Advertisements
Advertisements
प्रश्न
Discuss the continuity and differentiability of the
Advertisements
उत्तर
\[\text { Given: } f\left( x \right) = \left| x \right| + \left| x - 1 \right|\]
\[\left| x \right| = - x \text { for } x < 0\]
\[\left| x \right| = x \text { for }x > 0\]
\[\left| x - 1 \right| = - \left( x - 1 \right) = - x + 1 \text { for } x - 1 < 0 \text { or }x < 1\]
\[\left| x - 1 \right| = x - 1 \text { for } x - 1 > 0 \text { or }x > 1\]
Now,
\[f\left( x \right) = - x - x + 1 = - 2x + 1 x \in \left( - 1, 0 \right)\]
or
\[f\left( x \right) = x - x + 1 = 1 x \in \left( 0, 1 \right)\]
or
\[f\left( x \right) = x + x - 1 = 2x - 1 x \in \left( 1, 2 \right)\]
Now,
\[\text { LHL } = \lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^-} - 2x + 1 = 0 + 1 = 1\]
\[\text { RHL } = \lim_{x \to 0^+} f\left( x \right) = \lim_{x \to 0^+} 1 = 1\]
\[\text { Hence, at x = 0, LHL = RHL}\]
Again,
\[\text { LHL } = \lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^-} 1 = 1\]
\[\text { RHL} = \lim_{x \to 1^+} f\left( x \right) = \lim_{x \to 1^+} 2x - 1 = 2 - 1 = 1\]
\[\text { Hence, at x = 1, LHL = RHL}\]
Now,
\[f\left( x \right) = - x - x + 1 = - 2x + 1 x \in \left( - 1, 0 \right)\]
\[ \Rightarrow f'\left( x \right) = - 2 x \in \left( - 1, 0 \right)\]
\[or\]
\[f\left( x \right) = x - x + 1 = 1 x \in \left( 0, 1 \right)\]
\[ \Rightarrow f'\left( x \right) = 0 x \in \left( 0, 1 \right)\]
\[or\]
\[f\left( x \right) = x + x - 1 = 2x - 1 x \in \left( 1, 2 \right)\]
\[ \Rightarrow f'\left( x \right) = 2 x \in \left( 1, 2 \right)\]
Now,
\[\text { LHL }= \lim_{x \to 0^-} f'\left( x \right) = \lim_{x \to 0^-} - 2 = - 2\]
\[\text { RHL }= \lim_{x \to 0^+} f'\left( x \right) = \lim_{x \to 0^+} 0 = 0\]
\[\text { Since, at x = 0, LHL } \neq \text { RHL}\]
\[\text { Hence,} f\left( x \right) \text { is not differentiable at } x = 0\]
Again,
\[\text { LHL }= \lim_{x \to 1^-} f'\left( x \right) = \lim_{x \to 1^-} 0 = 0\]
\[\text { RHL } = \lim_{x \to 1^+} f'\left( x \right) = \lim_{x \to 1^+} 2 = 2\]
\[\text { Since, at } x = 1, \text { LHL } \neq \text { RHL }\]
\[\text { Hence }, f\left( x \right) \text { is not differentiable at } x = 1\]
APPEARS IN
संबंधित प्रश्न
Differentiate the function with respect to x.
sin (ax + b)
Differentiate the function with respect to x.
`(sin (ax + b))/cos (cx + d)`
Differentiate the function with respect to x.
`2sqrt(cot(x^2))`
Differentiate the function with respect to x.
`cos (sqrtx)`
Differentiate the function with respect to x:
(3x2 – 9x + 5)9
Differentiate the function with respect to x:
sin3 x + cos6 x
Differentiate the function with respect to x:
`(cos^(-1) x/2)/sqrt(2x+7)`, −2 < x < 2
Differentiate the function with respect to x:
`x^(x^2 -3) + (x -3)^(x^2)`, for x > 3
Find `dy/dx`, if y = 12 (1 – cos t), x = 10 (t – sin t), `-pi/2 < t < pi/2`.
If f(x) = |x|3, show that f"(x) exists for all real x and find it.
If sin y = xsin(a + y) prove that `(dy)/(dx) = sin^2(a + y)/sin a`
`"If y" = (sec^-1 "x")^2 , "x" > 0 "show that" "x"^2 ("x"^2 - 1) (d^2"y")/(d"x"^2) + (2"x"^3 - "x") (d"y")/(d"x") - 2 = 0`
Differential coefficient of sec (tan–1x) w.r.t. x is ______.
If u = `sin^-1 ((2x)/(1 + x^2))` and v = `tan^-1 ((2x)/(1 - x^2))`, then `"du"/"dv"` is ______.
| COLUMN-I | COLUMN-II |
| (A) If a function f(x) = `{((sin3x)/x, "if" x = 0),("k"/2",", "if" x = 0):}` is continuous at x = 0, then k is equal to |
(a) |x| |
| (B) Every continuous function is differentiable | (b) True |
| (C) An example of a function which is continuous everywhere but not differentiable at exactly one point |
(c) 6 |
| (D) The identity function i.e. f (x) = x ∀ ∈x R is a continuous function |
(d) False |
`sin sqrt(x) + cos^2 sqrt(x)`
sinn (ax2 + bx + c)
`tan^-1 (sqrt((1 - cosx)/(1 + cosx))), - pi/4 < x < pi/4`
`tan^-1 (secx + tanx), - pi/2 < x < pi/2`
`tan^-1 (("a"cosx - "b"sinx)/("b"cosx - "a"sinx)), - pi/2 < x < pi/2` and `"a"/"b" tan x > -1`
`sec^-1 (1/(4x^3 - 3x)), 0 < x < 1/sqrt(2)`
`tan^-1 ((sqrt(1 + x^2) + sqrt(1 - x^2))/(sqrt(1 + x^2) - sqrt(1 - x^2))), -1 < x < 1, x ≠ 0`
For the curve `sqrt(x) + sqrt(y)` = 1, `"dy"/"dx"` at `(1/4, 1/4)` is ______.
If k be an integer, then `lim_("x" -> "k") ("x" - ["x"])` ____________.
If `"f"("x") = ("sin" ("e"^("x"-2) - 1))/("log" ("x" - 1)), "x" ne 2 and "f" ("x") = "k"` for x = 2, then value of k for which f is continuous is ____________.
A function is said to be continuous for x ∈ R, if ____________.
If `y = (x + sqrt(1 + x^2))^n`, then `(1 + x^2) (d^2y)/(dx^2) + x (dy)/(dx)` is
`d/(dx)[sin^-1(xsqrt(1 - x) - sqrt(x)sqrt(1 - x^2))]` is equal to
If sin y = x sin (a + y), then value of dy/dx is
Let c, k ∈ R. If f(x) = (c + 1)x2 + (1 – c2)x + 2k and f(x + y) = f(x) + f(y) – xy, for all x, y ∈ R, then the value of |2(f(1) + f(2) + f(3) + ... + f(20))| is equal to ______.
A particle is moving on a line, where its position S in meters is a function of time t in seconds given by S = t3 + at2 + bt + c where a, b, c are constant. It is known that at t = 1 seconds, the position of the particle is given by S = 7 m. Velocity is 7 m/s and acceleration is 12 m/s2. The values of a, b, c are ______.
If f(x) = `{{:((sin(p + 1)x + sinx)/x,",", x < 0),(q,",", x = 0),((sqrt(x + x^2) - sqrt(x))/(x^(3//2)),",", x > 0):}`
is continuous at x = 0, then the ordered pair (p, q) is equal to ______.
If f(x) = `{{:(ax + b; 0 < x ≤ 1),(2x^2 - x; 1 < x < 2):}` is a differentiable function in (0, 2), then find the values of a and b.
If f(x) = `{{:(x^2"," if x ≥ 1),(x"," if x < 1):}`, then show that f is not differentiable at x = 1.
Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.
