Let f' (x) f(x)f" (x) f'(x)=0 , where f(x) is continuously derivable function with f'(x)≠0 and satisfies f(0)=1 and f'(0)=2, thenLimx→0f(x)−1x is

∴ |f' (x) f(x)f" (x) f'(x)|=0 ⇒ (f' (x))2−f (x)f" (x)=0⇒ (f' (x))2−f (x)f' (x)f2(x)=0 ⇒ −ddx (f' (x)f(x))=0 ⇒ f' (x)f(x)=c (c = constant)Putting x=0, c= f' (0)f(0)=21=2 ∴ In f(x)=2x+k⇒ f(x)=e2xek∴ f(0)=1 ⇒ek=1⇒ f(x)=e2x∴ Limx→0f(x)−1x=Limx→0e2x−1x=2(A) ∵ ∑r=1n4rn2+n+1 2 From sandwich theorem, given limit = 2(C) Given limit =Limn→∞∑r=1nrnn=∫01x dx==(23x3/2)01=23<2 (D) ∵ 2[x]=x+{x}=[x]+{x}+{x}⇒ [x]=2{x}∵ 0≤{x}<1 ⇒ 0≤{x}<2∴[x]=0,1∴{x}=0,12∴x=0,32No of solutions = 2

Let f' (x) f(x)f" (x) f'(x)=0 , where f(x) is continuously derivable function with f'(x)≠0 and satisfies f(0)=1 and f'(0)=2, thenLimx→0f(x)−1x is

∴ |f' (x) f(x)f" (x) f'(x)|=0 ⇒ (f' (x))2−f (x)f" (x)=0⇒ (f' (x))2−f (x)f' (x)f2(x)=0 ⇒ −ddx (f' (x)f(x))=0 ⇒ f' (x)f(x)=c (c = constant)Putting x=0, c= f' (0)f(0)=21=2 ∴ In f(x)=2x+k⇒ f(x)=e2xek∴ f(0)=1 ⇒ek=1⇒ f(x)=e2x∴ Limx→0f(x)−1x=Limx→0e2x−1x=2(A) ∵ ∑r=1n4rn2+n+1 2 From sandwich theorem, given limit = 2(C) Given limit =Limn→∞∑r=1nrnn=∫01x dx==(23x3/2)01=23<2 (D) ∵ 2[x]=x+{x}=[x]+{x}+{x}⇒ [x]=2{x}∵ 0≤{x}<1 ⇒ 0≤{x}<2∴[x]=0,1∴{x}=0,12∴x=0,32No of solutions = 2

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Let $|\begin{array}{l} f' (x) f (x) \\ f " (x) f' (x) \end{array}| = 0$ , where $f (x)$ is continuously derivable function with $f^{'} (x) \neq 0$ and satisfies $f (0) = 1 a n d f^{'} (0) = 2, t h e n \underset{x \to 0}{L i m} \frac{f (x) - 1}{x}$ is

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greater than $\underset{n \to \infty}{L i m} (\frac{\sqrt{1} + \sqrt{2} + \sqrt{3} ...... + \sqrt{n}}{n \sqrt{n}})$

less than $\underset{n \to \infty}{L i m} (\frac{\sqrt{1} + \sqrt{2} + \sqrt{3} ...... + 2 \sqrt{n}}{n \sqrt{n}})$

equal to $\underset{n \to \infty}{L i m} \sum_{r = 1}^{n} \frac{4 r}{n^{2} + r + 1}$

equal to number of solutions of the equation $2 [x] = x + {x}$
[ Note : Where [ ] and { } represent greatest integer and fractional part function respectively.]

answer is A, B, C, D.

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Detailed Solution

$∴ | \begin{array}{l} f' (x) f (x) \\ f " (x) f' (x) \end{array} | = 0 \Rightarrow {(f' (x))}^{2} - f (x) f " (x) = 0$
$\Rightarrow \frac{{(f' (x))}^{2} - f (x) f' (x)}{f^{2} (x)} = 0 \Rightarrow \frac{- d}{d x} (\frac{f' (x)}{f (x)}) = 0$
$\Rightarrow \frac{f' (x)}{f (x)} = c$ (c = constant)
Putting $x = 0, c = \frac{f' (0)}{f (0)} = \frac{2}{1} = 2$
$\begin{array}{l} ∴ I n f (x) = 2 x + k \\ \Rightarrow f (x) = e^{2 x} e^{k} \\ ∴ f (0) = 1 \Rightarrow e^{k} = 1 \Rightarrow f (x) = e^{2 x} \\ ∴ \underset{x \to 0}{L i m} \frac{f (x) - 1}{x} = \underset{x \to 0}{L i m} \frac{e^{2 x} - 1}{x} = 2 \\ (A) ∵ \frac{\sum_{r = 1}^{n} 4 r}{n^{2} + n + 1} < \sum_{r = 1}^{n} \frac{4 r}{n^{2} + r + 1} < \frac{\sum_{r = 1}^{n} 4 r}{n^{2} + 1 + 1} \end{array}$
From sandwich theorem, given limit = 2
$(B) ∵$ Given limit $= \underset{r = 1}{L i m} \sum_{r = 1}^{4 n} \frac{\sqrt{r}}{n \sqrt{r}} = \int_{0}^{4} \sqrt{x} d x$
$= {(\frac{2}{3} x^{3 / 2})}_{0}^{4} = \frac{16}{3} > 2$
From sandwich theorem, given limit = 2
$(C)$ Given limit $= \underset{n \to \infty}{L i m} \sum_{r = 1}^{n} \frac{\sqrt{r}}{n \sqrt{n}} = \int_{0}^{1} \sqrt{x} d x = = {(\frac{2}{3} x^{3 / 2})}_{0}^{1} = \frac{2}{3} < 2$
$(D) ∵ 2 [x] = x + {x}$
$\begin{array}{l} = [x] + {x} + {x} \\ \Rightarrow [x] = 2 {x} \\ ∵ 0 \leq {x} < 1 \Rightarrow 0 \leq {x} < 2 \\ ∴ [x] = 0, 1 \\ ∴ {x} = 0, \frac{1}{2} \\ ∴ x = 0, \frac{3}{2} \end{array}$
No of solutions = 2