Let f ( x ) = 7 x 10 + 9 x 8 ( 1 + x 2 + 2 x 9 ) 2 d x , x 0 , lim x 0 f ( x ) = 0 and f ( 1 ) = 1 4 . If A = [ 0 0 1 1 4 f ( 1 ) 1 2 4 1 ] and B = adj ( adj A ) be such that | B | = 81 , then 2 is equal to [2026]

Question

Let $f (x) = \int \frac{7 x^{10} + 9 x^{8}}{{(1 + x^{2} + 2 x^{9})}^{2}} d x, x > 0,$ $\lim_{x \to 0} f (x) = 0$ and $f (1) = \frac{1}{4}$ .

If $A = [\begin{matrix} 0 & 0 & 1 \\ \frac{1}{4} & f' (1) & 1 \\ α^{2} & 4 & 1 \end{matrix}]$ and $B = adj (adj A)$ be such that $| B |$ $= 81$ , then $α^{2}$ is equal to [2026]

Smart Achievers · Accepted Answer

(2)

$f (x) = \int \frac{(\frac{7}{x^{8}} + \frac{9}{x^{10}})}{{(\frac{1}{x^{9}} + \frac{1}{x^{7}} + 2)}^{2}} d x$

$Put t = \frac{1}{x^{9}} + \frac{1}{x^{7}} + 2 \Rightarrow \frac{d t}{d x} = \frac{- 9}{x^{10}} - \frac{7}{x^{8}}$

$f (x) = \int \frac{- d t}{t^{2}} = \frac{1}{t} + C$

$f (x) = \frac{1}{\frac{1}{x^{9}} + \frac{1}{x^{7}} + 2} + C$

$= \frac{x^{9}}{1 + x^{2} + 2 x^{9}} + C$

$Given f (1) = \frac{1}{4} = \frac{1}{4} + C \Rightarrow C = 0$

$f (x) = \frac{x^{9}}{1 + x^{2} + 2 x^{9}}$

$f' (x) = \frac{(1 + x^{2} + 2 x^{9}) - 9 x^{8} - x^{9} (2 x + 18 x^{8})}{{(1 + x^{2} + 2 x^{9})}^{2}}$

$f' (x) = \frac{36 - 20}{16} = 1$

$A = (\begin{matrix} 0 & 0 & 1 \\ 4 & 1 & 1 \\ α^{2} & \frac{1}{4} & 1 \end{matrix})$

$| A |$ $=$ $| 1 - α^{2} |$ $= 3$

$1 - α^{2} = 3, - 3 \Rightarrow α^{2} = - 2, 4$

$Value of α^{2} = 4$

$B = adj (adj A)$

$| B |$ $= 81 =$ $| A |^{4}$ $\Rightarrow$ $| A |$ $= 3$

Solution

1 1

2 4

3 2

4 3

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Solution

Q. Let f(x)=∫7x10+9x8(1+x2+2x9)2dx, x>0, limx→0f(x)=0 and f(1)=14. If A=[00114f'(1)1α241] and B=adj(adjA) be such that |B|=81, then α2 is equal to [2026]

1 1

2 4

3 2

4 3

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