MCQ

MCQ 11 Mark

The minimum value of the expression $3^x+3^{1-x}$, $x \in R$, is

A
$0$
B
$\frac{1}{3}$
C
3
✓
$2 \sqrt{3}$

Answer

Correct option: D.

$2 \sqrt{3}$

(D) $2 \sqrt{3}$
Explanation : We know that, A.M. $\geq$ G.M. for positive numbers.
$
\begin{array}{ll}
\text { Therefore, } & \frac{3^x+3^{1-x}}{2} \geq \sqrt{3^x \cdot 3^{1-x}} \\
\Rightarrow & \frac{3^x+3^{1-x}}{2} \geq \sqrt{3^x \cdot \frac{3}{3^x}} \\
\Rightarrow & 3^x+3^{1-x} \geq 2 \sqrt{3}
\end{array}
$

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MCQ 21 Mark

Let $S$ be the sum, $P$ be the product and $R$ be the sum of the reciprocals of 3 terms of a G.P. then $P^2 R^3$ : $S^3$ in equal to

✓
$1: 1$
B
(common ratio) ${ }^n: 1$
C
$(\text { (first term })^2:(\text { common ratio })^2$
D
none of these

Answer

Correct option: A.

$1: 1$

(A) $1: 1$
Explanation : Let the three terms of G.P. be $\frac{a}{r}, a, a r$.
Then,
$
\begin{aligned}
S & =\frac{a}{r}+a+a r=\frac{a\left(r^2+r+1\right)}{r} \\
P & =a^3, R=\frac{r}{a}+\frac{1}{a}+\frac{1}{a r} \\
& =\frac{1}{a}\left(\frac{r^2+r+1}{r}\right) \\
\frac{P^2 R^3}{S^3} & =\frac{a^6 \cdot \frac{1}{a^3}\left(\frac{r^2+r+1}{r}\right)^3}{a^3\left(\frac{r^2+r+1}{r}\right)^3}=1
\end{aligned}
$
Therefore, the ratio is $1: 1$.

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MCQ 31 Mark

In a G.P. of even number of terms, the sum of all terms is 5 times the sum of the odd terms. The common ratio of the G.P. is

A
$\frac{-4}{5}$
B
$\frac{1}{5}$
✓
4
D
none of these

Answer

Correct option: C.

(C) 4
Explanation : Let us consider a G.P. $a, a r, a r^2, \ldots . .$. with $2 n$ terms
We have,
$\frac{a\left(r^{2 n}-1\right)}{r-1}=\frac{5 a\left[\left(r^2\right)^n-1\right]}{r^2-1}$
(Since common ratio of odd terms will be $r^2$ and number of terms will be $n$ )
$
\begin{aligned}
\Rightarrow& & \frac{a\left(r^{2 n}-1\right)}{(r-1)} & =5 \frac{a\left(r^{2 n}-1\right)}{\left(r^2-1\right)} \\
\Rightarrow & & a(r+1) & =5 a \\
\Rightarrow & & r & =4
\end{aligned}
$

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MCQ 41 Mark

In a G.P. of positive terms, if any term in equal to the sum of the next two terms. Then the common ratio of the G.P. is

✓
$\frac{(\sqrt{5}-1)}{2}$
B
$\frac{\sqrt{5}+1}{2}$
C
$\frac{-\sqrt{5}-1}{2}$
D
$\frac{ \pm \sqrt{5}-1}{2}$

Answer

Correct option: A.

$\frac{(\sqrt{5}-1)}{2}$

(A) $\frac{(\sqrt{5}-1)}{2}$
Explanation : Given,
$
\begin{array}{rlrl}
t_n & =t_{n+1}+t_{n+2} \\
\Rightarrow & a r^{n-1} =a r^n+a r^{n+1} \\
\Rightarrow & 1 =r+r^2 \\
\Rightarrow & r^2+r-1 =0 \\
\therefore & r =\frac{-1 \pm \sqrt{5}}{2}
\end{array}
$
Since, $r>0$ therefore, $r=\frac{\sqrt{5}-1}{2}$

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MCQ 51 Mark

If the sum of $n$ terms of an A.P. is given by $S_n=3 n$ $+2 n^2$, then the common difference of the A.P. is

A
3
B
2
C
6
✓
4

Answer

Correct option: D.

(D) 4
Explanation : Given,
$
\begin{aligned}
Given,\quad S_n & =3 n+2 n^2 \\
S_1 & =3(1)+2(1)^2=5 \\
S_2 & =3(2)+2(2)^2=14 \\
Since, \quad S_1 & =a_1=5 \\
and\quad S_2-S_1 & =a_2
\end{aligned}
$
$
\therefore \quad a_2=14-5=9
$
Hence, common difference, $d=a_2-a_1=9-5=4$.
Note: If $S_n=a n^2+b n$ where $a$ and $b$ are constant then common difference of an A.P. $=2 \times$ Coefficient of $x^2$

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MCQ 61 Mark

The $10^{\text {th }}$ common term between the series $3+$ $7+11+$ _________________ and $1+6+11+\ldots$. Is

✓
191
B
193
C
211
D
None of these

Answer

Correct option: A.

191

(A) 191
Explanation : The first common term is 11 . Now, the next common term is obtained by adding LCM of the common difference of 4 and 5, i.e,. 20 . Therefore, $10^{\text {th }}$ common term $=T_{10}$ of the A.P. whose first term, $a=11$ and common difference, $d=20$
$
T_{10}=a+9 d=11+9(20)=191
$

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MCQ 71 Mark

In an AP, the $p^{\text {th }}$ term is $q$ and $(p+q)^{\text {th }}$ term is 0 . Then the $q^{\text {th }}$ term is

A
$-p$
✓
p
C
p+q
D
p-q

Answer

Correct option: B.

(B) p
Explanation : Let $a, d$ be the first term and common difference, respectively,
$
\begin{array}{l}
Therefore,\quad q=a+(p-1) d \quad \ldots(i)\\
and \quad 0=a+(p+q-1) d\quad \ldots(ii)
\end{array}
$
Subtracting (i) from (ii), we get
$
q d=-q \text { or } d=-1
$
Substituting in (i), we get,
$
\begin{aligned}
a & =q-(p-1)(-1) \\
a & =q+p-1 \\
Now,\quad T_q & =a+(q-1) d \\
& =(q+p-1)+(q-1)(-1) \\
& =q+p-1-q+1 \\
& =p
\end{aligned}
$

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MCQ 81 Mark

A sequence may be defined as a

A
relation, whose range $\leq N$ (natural number)
B
function, whose range $\leq N$
✓
function, whose domain $\leq N$
D
progression have real values

Answer

Correct option: C.

function, whose domain $\leq N$

(C) function, whose domain $\leq N$
Explanation : A sequence is a function $f: N \rightarrow X$ having domain $\leq N$.

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Applied Maths MCQ

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