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

17. Find the relationship between a and b so that the function $f$ defined by

$f (x) =$ ${\begin{matrix} λ (x^{2} - 2 x), & if x \leq 0 \\ 4 x + 1, & if x > 0 \end{matrix}$ continuous at $x$ = 0? What about continuity at $x$ = 1?

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alok kumar singh | Contributor-Level 10

4 months ago

17. Given, f (x) = ${\begin{matrix} λ (x^{2} - 2 x) & if x | ? | 0 \\ 4 x + 1 & if x > 0 . \end{matrix}$

For continuity at x = 0,

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

$\underset{x \to 0^{-}}{l i m}$ $λ (x^{2} - 2 x)$ = $\underset{x \to 0^{+}}{l i m}$ 4x + 1 = $λ (0^{2} - 2.0)$

0 = 1 = 0 which is not true

Hence, f is not continuous for any value of $λ .$

For x = 1,

$\underset{x \to 1}{l i m}$ f (x) = f (1).

$\underset{x \to 1}{l i m}$ 4x + 1 = 4 (1) + 1

$\Rightarrow$ 4 + 1 = 4 + 1

$\Rightarrow$ 5 = 5.

So, f is continuous at x = 1 value of $λ$

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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

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

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

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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, $α = \int_{0}^{1} f (t) d t + 1$

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option (D) satisfies

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alok kumar singh

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

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

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

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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

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

put x = 0, 2, 5

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

$? f' (3) = 0$

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17. Find the relationship between a and b so that the function $f$ defined by

$f (x) =$ ${\begin{matrix} λ (x^{2} - 2 x), & if x \leq 0 \\ 4 x + 1, & if x > 0 \end{matrix}$ continuous at $x$ = 0? What about continuity at $x$ = 1?

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17. Find the relationship between a and b so that the function f defined by f(x)= {λ(x2−2x), if x≤04x+1, if x>0 continuous at x = 0? What about continuity at x = 1?

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17. Find the relationship between a and b so that the function $f$ defined by

$f (x) =$ ${\begin{matrix} λ (x^{2} - 2 x), & if x \leq 0 \\ 4 x + 1, & if x > 0 \end{matrix}$ continuous at $x$ = 0? What about continuity at $x$ = 1?