Saved Bookmarks
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 7251. |
1+i ka whole sqare/ 3-1 |
| Answer» (1+i)^2/3-1=1+(i)^2 +2×1×i/2=1-1+2i/2=2i/2=i | |
| 7252. |
If acos2a+bsin2a=c prove that tana=2b/a+c |
| Answer» | |
| 7253. |
Define signum function |
| Answer» | |
| 7254. |
find the ratio in the line joining the point(4,4,-10)and(-2,2,4) is divided by x-y plane |
| Answer» | |
| 7255. |
Prove that cos7x-cos8x/1+2cos5x=cos2x-cos3x |
| Answer» | |
| 7256. |
Class 6th mathematics school for n c e r t |
| Answer» | |
| 7257. |
Class 6th mathematics school for ncrt |
| Answer» | |
| 7258. |
2%8 |
| Answer» | |
| 7259. |
Limites and derivative |
| Answer» | |
| 7260. |
tan(πcosA) =cot(πsinA) Prove that cos(A-π/4)=+-1/2√2 |
| Answer» | |
| 7261. |
Find the derivative of f(x) from the first principal, where f(x) is xsinx |
| Answer» Let f(x) =xsinx. =df(x)/dx=limf(x+h)-f(x)/h=lim(x+h)-xsinx/h = lim(x+h)(sinxcosh+cosxsinh)-xsinx/h [•.•sin(x+y)=sinxcosy+cosxsiny]=lim xsinxcosh-xsinx+xcosxsinh+h(sinxcosh+sinhcosx)/h=lim xsinx(cosh-1)+xcosxsinh+h(sinxcosh+sinhcosx)/h. =lim xsinx(cosh-1)/h+lim xcosx sinh/h+lim(sinxcosh+sinhcosx)=xcosx+sinx. Here lim in the complete solution.. h-0 | |
| 7262. |
What is meaning of €or£ |
| Answer» | |
| 7263. |
Find the derivatives of sin (x+1) |
| Answer» Cos(x+1) | |
| 7264. |
lim tanx - sinx÷sin^3xx-0 |
| Answer» | |
| 7265. |
Solve (1+x+x×x)whole power 6 |
| Answer» | |
| 7266. |
If eccentricity of conjugate hyperbola of hyperbola root(x-1)^2 |
| Answer» | |
| 7267. |
Draw quadrilateral |
| Answer» | |
| 7268. |
12 /21 |
| Answer» 4/7... | |
| 7269. |
tan^3x -x/5x-sinx =1/2 |
| Answer» | |
| 7270. |
Solve tan thita + tan 2 thita + tan thita . tan 2 thita = 1 |
| Answer» Hence proved | |
| 7271. |
Sin^[email\xa0protected]/[email\xa0protected] + cos^[email\xa0protected]/[email\xa0protected]@ |
| Answer» | |
| 7272. |
(1+x)\'2n+1 |
| Answer» | |
| 7273. |
Cos20°cos40°cos60°cos80°=1÷16 |
| Answer» | |
| 7274. |
TanA +tan (60°+A) - tan (60°-A) = 3tan 3A prove it |
| Answer» | |
| 7275. |
Find the equation of directrix and axis of parabola of 3x²=8y |
| Answer» x²=4ay3x²=8yx²=8y÷34a=8÷3a=2÷3Symmetric about x -axisEqn of directrix : y= -ay= -2÷3Axis (0,a) (0, 2÷3) | |
| 7276. |
Find limit of x10-1024/x5+32 |
| Answer» | |
| 7277. |
Show that A (a,b,c),B (b,c,a), C (c,a,b) are the vertices of an equilateral triangle |
|
Answer» It\'s very easy to solve this problem.Firstly using distance formula find distance AB,BC&AC. Firstly, find the distances of AB, BC, and AC, then we get the all three distances are equal then, given triangle is equilatral triangle |
|
| 7278. |
Prove that -1=1 |
|
Answer» But it isn\'t possible. Since square of -1=square of 1.so in other words they are equal. |
|
| 7279. |
x^2 + 1=0 what is value of x? |
| Answer» X=i | |
| 7280. |
Find the Value of n such thatnp5=42 np5 |
| Answer» | |
| 7281. |
1+3+9.........3n-1 |
| Answer» Explain question ? | |
| 7282. |
Find the derivate of 10 x from fist principle |
|
Answer» Since F(x)=f(x+h)-f(x)/hTherefore f(x)=10(x+h)-10x/hF(x)= 10x+10h-10x/hF(x)=10h/hF(x)= 10 10 |
|
| 7283. |
(A-B)+A=(A-B)\' |
| Answer» | |
| 7284. |
Derivative of sinx |
|
Answer» d/d(x)sinx= cosx cosx |
|
| 7285. |
Find the coordinates of the foot of thebperpendicular from the point(-1,3) to the line 3x-4y-16=0 |
| Answer» | |
| 7286. |
Derivative of xtanx \\secx±tanx |
| Answer» It should be done by yourself @ ak rocks | |
| 7287. |
find the probabality of 4 digit divisibile by 3 repition is not allowed |
| Answer» | |
| 7288. |
Formula of Sin 6A |
| Answer» | |
| 7289. |
Cos4x=1-8sin"xcos" x. ( " squre) |
| Answer» 4x=2×2x let 2x=∆ now, cos2∆=1-2sin"∆put value of ∆=2xcos4x =1-2(sin 2x)". sin2x =2sinxcosx cos4x =1-2(2sinxcosx)" cos4x =1-2(4sin"xcos"x) cos4x =1-8sin"xcos"x | |
| 7290. |
Reduce eq in normal form3x-4y+4=0 |
|
Answer» Normal form of Line 3x-4y=-4 is , x cos k + y Sin k = p, where p is length of perpendicular from origin to that line and k is angle made by that perpendicular with positive x-axis. As both forms represents same line, Alternately we can consider them as two different coinciding lines. Hence we can write 3/(cos k) = -4/(sin k) = -4/pFrom this equation we can determine cos k and sin k and then value of p by using identity sqr(Sin k) + sqr(cos k)= 1. You can also determine tan k. As per my calculation tan k = -4/3 therefore angle k = tan inverse (-4/3) , angle k will be greater than 90 degree as tan k is negative. Also p = 4/5.. We can directly substitute cos k and sin k values in normal form x cos k + y Sin k = p, without knowing actual value of k. x cos $ + y sin $ = p is normal form of same line 3x - 4y = - 4 , where p is perpendicular distance of line from origin and $ is angle made by that perpendicular with positive X axis. As both equation represents same line we can consider them as two different lines which are "coinciding" . Hence we can write (3/ cos $)=(-4/sin $)=(-4/p) , from these ratios we can find cos $, sin $ and hence tan $ and angle $. p can also determined using identity sqr(sin $) + sqr(cos $) = 1. After knowing value of p and $ , we can substitute in normal form equation.. |
|
| 7291. |
Solve the equation a/ax-1 + b/bx-1= a+b |
| Answer» | |
| 7292. |
Infinity is the arrival or departure number of math. It is true or false |
| Answer» | |
| 7293. |
What is the value of sin 30 + sin 80 |
| Answer» | |
| 7294. |
Sin 10 degree? |
| Answer» | |
| 7295. |
Where is used limit and derivative .sir |
| Answer» In mathematical operations. | |
| 7296. |
\xa0Evaluate lim x->pi/4 sinx-cosx/ x-pi/4 |
| Answer» | |
| 7297. |
Prove that, cos 6x =32cos^6 x -48 cos^4 x +18 cos^2 x -1. |
| Answer» Ftgt | |
| 7298. |
Prove that : Tan x+ tan (x + 10/3) + tan (x +2π/3) =3 |
| Answer» | |
| 7299. |
Application of reflective property of ellipse for light |
| Answer» | |
| 7300. |
What is the pattern of exam..chapter wise marks schedule please |
| Answer» | |