \text { satisfies the differential equation } \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=0$
As per the given question,
$\text { Verify that } y=\frac{c-x}{1+c x} \text { is a solution of the differential equation }\left(1+x^{2}\right) \frac{d y}{d x}+\left(1+y^{2}\right)=0 \text {. }$
As per the given question,
$\text { Verify that } A x^{2}+B y^{2}=1 \text { is a solution of the differential equation } x\left\{y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}\right\}=y \frac{d y}{d x}$
As per the given question,
$\text { Verify that } y=e^{-x}+A x+B \text { is a solution of the differential equation } e^{x}\left(\frac{d^{2} y}{d x^{2}}\right)=1$
As per the given question,
$\text { Verify that } y=\frac{a}{x}+b \text { is a solution of the differential equation } \frac{d^{2} y}{d x^{2}}+\frac{2}{x}\left(\frac{d y}{d x}\right)=0$
As per the given question,
$\text { Verify that } y=a e^{b x} \text { is a solution of the differential equation } \frac{d^{2} y}{d x^{2}}=\frac{1}{y}\left(\frac{d y}{d x}\right)^{2}$
As per the given question,
Verify that $y=c e^{\tan ^{-1} x}$ is a solution of the differential equation $\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x}={0}$
As per the given question,
Verify that $y^{2}\;=\;4a\;(x + a)$ is a solution of the differential equation $y\left\{1-\left(\frac{d y}{d x}\right)^{2}\right\}=2 x \frac{d y}{d x}$
As per the given question,
Show that $y\; =\;e^{x} (A\;cos\;x + B\;sin\;x)$ is the solution of the differential equation $d^{2}y/dx^{2}-2\;dy/dx+2y\;=\;0$
As per the given question,
Verify that $y\;=\;ae^{2x}\;+\;be^{- x}$ is the general solution of the differential equation $d^{2}y/dx^{2}\;-\;dy/dx\;-\;2y=0$
As per the given question,
Verify that $y\;=\;A\;cos\;2x\;-\;B\;sin\;2x$ is the general solution of the differential equation $d^{2}y/dx^{2}\;+\;4y\;=\;0$.
As per the given question,
Verify that $y\;=\;e^{x} \;(A \;cos x \;+ B \;sin x)$ is the general solution of the differential equation $d^{2}y/dx^{2} \;- \;2dy/dx \;+ \;2y \;= \;0$
As per the given question,
Verify that $y\;=\;(a\;+\;bx)\;e^{2x}$ is the general solution of the differential equation $d^{2}y/dx^{2}\;-\;4dy/dx\;+\;4y=\;0$
As per the given question,
Verify that $y\;=\;e^{m\;cos^{-1}x}$ is a solution of the differential equation $(1\;-\;x^{2})d^{2}y/dx^{2}\;-x\;dy/dx\;-\;m^{2}y=0$
As per the given question,
Verify that $y\;=\;e^x\;cos\;bx$ is a solution of the differential equation $d^2 y/dx^2-2\;dy/dx\;+\; 2y=0$
As per the given question,
Verify that $x^2\;=\;2y^2\;log \;y$ is a solution of the differential equation $(x^2\;+\;y^2)\;dy/dx-\;xy\;=\;0$
As per the given question,
$y=\int _{0}^{\pi }\log \left( 1+\cos x \right)dx…..(1)$
Use king theorem of definite integral \[\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx\] \[y=\int _{0}^{\pi }\log (1+\cos (\pi -x))dx\] \[y=\int _{0}^{\pi }\log (1-\cos x)dx........(2)\] Adding...
$y=\int _{0}^{\pi }x\log \sin xdx….(1)$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }\left( \pi -x \right)\log \sin \left( \pi -x \right)dx$ $y=\int _{0}^{\pi }\pi \log \sin...
$y=\int _{0}^{\pi /2}\log \left( 2\sin x\cos x \right)dx$
$y=\int _{0}^{\pi /2}\log 2+\log \sin x+\log \cos xdx$ Let, $I=\int _{0}^{\pi /2}\log \sin xdx.....(1)$ Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)$ $I=\int...
Let$\sin x+\cos x=t$
$\Rightarrow \cos x-\sin xdx=dt$ At $x=0,t=1$ At $x=\pi /2,t=1$ $y=\int _{1}^{1}-\log tdt$ We know that when upper and lower limit in definite integral is equal then value of integration is zero....
$y=\int _{0}^{\pi }{{\sin }^{2m}}x{{\cos }^{2m+1}}xdx…..(1)$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }{{\sin }^{2m}}\left( \pi -x \right){{\cos }^{2m+1}}\left( \pi -x \right)dx$ $y=-\int _{0}^{\pi...
$y=\int _{0}^{\pi }{{\sin }^{2}}x{{\cos }^{3}}xdx…..(1)$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }{{\sin }^{2}}\left( \pi -x \right){{\cos }^{3}}\left( \pi -x \right)dx$ $y=-\int _{0}^{\pi...
$y=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx…….(1)$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{a}\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx.......(2)$ Adding equation (1) and (2) $2y=\int...
Let $x=a\sin t$
$\Rightarrow dx=a\cos tdt$ At $x=0,t=0$ At $x=a,t=0$ $y=\int _{0}^{\pi /2}\frac{a\cos t}{a\sin t+\sqrt{{{a}^{2}}-{{a}^{2}}{{\sin }^{2}}t}}dt$ $y=\int _{0}^{\pi /2}\frac{\cos t}{\sin t+\cos t}dt$...
$y=\int _{0}^{\infty }\frac{x}{(1+x)(1+{{x}^{2}})}dx$
Let, $x=\tan t$ $\Rightarrow dx={{\sec }^{2}}tdt$ At $x=0,t=0$ At $x=\infty ,t=\pi /2$ $y=\int _{0}^{\pi /2}\frac{\tan t}{(1+\tan t)(1+{{\tan }^{2}}t)}{{\sec }^{2}}tdt$ $y=\int _{0}^{\pi...
$y=\int _{0}^{\pi /2}\log \frac{{{\cos }^{2}}x}{\sin 2x}dx$
$y=\int _{0}^{\pi /2}\log \frac{{{\cos }^{2}}x}{2\sin x\cos x}dx$ $y=\int _{0}^{\pi /2}\log \left( \frac{1}{2}\cot x \right)dx.....(1)$ Use king theorem of definite integral $\int...
$y=\int _{0}^{\pi }\frac{x}{1+{{\sin }^{2}}x}dx…..(1)$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }\frac{\left( \pi -x \right)}{1+{{\sin }^{2}}\left( \pi -x \right)}dx$ $y=\int _{0}^{\pi...
Find the value of $k$ for which the points $A(1,-1), B(2, k)$ and $C(4,5)$ are collinear.
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Prove that $\left|\begin{array}{ccc} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{array}\right|=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$
Solution: Operating $\mathrm{C}_{1} \rightarrow \mathrm{aCl}_{1}$ $=\frac{1}{a}\left|\begin{array}{ccc} a^{2} & b-c & c+b \\ a^{2}+a c & b & c-a \\ a^{2}-a b & b+a & c...
Solve the following equations: $\left|\begin{array}{ccc} x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2 \end{array}\right|=0$
Solution: Expanding with R1 $\begin{array}{l} 0=x\left(-3 x^{2}-6 x-2 x^{2}+6 x\right)+6(2 x+4+3 x-9)-1(4 x-9 x) \\ 0=x\left(-5 x^{2}\right)+6(5 x-5)-1(-5 x) \\ 0=-5 x^{3}+30 x-30+5 x \\ 0=-5...
Solve the following equations: $\left|\begin{array}{lll} \mathrm{x} & 3 & 7 \\ 2 & \mathrm{x} & 2 \\ 7 & 6 & \mathrm{x} \end{array}\right|=0$
Solution: Operating $R 1 \rightarrow R 1+R_{2}+R_{3}$ $0=\left|\begin{array}{ccc} x+9 & x+9 & x+9 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right|$ Taking $(x+9)$ common from...
Solve the following equations: $\left|\begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array}\right|=0$
Solution: Operating $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ $0=\left|\begin{array}{ccc} x+9 & 3 & 5 \\ x+9 & x+2 & 5 \\ x+9 & 3 & x+4 \end{array}\right|$ Taking $(x+9)$ common...
Solve the following equations: $\left|\begin{array}{ccc} 3 x-8 & 3 & 3 \\ 3 & 3 x-8 & 3 \\ 3 & 3 & 3 x-8 \end{array}\right|=0$
Solution: $\text { Operating } \mathrm{Cl} \rightarrow \mathrm{Cl}+\mathrm{C}_{2}+\mathrm{C}_{3}$ $0=\left|\begin{array}{ccc} 3 x-8+3+3 & 3 & 3 \\ 3+3 x-8+3 & 3 x-8 & 3 \\ 3+3+3 x-8...
Solve the following equations: $\left|\begin{array}{ccc} x+a & b & c \\ a & x+b & c \\ b & b & x+c \end{array}\right|=0$
Solution: Operating $\mathrm{Cl} \rightarrow \mathrm{Cl}+\mathrm{C}_{2}+\mathrm{C}_{3}$ $\left|\begin{array}{ccc} x+a+b+c & b & c \\ x+a+b+c & x+b & c \\ x+a+b+c & b & x+c...
Solve the following equations: $\left|\begin{array}{lll} 1 & x & x^{3} \\ 1 & b & b^{3} \\ 1 & c & x^{3} \end{array}\right|=0$
Solution: Operating $R 1 \rightarrow R 1^{-} R_{2}, R_{2 \rightarrow R 2}-R_{3}$ $\left|\begin{array}{ccc} 0 & x-b & x^{3}-b^{3} \\ 0 & b-c & b^{3}-c^{3} \\ 1 & c & c^{3}...
Show that $x=2$ is a root of the equation $\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & 2+x\end{array}\right|=0$.
Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}$ $\begin{aligned} 0 &=\left|\begin{array}{ccc} x-2 & -6+3 x & -1-x+3 \\ 2 & -3 x & x-3 \\ -3 & 2 x & 2+x...
Without expanding the determinant, prove that: $\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|=\left|\begin{array}{lll} 1 & b c & b+c \\ 1 & c a & c+a \\ 1 & a b & a+b \end{array}\right|$
Solution: $\text { Operating } R 1 \rightarrow R 1^{-} R_{3}, R_{2 \rightarrow R_{2}-R 3}$ $\begin{array}{l} \left|\begin{array}{ccc} 0 & \mathrm{a}-\mathrm{c} &...
$y=\int _{0}^{\pi }\frac{x\sin x}{1+\sin x}dx……(1)$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }\frac{\left( \pi -x \right)\sin \left( \pi -x \right)}{1+\sin \left( \pi -x \right)}dx$ $y=\int...
Without expanding the determinant, prove that: $\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|$
Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$ $\rightarrow\left|\begin{array}{ccc} 0 & \mathrm{a}-\mathrm{b} & \mathrm{b} c-\mathrm{ac} \\ 0 &...
$t=\int _{0}^{\pi }\frac{x\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\cos x}dx$
$y=\int _{0}^{\pi }\frac{x\sin x}{1+{{\cos }^{2}}x}dx......(1)$ Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }\frac{\left( \pi -x \right)\sin...
$y=\int _{0}^{\pi /2}\frac{{{\cos }^{2}}x}{\sin x+\cos x}dx…..(1)$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi /2}\frac{{{\cos }^{2}}\left( \frac{\pi }{2}-x \right)}{\sin \left( \frac{\pi }{2}-x \right)+\cos...
Prove that $\left|\begin{array}{lll}1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3}\end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)$
Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$ $\begin{array}{l} =\left|\begin{array}{ccc} 0 & \mathrm{a}^{2}+\mathrm{bc}-\mathrm{b}^{2}-\mathrm{ac} &...
$y=\int _{0}^{\pi }\frac{x\tan x}{\sec x\cos ecx}dx…….(1)$
Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }\frac{\left( \pi -x \right)\tan \left( \pi -x \right)}{\sec \left( \pi -x \right)\cos ec\left(...
If $x \neq y \neq z$ and $\left|\begin{array}{lll}x & x^{3} & x^{4}-1 \\ y & y^{3} & y^{4}-1 \\ z & z^{3} & z^{4}-1\end{array}\right|=0$, prove that $x y z(x y+y z+z x)=(x+y+z)$
Solution: By properties of determinants, we can split the given determinant into 2 parts $\rightarrow 0=\left|\begin{array}{lll} \mathrm{x} & \mathrm{x}^{3} & \mathrm{x}^{4} \\ \mathrm{y}...
$y=\int _{0}^{2}x\sqrt{2-x}dx$
$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{2}(2-x)\sqrt{x}dx$ $y=\int _{0}^{2}2{{x}^{\frac{1}{2}}}-{{x}^{\frac{1}{2}}}dx$ $y=\left(...
Using properties of determinants prove that: $\left|\begin{array}{ccc} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|=-2$
Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$ $\begin{array}{l} =\left|\begin{array}{ccc} (a+1)(a+2)-(a+2)(a+3) & a+2-a-3 & 0 \\ (a+2)(a+3)-(a+3)(a+4)...