Home
Class 12
MATHS
Let g(x)|f(x+c)f(x+2c)f(x+3c)f(c)f(2c)f(...

Let `g(x)|f(x+c)f(x+2c)f(x+3c)f(c)f(2c)f(3c)f^(prime)(c)f^(prime)(2c)f^(prime)(3c)|,` where `c` is constant, then find `(lim)_(xvec0)(g(x))/x`

Text Solution

Verified by Experts

The correct Answer is:
0
`g(x)=|{:(f(x+c),f(x+2c),f(x+3c)),(f(c),f(2c),f(3c)),(f'(c),f'(2c),f'(3c)):}|`
`therefore" "g(0)=0`
`therefore" "underset(xrarr0)lim(g(x))/(x)" "((0)/(0)from)`
`=underset(xrarr0)lim(g'(x))/(1)" (using L' Hopital rule)"`
`g'(0)`
`"Now, "g'(x)|{:(f'(x+c),f'(x+2c),f'(x+3c)),(f(c),f(2c),f(3c)),(f'(c),f'(2c),f'(3c)):}|`
`therefore" "g'(0)=|{:(f'(c),f'(2c),f'(2c)),(f(c),f(2c),f(3c)),(f'(c),f'(2c),f'(3c)):}|=0`
`therefore" "underset(xrarr0)lim(g(x))/(x)=0`
Promotional Banner

Topper's Solved these Questions

  • DIFFERENTIATION

    CENGAGE|Exercise Exercise 3.8|15 Videos

  • DIFFERENTIATION

    CENGAGE|Exercise Exercise 3.9|14 Videos

  • DIFFERENTIATION

    CENGAGE|Exercise Exercise 3.6|8 Videos

  • DIFFERENTIAL EQUATIONS

    CENGAGE|Exercise Question Bank|7 Videos

  • DOT PRODUCT

    CENGAGE|Exercise DPP 2.1|15 Videos

Similar Questions

Explore conceptually related problems

If g(x)=|(f(x+c),f(x+2c),f(x+3c)),(f(c),f(2c),f(3c)),(f'(c),f'(2c),f'(3c))|, where c is a constant, then lim_(x rarr0)(g(x))/(x) is equal to

If y=f(x^3),z=g(x^5),f^(prime)(x)=tanx ,a n dg^(prime)(x)=secx , then find the value of (lim)_(xvec0)(((dy)/(dz)))/x

Let g(x) be the inverse of an invertible function f(x), which is differentiable for all real xdot Then g^('')(f(x)) equals. (a) -(f^('')(x))/((f^'(x))^3) (b) (f^(prime)(x)f^('')(x)-(f^(prime)(x))^3)/(f^(prime)(x)) (c) (f^(prime)(x)f^('')(x)-(f^(prime)(x))^2)/((f^(prime)(x))^2) (d) none of these

Given ("lim")_(xvec0)(f(x))/(x^2)=2,w h e r e[dot] denotes the greatest integer function, then (a) ("lim")_(xvec0)[f(x)]=0 (b) ("lim")_(xvec0)[f(x)]=1 (c) ("lim")_(xvec0)[(f(x))/x] does not exist (d) ("lim")_(xvec0)[(f(x))/x] exists

If f(x-y),f(x)f(y),a n df(x+y) are in A.P. for all x , y ,a n df(0)!=0, then (a) f(4)=f(-4) (b) f(2)+f(-2)=0 (c) f^(prime)(4)+f^(prime)(-4)=0 (d) f^(prime)(2)=f^(prime)(-2)

Suppose fa n dg are functions having second derivative f'' and g' ' everywhere. If f(x)dotg(x)=1 for all xa n df^(prime)a n dg' are never zero, then (f^('')(x))/(f^(prime)(x))-(g^('')(x))/(g^(prime)(x))e q u a l (a) (-2f^(prime)(x))/f (b) (2g^(prime)(x))/(g(x)) (c) (-f^(prime)(x))/(f(x)) (d) (2f^(prime)(x))/(f(x))

If f is a continuous function on [0,1], differentiable in (0, 1) such that f(1)=0, then there exists some c in (0,1) such that cf^(prime)(c)-f(c)=0 cf^(prime)(c)+cf(c)=0 f^(prime)(c)-cf(c)=0 cf^(prime)(c)+f(c)=0

Let f(x)=|x-1|dot Then (a) f(x^2)=(f(x))^2 (b) f(x+y)=f(x)+f(y) (c) f(|x|)-|f(x)| (d) none of these

Evaluate: inte^x(f(x)+f^(prime)(x))dx=e^xf(x)+C

The function f: RvecR satisfies f(x^2)dotf^(x)=f^(prime)(x)dotf^(prime)(x^2) for all real xdot Given that f(1)=1 and f^(1)=8 , then the value of f^(prime)(1)+f^(1) is 2 b. 4 c. 6 d. 8