Most Important Selected Qs for JEE AdvancedMathematicsContinuity and Differentiability
Let a function is defined as f(x)= cases a(1-x x)+b x+5 x^2 & if x 0 cases where a, b, c and d be constants, if f(x) is continuous at x=0 , then: [Note: [k] denotes greatest integer function less than or equal to k .]
Options
- A_ x 0 e^ b x -1 x =-4
- Bnumber of points of discontunity of g(x)=[c-3 a x] in [0, ] is 5 .
- Cthe value of definite integral _ 1+a ^ -b d x x^2+16 = 4 .
- Dthe value of 2 e^d+7 c-3 a-5 b is equal to 29 .
Correct answer
D. the value of 2 e^d+7 c-3 a-5 b is equal to 29 .
Step-by-step solution
f (0⁻ )=f(0)=f (0⁺ )f (0⁻ )=3 _ h 0 a(1-h h)+b h+5 h^2 =3 _ h 0 a [1-h (h- h^3 6 . . ) ]+b (1- h^2 2 ! + h^4 4 ! + . )+5 h^2 =3 _ h 0 (a+b+5)+h^2 (-a- b 2 )+ . h^2 =3a+b+5=0 and a+ 1 2 =-3-3+ b 2 =-5 ; b 2 =-2 b=-4 ; a=-1 Now, f (0⁺ )=3 aligned & _ h 0 (1+ c h+d h^3 h^2 )^ 1 / h =3 & C=0 & _ h 0 (1+d h)^ 1 / h =3 & e^d=3 aligned array ll & d= 3 & 2 e^d+7 c-3 a-5 b=2(3)+7(0)-3(-1)-5(-4)=6+3+20=29 array