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Its the third option !

this thing has come to ur mind now dear..lolz .

what are the choices ?

shouldn't the area be 10 pie ..cuz its an ellipse with centre ( 4/7 , - 89 / 42 ) ...as the determinant is not equal to zero and ab - h (sq) > 0 .

HAPPY BIRTHDAY INDEED KASI !

Its easy ...in any kind of maximum or minima question ....differentiate the given function f(x) with respect to its parameter ( which is x here ) .
so u get f'(x) .. Now equate f'(x) to zero ..u will get a quadratic equation in cos x .

f(x) = sinx( 1 + cosx)

f'(x) = 2 cos (sq) x + cos x - 1

Now ...f'(x) =1

=> (cos x +1 ) ( 2 cos x - 1 ) =0

so x= 2n pie + 0 ans x = 2n pie + pie / 3 where n belongs to the set of integers (using general solution of cos x = cos alpha )

putting some of the value of n .........u will still find pie / 3 as being the source of greatest value of f(x) ...so just pie / 3 in f(x) and get max value of f(x) .

(3 x root 3) / 4

Just make a right angle triangle ...take the values of the sides as given in the question .....now as u shall know ....

tan theta = V(sq) / R g for a safe turn

=> h / b = V(sq) / R g

=> h = V(sq) b / R g

1st one is more acidic and the second one is more basic .........cuz of presence of methyl gp in the second compound which is quite a strong +I gp and makes the compund more basic, soluble and less tendency to act as a leaving gp .

Certainly the first one .....because of the more -I effect .

yeah its true ...and its +ve only if the rotation is done anticlockwise . Otherwise ....if its done clockwise ...then - ve charge accumulates at the centre and +ve charge accumulates at the rim ....thereby manking the rim be at a higher potential than that of the cetre . In that case the moment will be -ve .

h = V² b / R g

Its 2 pie * R * $\lambda$ * X / [ 4 pie ebsilon knot ( X ² + R²) ]

.........for X >>>R ..it becomes 2 pie * R * $\lambda$ / 4 pie epsilon knot X²

^.........for X<<<R ..it becomes 2 pie * $\lambda$ / 4 pie epsilon knot R

Hii Chandana ...well ....this is natural ..I also used to face such kind of difficulty ..see...suppose u have mass m1 and mass m2 ..and they are tied to the string which passes through a frictionless pulley then ...

the tension in the string is given by T = (2g x m1 x m2 ) / ( m1 + m2 )

the acceleration is given by a = (m2 - m1 ) g / ( m2 + m1 ) ( assuming m2 > m1 )

( if the answer comes negetive ..its all right ...the thing is ..just the supposition of direction is opposite ...your teacher may have a different supposition ....{so take the mod value } )

thrust acting on the pulley ..i.e. reaction at the pulley is given by R = (4 g x m1 x m2 ) / ( m1 + m2 )

BeH2 ...cuz all compounds of Be are covalent only ...Be forms covalent compounds only .