SECTION - B [MATHS - NUMERIC] - JEE STD 11 Science Questions

Question 14 Marks

Let $\mathrm{S}=\mathrm{x}: \cos ^{-1} \mathrm{x}=\pi+\sin ^{-1} \mathrm{x}+\sin ^{-1} 2 \mathrm{x}+1$
Then $\sum_{x \in S}(2 x-1)^{2}$ is equal to $\qquad$ -

Answer

(5)
Sol. $\cos ^{-1} x=\pi+\sin ^{-1} x+\sin ^{-1}(2 x+1)$
$2 \cos ^{-1} x-\sin ^{-1}(2 x+1)=\frac{3 \pi}{2}$
$2 \alpha-\beta=\frac{3 \pi}{2}$ where $\cos ^{-1} x=\alpha, \sin ^{-1}(2 x+1)=\beta$
$2 \alpha=\frac{3 \pi}{2}+\beta$
$\cos 2 \alpha=\sin \beta$
$2 \cos ^{2} \alpha-1=\sin \beta$
$2 x^{2}-1=2 x+1$
$\mathrm{x}^{2}-\mathrm{x}-1=0$
$\Rightarrow \mathrm{x}=\frac{1-\sqrt{5}}{2},\left\{\mathrm{x}=\frac{1+\sqrt{5}}{2}\right.$ rejected $\}$
$\therefore 4 \mathrm{x}^{2}-4 \mathrm{x}=4$
$(2 x-1)^{2}=5$

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Question 24 Marks

The number of 6 -letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is 4 __________.

Answer

(1405)
Sol. (i) Single letter is used , then no. of words $=5$
(ii) Two distinct letters are used, then no. of words
${ }^{5} \mathrm{C}_{2} \times\left(\frac{6!}{2!4!} \times 2+\frac{6!}{3!3!}\right)=10(30+20)=500$
(iii) Three distinct letters are used, then no. of words
${ }^{5} \mathrm{C}_{3} \times \frac{6!}{2!2!2!}=900$
Total no. of words $=1405$

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Question 34 Marks

Let $[t]$ be the greatest integer less than or equal to $t$. Then the least value of $p \in N$ for which
$\lim _{x \rightarrow 0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots . .+\left[\frac{p}{x}\right]\right)-x^{2}\left(\left[\frac{1}{x^{2}}\right]+\left[\frac{2^{2}}{x^{2}}\right]+\ldots .+\left[\frac{9^{2}}{x^{2}}\right]\right)\right) \geq 1$
is equal to ___________ .

Answer

(24)
Sol. $\lim _{x \rightarrow 0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots . .+\left[\frac{p}{x}\right]\right)-x^{2}\left(\left[\frac{1}{x^{2}}\right]+\left[\frac{2^{2}}{x^{2}}\right]+\left[\frac{9^{2}}{x^{2}}\right]\right)\right) \geq 1$
$
(1+2+\ldots \ldots+p)-\left(1^{2}+2^{2}+\ldots .9^{2}\right) \geq 1
$
$\frac{\mathrm{p} \mathrm{p}+1}{2}-\frac{9.10 .19}{6} \geq 1$
$\mathrm{p}(\mathrm{p}+1) \geq 572$
Least natural value of $p$ is 24

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Question 44 Marks

Let $S=m \in Z: A^{m^{2}}+A^{m}=3 I-A^{-6}$, where $A=\left[\begin{array}{cc}2 & -1 \\ 1 & 0\end{array}\right]$. Then $n(S)$ is equal to ____________ .

Answer

(2)
Sol. $\quad A=\left[\begin{array}{cc}2 & -1 \\ 1 & 0\end{array}\right]$
$A^{2}=\left[\begin{array}{ll}3 & -2 \\ 2 & -1\end{array}\right], A^{3}=\left[\begin{array}{ll}4 & -3 \\ 3 & -2\end{array}\right], A^{4}=\left[\begin{array}{ll}5 & -4 \\ 4 & -3\end{array}\right]$
and so on
$A^{6}=\left[\begin{array}{ll}7 & -6 \\ 6 & -5\end{array}\right]$
$A^{\mathrm{m}}=\left[\begin{array}{cc}\mathrm{m}+1 & -\mathrm{m} \\ \mathrm{m} & -\mathrm{m}+1\end{array}\right]$,
$A^{m^{2}}=\left[\begin{array}{cc}m^{2}+1 & -m^{2} \\ m^{2} & -\left(m^{2}-1\right)\end{array}\right]$
$A^{m^{2}}+A^{m}=3 I-A^{-6}$
$\left[\begin{array}{cc}m^{2}+1 & -m^{2} \\ m^{2} & -\left(m^{2}-1\right)\end{array}\right]+\left[\begin{array}{cc}m+1 & -m \\ m & -m+1\end{array}\right]$
$=3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]-\left[\begin{array}{ll}-5 & 6 \\ -6 & 7\end{array}\right]$
$=\left[\begin{array}{ll}8 & -6 \\ 6 & -4\end{array}\right]$
$=\mathrm{m}^{2}+1+\mathrm{m}+1=8$
$=\mathrm{m}^{2}+\mathrm{m}-6=0 \Rightarrow \mathrm{~m}=-3,2$
$\mathrm{n}(\mathrm{s})=2$

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Question 54 Marks

Let $\mathrm{f}:(0, \infty) \rightarrow \mathrm{R}$ be a twice differentiable function. If for some $a \neq 0, \int_{0}^{1} \mathrm{f}(\lambda \mathrm{x}) \mathrm{d} \lambda=\operatorname{af}(\mathrm{x})$, $f(1)=1$ and $f(16)=\frac{1}{8}$, then $16-f^{\prime}\left(\frac{1}{16}\right)$ is equal to __________ .

Answer

(112)
Sol. $\int_{0}^{1} f(\lambda x) d \lambda=a f(x)$
$\lambda \mathrm{x}=\mathrm{t}$
$\mathrm{d} \lambda=\frac{1}{\mathrm{x}} \mathrm{dt}$
$\frac{1}{x} \int_{0}^{x} f(t) d t=a f(x)$
$\int_{0}^{x} f(t) d t=\operatorname{axf}(x)$
$\mathrm{f}(\mathrm{x})=\mathrm{a}\left(\mathrm{x} \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})\right)$
$(1-a) f(x)=a . x f^{\prime}(x)$
$\frac{f^{\prime}(x)}{f(x)}=\frac{(1-a)}{a} \frac{1}{x}$
$\operatorname{lnf}(\mathrm{x})=\frac{1-\mathrm{a}}{\mathrm{a}} \operatorname{\ell n} \mathrm{x}+\mathrm{c}$
$\mathrm{x}=\mathrm{l}, \mathrm{f}(1)=\mathrm{l} \Rightarrow \mathrm{c}=0$
$\mathrm{x}=16, \mathrm{f}(16)=\frac{1}{8}$
$\frac{1}{8}=(16)^{\frac{1-a}{a}} \Rightarrow-3=\frac{4-4 a}{a} \Rightarrow a=4$
$f(x)=x^{-\frac{3}{4}}$
$f^{\prime}(x)=-\frac{3}{4} x^{-\frac{7}{4}}$
$\therefore 16-\mathrm{f}^{\prime}\left(\frac{1}{16}\right)$
$=16-\left(-\frac{3}{4}\left(2^{-4}\right)^{-7 / 4}\right)$
$=16+96=112$

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JEE SECTION - B [MATHS - NUMERIC]

SECTION - B [MATHS - NUMERIC]

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