हिंदी

If f(x)={mx2+nx<0nx+m0≤x≤1nx3+mx>1 For what integers m and n does limx→0f(x) and limx→1f(x) exist?

Advertisements
Advertisements

प्रश्न

if `f(x) = { (mx^2 + n, x < 0),(nx + m, 0<= x <= 1),(nx^3 + m, x > 1):}`

For what integers m and n does `lim_(x-> 0) f(x)` and `lim_(x -> 1) f(x)` exist?

योग
Advertisements

उत्तर

(i) At x = 0,

`lim_(x → 0^-) f(x) = lim_(x → 0^-) (mx^2 + n) = n`

`lim_(x → 0^+) f(x) = lim_(x → 0^+) (nx + m) = m`

⇒ m = n

(ii) At x = 1,

 `lim_(x → 1^-) f(x) = lim_(x → 1^-) (nx + m) = n + m = 2m, m ∈ R`

 `lim_(x → 1^+) f(x) = lim_(x → 1^+) (nx^3 + m) = n + m = 2m, m ∈ R`

∴ m = n, for n ∈ R

`lim_(x → 0) f(x) = m, m ∈ R`

`lim_(x → 1) f(x) = 2m, m ∈ R`

Hence, for `lim_(x → 0) f(x)` to exist, m = n must be there; `lim_(x → 1) f(x)` exists for any integer values ​​of m and n.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Limits and Derivatives - EXERCISE 12.1 [पृष्ठ २३९]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 11
अध्याय 12 Limits and Derivatives
EXERCISE 12.1 | Q 32. | पृष्ठ २३९

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find `lim_(x -> 1)` f(x), where `f(x) = {(x^2 -1, x <= 1), (-x^2 -1, x > 1):}`


Let a1, a2,..., an be fixed real numbers and define a function f ( x) = ( x − a1 ) ( x − a2 )...( x − an ).

What is `lim_(x -> a_1) f(x)` ? For some a ≠ a1, a2, ..., an, compute `lim_(x -> a) f(x)`


If f(x) = `{(|x| +  1,x < 0), (0, x = 0),(|x| -1, x > 0):}`

For what value (s) of a does `lim_(x -> a)`  f(x) exists?


\[\lim_{x \to 0} \frac{\sqrt{a^2 + x^2} - a}{x^2}\] 


\[\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\] 


\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] 


\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\] 


\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]


\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\] 


\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\] 


\[\lim_{x \to 4} \frac{2 - \sqrt{x}}{4 - x}\]


\[\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}\]


\[\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - 10}\] 


\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\] 

 


\[\lim_{x \to 0} \frac{a^x + a^{- x} - 2}{x^2}\]


\[\lim_{x \to 0} \frac{a^{mx} - 1}{b^{nx} - 1}, n \neq 0\]


\[\lim_{x \to 0} \frac{a^x + b^x - 2}{x}\]


\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\] 


\[\lim_{x \to 0} \frac{a^x + b^x + c^x - 3}{x}\] 


\[\lim_{x \to 2} \frac{x - 2}{\log_a \left( x - 1 \right)}\]


\[\lim_{x \to 0} \frac{a^x + b^ x - c^x - d^x}{x}\]


\[\lim_{x \to 0} \frac{e^x - 1 + \sin x}{x}\]


\[\lim_{x \to 0} \frac{e\sin x - 1}{x}\] 


\[\lim_{x \to 0} \frac{e^{2x} - e^x}{\sin 2x}\]


\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log \left( a - x \right)}{x}\]


\[\lim_{x \to 0} \frac{\log \left( 2 + x \right) + \log 0 . 5}{x}\]


\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]


\[\lim_{x \to 5} \frac{e^x - e^5}{x - 5}\]


`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`


\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\] 


\[\lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x}\] 


\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\] 


`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`


\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]


Write the value of \[\lim_{x \to - \infty} \left( 3x + \sqrt{9 x^2 - x} \right) .\]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×