$\begin{matrix} List- I & List-II (I) & Number of integral solutions of & (P) & 132 x + y + z = 1, x \geq - 4, y \geq - 4, z \geq - 4 is less than (Q) & 99 (I I) & Greatest term in the expression of \frac{4}{3 \sqrt{2}} {(1 + \frac{1}{\sqrt{2}})}^{12} is (R) & 120 (I I I) & If a_{1}, a_{2}, a_{3} . . . a_{100} are in H.P. then value of \sum_{i = 1}^{99} \frac{a_{i} a_{i + 1}}{a_{1} a_{100}} is - (S) & 100 (I V) & If 8 points out of 11 are in same straight line then number of triangles formed is less then & (T) & 125 \end{matrix}$

Which of the following is only INCORRECT combination?

(I) \to (P) (Q)

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(I I) \to (P)

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(I I I) \to (Q)

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(I V) \to (P) (R) (T)

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Solution

The correct option is A $(I) \to (P) (Q)$
$(I) \to$ $x + y + z = 1 \begin{matrix} x + 4 = x_{1} \geq 0 y + 4 = y_{1} \geq 0 z + 4 = z_{1} \geq 0 \end{matrix} ⎫ ⎪ ⎬ ⎪ ⎭ x_{1} + y_{1} + z_{1} = 13$
So number of solutions $=^{13 + 3 - 1} C_{3 - 1} =^{15} C_{2} = 105$

$(I I) {(1 + \frac{1}{\sqrt{2}})}^{12} \to | T_{r + 1} | \geq | T_{r} |^{12} C_{r} {(\frac{1}{\sqrt{2}})}^{r} \geq^{12} C_{r - 1} {(\frac{1}{\sqrt{2}})}^{r - 1} \frac{12 - r + 1}{r} \geq \sqrt{2} r \leq \frac{13}{\sqrt{2} + 1} r \leq 5.385$
So $r = 5$
$T_{6}$ is numerically greatest term.
$T_{6} =^{12} C_{5} {(\frac{1}{\sqrt{2}})}^{5} ∴ \frac{4}{3 \sqrt{2}} \times^{12} C_{5} \times {(\frac{1}{\sqrt{2}})}^{5} = 132$

$(I I I) \frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, . . . \frac{1}{a_{100}}$ in A.P.
$\frac{1}{a_{2}} - \frac{1}{a_{1}} = \frac{1}{a_{3}} - \frac{1}{a_{2}} = . . . = \frac{1}{a_{100}} - \frac{1}{a_{99}} \frac{a_{1} - a_{2}}{a_{1} a_{2}} = d \frac{1}{a_{1} a_{100}} \sum_{i = 1}^{99} a_{i} a_{i + 1} = \frac{1}{a_{1} a_{100}} (a_{1} a_{2} + a_{2} a_{3} + . . . + a_{99} a_{100}) = \frac{1}{a_{1} a_{100}} (\frac{a_{1} - a_{2}}{d} + \frac{a_{2} - a_{3}}{d} + . . . + \frac{a_{99} - a_{100}}{d}) = \frac{a_{1} - a_{100}}{a_{1} a_{100} \times d}$
Now $\frac{1}{a_{100}} = \frac{1}{a_{1}} + 99 d \Rightarrow d = \frac{a_{1} - a_{100}}{a_{1} a_{100}} \times \frac{1}{99} \Rightarrow \frac{a_{1} - a_{100}}{a_{1} a_{100} \times d} = 99$

$(I V)$ Number of triangles $=^{11} C_{3} -^{8} C_{3} = 109$

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Similar questions

\begin{matrix} List- I & List-II (I) & Number of integral solutions of & (P) & 132 x + y + z = 1, x \geq - 4, y \geq - 4, z \geq - 4 is less than (Q) & 99 (I I) & Greatest term in the expression of \frac{4}{3 \sqrt{2}} {(1 + \frac{1}{\sqrt{2}})}^{12} is (R) & 120 (I I I) & If a_{1}, a_{2}, a_{3} . . . a_{100} are in H.P. then value of \sum_{i = 1}^{99} \frac{a_{i} a_{i + 1}}{a_{1} a_{100}} is - (S) & 100 (I V) & If 8 points out of 11 are in smae straight line then number of triangles formed is less then & (T) & 125 \end{matrix}

Which of the following is only CORRECT combination?

\begin{matrix} List- I & List-II (I) & If {log}_{| cos x |} (1 + sin x) = 2, x \in [- 2 π, 2 π], & (P) & - 2 then the value(s) of x is/are (II) & If max | | x - 4 | - | x - 3 | | = a and & (Q) & - 1 min | x - 4 | + | x - 3 | = b, then a + b = (III) & If f (x^{3} - 6 x^{2} + 11 x - 3) = x - 1, then & (R) & 0 f (3) = (IV) & If x cos θ - y sin θ + z = 1 + cos ϕ & (S) & 1 x sin θ + y cos θ + z = 1 - sin ϕ x cos (θ + ϕ) - y sin (θ + ϕ) = z, then x^{2} + y^{2} + z^{2} = (T) & 2 (U) & 3 \end{matrix}

Which of the following is the only INCORRECT combination?

$\begin{matrix} C o l u m n I & C o l u m n 2 & C o l u m n 3 (I) & f (x) = t a n (2 t a n^{- 1} \sqrt{\frac{\sqrt{1 + \sqrt{x}} - 1}{\sqrt{1 + \sqrt{x}} + 1}}) & (i) & \int f (x) d x = \frac{4}{3} x^{\frac{3}{4}} + C & (P) & \int_{0}^{1} f (x) d x = \frac{4}{3} (I I) & f (x) = c o t (2 t a n^{- 1} \sqrt{\frac{\sqrt{1 + \sqrt{x}} - \sqrt[4]{x}}{\sqrt{1 + \sqrt{x}} + \sqrt[4]{x}}}) & (i i) & \int f (x) d x = \frac{4}{5} x^{\frac{5}{4}} + C & (Q) & \int_{0}^{1} f (x) d x = \frac{4}{5} (I I I) & f (x) ⎛ ⎜ ⎝ \frac{1 - t a n (\frac{1}{2} s i n^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}))}{1 + t a n (\frac{1}{2} s i n^{- 1} (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}))} ⎞ ⎟ ⎠ & (i i i) & \int f (x) d x = \frac{2}{3} x^{\frac{3}{4}} + C & (R) & \int_{0}^{1} f (x) d x = \frac{2}{3} (I V) & f (x) = \sqrt{x} t a n (2 t a n^{- 1} (\frac{\sqrt{\sqrt{1 + \sqrt{x} + 1}} - \sqrt{\sqrt{1 + \sqrt{x} - 1}}}{\sqrt{\sqrt{1 + \sqrt{x} + 1}} + \sqrt{\sqrt{1 + \sqrt{x} - 1}}})) & (i v) & \int f (x) d x = \frac{2}{5} x^{\frac{5}{4}} + C & (S) & \int_{0}^{1} f (x) d x = \frac{2}{5} \end{matrix}$

Which of the following options is only correct combination?

Q. Consider the curve

sin x + sin y = 1

, lying in the first quadrant , then

\begin{matrix} List- I & List-II (I) & lim x \to π / 2 \frac{d^{2} y}{d x^{2}} = & (P) & 0 (II) & lim x \to 0^{+} x^{3 / 2} \frac{d^{2} y}{d x^{2}} = & (Q) & 1 (III) & lim x \to 0^{+} x^{2} \frac{d^{2} y}{d x^{2}} = & (R) & \frac{1}{\sqrt{2}} (IV) & lim x \to π / 2 \frac{d y}{d x} = & (S) & \frac{1}{2 \sqrt{2}} (T) & \sqrt{2} (U) & 3 \end{matrix}

Which of the following is the only INCORRECT combination?

\begin{matrix} List - I & List - II (I) & Number of solutions of the equation & (P) & 0 e^{x} + e^{- x} = tan x \forall x \in [0, \frac{π}{2}) (II) & Number of solutions of the equations & (Q) & 1 x + y = \frac{2 π}{3} and cos x + cos y = \frac{3}{2} is (III) & Number of solutions of the equation & (R) & 2 cos x + 2 sin x = 1, x \in [0, 2 π) is (IV) & Number of solutions of the equation & (S) & Infinite (\sqrt{3} sin x + cos x)^{\sqrt{\sqrt{3} sin 2 x - cos 2 x + 2}} = 4 is \end{matrix}

Which of the following is only INCORRECT combination?

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