14. <math><mrow><mi>f</mi><mrow><mo>(</mo><mrow><mi>x</mi></mrow><mo>)</mo></mrow><mo>=</mo></mrow></math> <math><mrow><mrow><mo>{</mo><mrow><mtable><mtr columnalign="center"><mtd columnalign="center"><mn>2</mn><mi>x</mi><mo>,</mo><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext> if <mi>x</mi><mo><</mo><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext> if <mn>0</mn><mo>≤</mo><mi>x</mi><mo>≤</mo><mn>1</mn></mtd><mtd columnalign="center"><mn>4</mn><mi>x</mi><mo>,</mo><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext> if <mi>x</mi><mo>></mo><mn>1</mn></mtd></mtr></mtable></mrow></mrow></mrow></math>

14. Given f(x) = {2x, if x<00, if 0≤x≤14x, if x>1.For (c) = c < 0,f(c) = 2c.limx→c f(x) = limx→c 2x = 2c = f(c)So, f is continuous at x |<| 0For x = c > 1,f(c) = 4climx→c f(x) = limx→c 4x = 4c = f(c)So, f is continuous at x> 1.For x = 0L.H.L. = limx→0− f(x) = limx→0− . 2x = 2 (0) = 0R.H.L. = limx→0+ f(x) = limx→0+ . 0 = 0.f(0) = 0.∴ L.H.L. = R.H.L. = f(0).So, f is continuous at x = 0.For x = 1.L.H.L. = limx→1− f(x) = limx→1− . 0 = 0R.H.L. = limx→1+ f(x) = limx→1+ . 4x = 4 (1) = 4.∴ L.H.L. = R.H.L.So, f is discontinuous at x = 1.

Ask & Answer Question

Continuity and Differentiability Ncert Solutions Maths class 12th Maths Class 12th Continuity and Differentiability

14. $f (x) =$ ${\begin{matrix} 2 x, if x < 0 & 0, if 0 \leq x \leq 1 & 4 x, if x > 1 \end{matrix}$

5 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

4 months ago

14. Given f(x) = ${\begin{matrix} 2 x, if x < 0 & 0, if 0 \leq x \leq 1 & 4 x, if x > 1 . \end{matrix}$

For (c) = c < 0,

f(c) = 2c.

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 2x = 2c = f(c)

So, f is continuous at x $| < |$ 0

For x = c > 1,

f(c) = 4c

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 4x = 4c = f(c)

So, f is continuous at x> 1.

For x = 0

L.H.L. = $\underset{x \to 0^{-}}{l i m}$ f(x) = $\underset{x \to 0^{-}}{l i m}$ . 2x = 2 (0) = 0

R.H.L. = $\underset{x \to 0^{+}}{l i m}$ f(x) = $\underset{x \to 0^{+}}{l i m}$ . 0 = 0.

f(0) = 0.

∴ L.H.L. = R.H.L. = f(0).

So, f is continuous at x = 0.

For x = 1.

L.H.L. = $\underset{x \to 1^{-}}{l i m}$ f(x) = $\underset{x \to 1^{-}}{l i m}$ . 0 = 0

R.H.L. = $\underset{x \to 1^{+}}{l i m}$ f(x) = $\underset{x \to 1^{+}}{l i m}$ . 4x = 4 (1) = 4.

∴ L.H.L. $=$ R.H.L.

So, f is discontinuous at x = 1.

0 Upvotes 0 Downvotes

Similar Questions for you

Let f and g be twice differentiable even functions on (-2, 2) such that $f (\frac{1}{4}) = 0, f (\frac{1}{2}) = 0, f (1) = 1$ and $f (\frac{3}{4}) = 0, g (1) = 2$

Then, the minimum number of solutions of $f (x) g " (x) + f' (x) g' (x) = 0 i n (- 2, 2)$ is equal to……….

Vishal Baghel

f (x) is an even function

$f (- \frac{1}{4}) = f (- \frac{1}{2}) = f (\frac{1}{2}) = f (\frac{1}{4}) = 0$

So, f (x) has at least four roots in (-2, 2)

$g (\frac{- 3}{4}) = g (\frac{3}{4}) = 0$

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) $\cdot g " (x) = f' (x) \cdot g' (x) = 0$

It is same as number of roots of $\frac{d}{d x} (f (x) \cdot g' (x)) = 0$ will have atleast 4 roots in (-2, 2)

Let f be a real valued continuous function on [0, 1] and f(x) = $x + \int_{0}^{1} (x - t) f (t) d t .$ Then, which of the following points (x, y) lies on the curve y = f(x)?

Vishal Baghel

$f (x) = x + x \int_{0}^{1} f (t) d t - \int_{0}^{1} t_{0} f (t) d t$

Let $1 + \int_{0}^{1} f (t) d t = α$

$\int_{0}^{1} t f (t) d t = β$

So, f(x) = x

Now, $α = \int_{0}^{1} f (t) d t + 1$

$α = \int_{0}^{1} (a t - β) d t + 1$

$β = \int_{0}^{1} t \cdot f (t) d t$

$β = \frac{4}{13}, α = \frac{18}{13}$

f(x) = αx – b

$= \frac{18 x - 4}{13}$

option (D) satisfies

Let f : R -> R be a function defined by f(x) = ${(x - 3)}^{n_{1}} {(x - 5)}^{n_{2}}, n_{1}, n_{2} \in N .$ Then which of the following is NOT true?

alok kumar singh

$f' (x) = \frac{n_{1} \cdot f (x)}{x - 3} + n_{2} \cdot \frac{f (x)}{x - 5}$

$= \frac{f (x) \cdot (n_{1} + n_{2})}{(x - 3) (x - 5)} (x - \frac{(5 n_{1} + 3 n_{2})}{n_{1} + n_{2}})$

$f' (x) = {(x - 3)}^{n_{1} - 1} \cdot {(x - 5)}^{n_{2} - 1} \cdot (n_{1} + n_{2}) (x - \frac{(5 n_{1} + 3 n_{2})}{n_{1} + n_{2 l}})$

option (C) is incorrect, there will be minima.

Let f(x) be non-constant thrice differentiable function defined on $(- \infty, \infty)$ such that f(x) = f(6 – x) and f’(0) = 0 = f’(2) = f’(5). If ‘n’ is the minimum number of roots of (f”(x))² + f(x)f”(x) = 0 in the interval x $\in$ [0, 6] then sum of digits of n equals

Vishal Baghel

f (x) = f (6 – x) Þ f' (x) = -f' (6 – x) …. (1)

put x = 0, 2, 5

f' (0) = f' (6) = f' (2) = f' (4) = f' (5) = f' (1) = 0

and from equation (1) we get f' (3) = -f' (3)

$? f' (3) = 0$

So f' (x) = 0 has minimum 7 roots in $x ? [0, 6] ? f'' (x)$ has min 6 roots in $x ? [0, 6]$

h (x) = f' (x) . f' (x)

h' (x) = (f' (x)² + f' (x) f' (x)

h (x) = 0 has 13 roots in x $?$ [0, 6]

h' (x) = 0 has 12 roots in x $?$ [0, 6]

Number of points of non-differentiability of f(x) = $| | x | - 1 | + | c o s π x |;$ -2 < x < 2 is

alok kumar singh

$I n x \in (- 2, 2)$

|x|- 1| is not differentiable at x = -1, 0, 1

|cospx| is not differentiable at x = $\frac{3}{2}, - \frac{1}{2}, \frac{1}{2}, \frac{3}{2}$ -

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

14. $f (x) =$ ${\begin{matrix} 2 x, if x < 0 & 0, if 0 \leq x \leq 1 & 4 x, if x > 1 \end{matrix}$

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

14. f(x)= {2x, if x<00, if 0≤x≤14x, if x>1

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

14. $f (x) =$ ${\begin{matrix} 2 x, if x < 0 & 0, if 0 \leq x \leq 1 & 4 x, if x > 1 \end{matrix}$