BITSAT!!

@ Mr.Kalyan
check post #21 by ankit[1]

afta dividin thruout by x do substitue d limit :P

4) If x² − x +1 = 0 then value of x³ⁿ is :

x = -ω , - ω²

(a) -1,1

is correct

oyeeeeee answer toh orally aata hai yeh sab kya hai :P

Is da ans to the 4th Q a??
x is e^i'pi/3 or e^-i'pi/3

x^3 = e^i'pi or e^-i'pi both meaning the same

So x^3n= e^i'npi = 1, -1

yaar itna nautanki kyun karte ho .. u know the ray will pas thro' focus so (a,0) is satisfied .. put it in the 3 options only one is satisfying :)

limitz wala u can also do like first rationalize
so ur q is
lim x→infi (-x)/(x+√x^2+x

so nw divide thruout by x
u get ur answer as -1/2

How can you say that the ray passes through the focus.
This mirror is parabolic, and not spherical.

We need to take the normal at (b²/4a,b)

And make <i = <r

We get the slope of the reflected ray
And equation thru slope point form.of line

\int \frac{xsin^{-1}xdx}{\sqrt{1-x^2}} = \int x \;d\left(\frac{(sin^{-1}x)^2}{2} \right) = x\frac{\left(sin^{-1}x \right)^2}{2}-\int \frac{(sin^{-1})^2}{2}}dx Got the answer....but is this step right (I'm asking for all problems which end up like this)??

Ani ......
yaar do this question b parts

iss mein mushkil kya hai

their method is not wrong but is difficult to understand (u can verify both the results )

take x/√1-x² as the 2nd f(x) and sin^-1x as the 1st f(x)
and substitute x=cos@ and dx=-sin@d@ to solve it and u will reach to ur answer [1]

k wich q goin on....... ?

3)

\int \frac{dx}{sin^6x + cos^6x}

(a)tan^{-1}(tanx-cotx)+ c

(b) -tan^{-1}(-2cot2x)+c

(c) log(tanx-cotx)+c

(d) log(cotx-tanx)+c

i couldn't understand the options for this question:

4) If x² − x +1 = 0 then value of x³ⁿ is :

(a) -1,1

(b)1,0

(c)-2,2

(d)0,2

hey aragon is the equation x²-x+1..[1][1][1].......

oops....sorry yes, prajith.

SORRY EVERYONE

2)
limx-->∞ x-√(x+1/2)²-1/4
now i hope its clear

√(x+1/2)²-1/4 --------> x+1/2 as x tends to ∞

so answer must be -1/2

Sorry mani....i'm weak in limits

Why shd √(x+1/2)²-1/4 ----> ∞ as x ----> ∞ ???

That is the question as provided in the paper

2)

\lim_{x\rightarrow \infty}\left[x-\sqrt{x^2+x} \right]

(a)1/2

(b)1

(c)-1/2

(d)0

is the answer for the question 4......0,2............

3)
HINT :

sin⁶x+cos⁶x=(sin²x)³+(cos²x)³

=1-3sin²xcos²xdx

divide numerator and denominator by cos⁴x
convert it into tan and sec functions
do the required substitutions and i hope u will be able to olve it
in case of doubts
do ask again[1]

itz b or d thats sure...........