We have, ${e^{ – 1}} = 1 + ( – 1) + \frac{{{{( – 1)}^2}}}{{2!}} + \frac{{{{( – 1)}^3}}}{{3!}} + \frac{{{{( – 1)}^4}}}{{4!}} + ……….$
The given integral = $\int {{{({e^{ – 1}})}^x}dx} = \int {{e^{ – x}}dx} = – {e^{ – x}} + C$
Area Bounded by Tangent to $xy=c^2$ & x-y Axis
The area of triangle OAB bounded by the tangent to $xy=c^2$ in the 1st quadrant, x-axis & y-axis is (refer figure):
(1) Maximum if P is the midpoint of AB
(2) Increases as P moves downwards or upwards
(3) Constant
(4) Independent of c
Solution
$xy=c^2$ is rectangular hyperbola. The equation of tangent in parametric form at some point P $(ct, \frac {c}{t})$ is given by,
$\frac {x}{t}+yt=2c$
At point A, $x=2ct=OA$
At point B, $y=\frac {2c}{t}=OB$
Area of $\Delta OAB$ = $\frac {1}{2}.OA.OB$ = $\frac {1}{2}.2ct.\frac {2c}{t}$ = $2c^2$
Hence, Option (3).
2x-5y+z=3, x+y+4z=5, x+3y+6z=1
The equation of the plane containing the line 2x-5y+z=3, x+y+4z=5 and parallel to the plane x+3y+6z=1 is
(1) x+3y+6z=-7
(2) x+3y+6z=7
(3) 2x+6y+12z=-13
(4) 2x+6y+12z=13
Solve for $x \in \left[ {0,\frac{\pi }{2}} \right]$$8{\sin ^2}x + 2\cos 3x + 6\cos x = 7$
We have, $8(1 – {\cos ^2}x) + 2(4{\cos ^3}x – 3\cos x) + 6\cos x = 7$
$ \Rightarrow 8{\cos ^3}x – 8{\cos ^2}x + 1 = 0$
Let $cos x = t$
So, $8{t^3} – 8{t^2} + 1 = 0$ Continue reading Solve for $x \in \left[ {0,\frac{\pi }{2}} \right]$
$8{\sin ^2}x + 2\cos 3x + 6\cos x = 7$
Prove That,$\cos \frac{\pi }{7}-\cos \frac{{2\pi }}{7}+\cos \frac{{3\pi}}{7}=\frac{1}{2}$
L.H.S.=$\cos \frac{\pi }{7} – \cos \frac{{2\pi }}{7} + \cos \frac{{3\pi }}{7}$
= $\cos \frac{\pi }{7} + \cos \frac{{3\pi }}{7} – \cos \left( {\pi – \frac{{5\pi }}{7}} \right)$
= $\cos \frac{\pi }{7} + \cos \frac{{3\pi }}{7} + \cos \frac{{5\pi }}{7}$
= $\cos \left( {\frac{\pi }{7} + \frac{{3 – 1}}{2}.\frac{{2\pi }}{7}} \right).\frac{{\sin \left( {3.\frac{{2\pi /7}}{2}} \right)}}{{\sin \left( {\frac{{2\pi /7}}{2}} \right)}}$ Continue reading Prove That,
$\cos \frac{\pi }{7}-\cos \frac{{2\pi }}{7}+\cos \frac{{3\pi}}{7}=\frac{1}{2}$
Ladder, Cube & Wall Problem
A ladder of length $l=2\sqrt 6$ unit is resting against a wall just touching the cube of edge 1 unit as shown in the figure. Find h above the cube.
Solution
We have, $\tan \theta = h$ and $\sin \theta = \frac{{h + 1}}{l}$
So, $2\sqrt 6 \sin \theta = 1 + \tan \theta $ Continue reading Ladder, Cube & Wall Problem
$\frac{{{x^4} – {x^2}}}{{{x^6} + 2{x^3} – 1}}$
$\begin{array}{l}y = \frac{{{x^4} – {x^2}}}{{{x^6} + 2{x^3} – 1}},x > 1\\{y_{\max }} = ?\end{array}$