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Most Important Selected Qs for JEE AdvancedMathematicsApplication of Derivatives

Match the following array |c|l|c|l| & Column I & & Column II (A) & The function f(x)=-x^4 e⁻² is strictly decreasing in & (p) & (1 / 3, ) (B) & array l The centre of circle passing through the points (0,0), (1,0) and touching the circle x^2+y^2=9 is array & (q) & (0,4) (C) & array l The real values of a for which the point (-2 a, a+1) lies in the smaller region bounded by the circle x^2+y^2=4 and the parabola y^2=4 x

Options

  1. A( A ) ( q ) ;( B ) ( s ) ;( C ) ( p ) ;( D ) ( r )
  2. B( A ) ( p ) ;( B ) ( q ) ;( C ) ( r ) ;( D ) ( s )
  3. C( A ) ( q,r ) ;( B ) ( r ) ;( C ) ( s ) ;( D ) ( p,r )
  4. D( A ) ( r ) ;( B ) ( p ) ;( C ) ( s ) ;( D ) ( q )

Correct answer

C. ( A ) ( q,r ) ;( B ) ( r ) ;( C ) ( s ) ;( D ) ( p,r )

Step-by-step solution

(A) f^ (x) 0 +x^1 c^ -x -4 x^3 c^ -x 0x^3 e^ -x (x-4) 0 (B) c =0, ~g =- 1 2 as x ^2+ y ^2+2 gx +2 fy + c =0 passes through (0,0),(1,0) and given touches x^2+y^2=9 (C) If the point (-2 a , a +1) lies in shaded region aligned & 4 a^2+(a+1)^2-4 0 and (a+1)^2-4(-2 a) 0 & 5 a^2+2 a-3 0 and a^2+10 a+1 0 a (-1,-5+2 6 ) aligned aligned & (D) 2 - ⁻¹ 6 x 1+9 x^2 =- 2 +2 ⁻¹ 3 x ⁻¹ 6 x 1+9 x^2 = -2 ^1 3 x & ⁻¹ [ 2(3 x) 1+(3 x)^2 ]= -2 ⁻¹ 3 x 3 x 1 aligned

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