If ${log}_{x + 1} (4 x^{3} + 9 x^{2} + 6 x + 1) + {log}_{4 x + 1} (x^{2} + 2 x + 1) = 5,$ then the number of solution is

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${log}_{x + 1} (4 x^{3} + 9 x^{2} + 6 x + 1) + {log}_{4 x + 1} (x^{2} + 2 x + 1) = 5$
We know that,
$x + 1 > 0, x + 1 \neq 1 \Rightarrow x > - 1, x \neq 0 4 x + 1 > 0, 4 x + 1 \neq 1 \Rightarrow x > - \frac{1}{4}, x \neq 0 ∴ x \in (- \frac{1}{4}, \infty) - {0} \dots (1)$

Now,
${log}_{x + 1} (4 x^{3} + 9 x^{2} + 6 x + 1) + {log}_{4 x + 1} (x^{2} + 2 x + 1) = 5 \Rightarrow {log}_{x + 1} [(x + 1) (4 x^{2} + 5 x + 1)] + {log}_{4 x + 1} (x + 1)^{2} = 5 \Rightarrow {log}_{x + 1} [(x + 1)^{2} (4 x + 1)] + 2 {log}_{4 x + 1} (x + 1) = 5 \Rightarrow 2 {log}_{x + 1} (x + 1) + {log}_{x + 1} (4 x + 1) + 2 {log}_{4 x + 1} (x + 1) = 5 \Rightarrow {log}_{x + 1} (4 x + 1) + \frac{2}{{log}_{x + 1} (4 x + 1)} = 3$

Assuming ${log}_{x + 1} (4 x + 1) = t$
$\Rightarrow t + \frac{2}{t} = 3 \Rightarrow t^{2} - 3 t + 2 = 0 \Rightarrow (t - 1) (t - 2) = 0 \Rightarrow t = 1, 2$

Now,
$t = 1 \Rightarrow {log}_{x + 1} (4 x + 1) = 1$
$\Rightarrow 4 x + 1 = x + 1 \Rightarrow x = 0$
Which is not possible [from equation (1)]

$t = 2 \Rightarrow {log}_{x + 1} (4 x + 1) = 2$
$\Rightarrow 4 x + 1 = (x + 1)^{2} \Rightarrow x^{2} - 2 x = 0 \Rightarrow x (x - 2) = 0 \Rightarrow x = 2 (∵ x \neq 0)$

Hence, the number of solution is $1$ .

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\begin{matrix} List - I & List - II (I) & If A and B are two sets with elements 4 and 3, & (P) & 1 respectively. Then the number of onto function from A to B is (II) & The number of points in the interval [- \sqrt{13}, \sqrt{13}] & (Q) & 3 at which f (x) = sin (x^{2}) + cos (x^{2}) attains its maximum value, is (III) & In a Δ X Y Z, y^{2} sin (2 Z) + z^{2} sin (2 Y) = 2 y z, & (R) & 4 where y = 15, z = 8 . Then the length of inradius is (IV) & {log}_{x + 1} (4 x^{3} + 9 x^{2} + 6 x + 1) + {log}_{4 x + 1} (x^{2} + 2 x + 1) & (S) & 8 = 5, then the number of solution is (T) & 24 (U) & 36 \end{matrix}

\begin{matrix} List - I & List - II (I) & If A and B are two sets with elements 4 and 3, & (P) & 1 respectively. Then the number of onto function from A to B is (II) & The number of points in the interval [- \sqrt{13}, \sqrt{13}] & (Q) & 3 at which f (x) = sin (x^{2}) + cos (x^{2}) attains its maximum value, is (III) & In a Δ X Y Z, y^{2} sin (2 Z) + z^{2} sin (2 Y) = 2 y z, & (R) & 4 where y = 15, z = 8 . Then the length of inradius is (IV) & {log}_{x + 1} (4 x^{3} + 9 x^{2} + 6 x + 1) + {log}_{4 x + 1} (x^{2} + 2 x + 1) & (S) & 8 = 5, then the number of solution is (T) & 24 (U) & 36 \end{matrix}

\begin{matrix} List- I & List-II (I) & {log}_{x + 1} (4 x^{3} + 9 x^{2} + 6 x + 1) + {log}_{4 x + 1} (x^{2} + 2 x + 1) & (P) & - 1 = 5, then the number of solution is (II) & If f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {(\frac{3}{4})}^{\frac{tan 4 x}{tan 3 x}}, & 0 < x < \frac{π}{2} a + 2, & x = \frac{π}{2} (1 + | cot x |)^{a b | tan x |}, & \frac{π}{2} < x < π \end{matrix} & (Q) & 0 is continuous at x = \frac{π}{2}, then a + b = (III) & If I = 3 \int - 1 \frac{1}{x^{2}} d x, then | 3 I | = & (R) & 1 (IV) & If x-z plane divides the line joining of the points & (S) & 2 P (3, a, - 4) and Q (2, - \frac{2}{3}, 5) in the ratio 3 : 1, then, a = (T) & 3 (U) & Does not exist \end{matrix}

\sqrt{9 x^{2} + 6 x + 1} < 5 - x

{log}_{(x + 1)} (x^{2} - 2 x) > 0,

x \in

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