I think u shud do ARIHANT DIFFERENTIAL CALCULUS, it is good for present day level of JEE

Just Apply the L-Hospital rule twice

  by applying it once, u get

 Lim_{x-> 0}( - (m+2) sin(m+2)x  + msinx)/ (- (m+4)sin(m+4)x + (m+2) sin(m+2)x )

again applying L-Hospital,

Lim_{x-> 0}(-(m+2)²cos(m+2)x + m²cosmx )/ ( -(m+4)²cos(m+4)x + (m+2)²cos(m+2)x)

Putting limits, we get

(m² - (m+2)²)/( (m+2)² - (m+4)² )

=>  (m+1)/ ( m+3)

Rate me..........

wats ur problem.....................

 (99)¹⁰⁰ > (100)⁹⁹

Because,

(100)⁹⁹ = (99+1)⁹⁹

= ⁹⁹C₀ (99)⁹⁹ + ⁹⁹C₁(99)⁹⁸ + ....................... + 1

= (99)⁹⁹ + (99).(99)⁹⁸ +............... + 1

= 2.(99)⁹⁹ + terms relatively much smaller terms

and this is much smaller than 99.(99)⁹⁹ i.e (99)¹⁰⁰

You willl top JEE

is it 2⁹ i.e 512....................

In BITSAT there are 150 ques. P-40, C-40, M-45, Eng& logical-25,

Each ques is of three marks

but if u attempt all ( without leaving any), you will be given 12 additional ques. if time permits,

means u can score even 486 out of 450, 36 are bonus..........

In this

f(x) = 1+ 2x² + 2²x⁴ +.................... + 2¹⁰x²⁰

f¹(x) = 4x + 4.2²x³ +................+ 20.2¹⁰x¹⁹

       = x(4 +4.2²x² + ...............+ 20.2¹⁰x¹⁹)

for f¹(x)=0

either x=0

or

4 + 4.2²x² +.................. + 20.2¹⁰x¹⁹ =0,

This equation has all terms with even powers of x and also all coff. are positive, thats why it has all imaginary roots.

therefore, there is only one val x=0 for which there may be max or min,

Now, at x=0 F²(x)= 4,

that's why there is only one minima..........