3 Marks Question

Question 13 Marks

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer

$\text {AP=AS----(i)}$
$\text {BP=BQ ---- (ii) }$
$\text {CR=CQ-----(iii)}$
$\text {DR=DS ---- (vi) }$
Adding $\text{(i), (ii), (iii) (iv)}$
$\text {AP+BP+CR+DR=AS+BQ+CQ+DS}$
$\Rightarrow \text {AB+CD=AD+BC}$
But $\text {A B C D}$ is a parallelogram $\Rightarrow \text {A B=C D}$ and $\text {A D=B C}$
$\therefore 2 AB=2 AD \text { or } AB=AD$
Hence, $\text {A B C D}$ is a rhombus.

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Question 23 Marks

Prove that $\frac{2-\sqrt{3}}{5}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.

Answer

Assuming $\frac{2-\sqrt{3}}{5}$ to be a rational number.
$\Rightarrow \frac{2-\sqrt{3}}{5}=\frac{p}{q}$, where p and q are integers $\& q \neq 0$
$
\Rightarrow \sqrt{3}=\frac{2 q-5 p}{q}
$
Here RHS is rational but LHS is irrational.
Therefore our assumption is wrong.
Hence $\frac{2-\sqrt{3}}{5}$ is an irrational number.

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Question 33 Marks

If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2+x-2$, then find the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$.

Answer

Here $\alpha+\beta=-1$ and $\alpha \beta=-2$
$
\begin{aligned}
\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta} \\
=\frac{(-1)^2-2(-2)}{-2}=-\frac{5}{2}
\end{aligned}
$

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Question 43 Marks

Find the zeroes of the polynomial $4 x^2+4 x-3$ and verify the relationship between zeroes and coefficients of the polynomial.

Answer

$
\begin{aligned}
P(x) & =4 x^2+4 x-3 \\
& =(2 x+3)(2 x
\end{aligned}
$
$\therefore$ Zeroes of the polynomial are $\frac{-3}{2}, \frac{1}{2}$
Sum of Zeroes $=\frac{-3}{2}+\frac{1}{2}=\frac{-3+1}{2}=-1=\frac{-4}{4}=\frac{-(\text { coefficient of } x)}{\left(\text { coefficient of } x^2\right)}$
Product of Zeroes $=\frac{-3}{2} \times \frac{1}{2}=\frac{-3}{4}=\frac{\text { constant term }}{\text { coefficient of } x^2}$

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Question 53 Marks

Rehana went to a bank to withdraw $₹ 2,000 $. She asked the cashier to give her $₹ 50$ and $₹ 100$ notes only. Rehana got $25$ notes in all. Find how many notes of $₹ 50$ and $₹ 100$ did she receive.

Answer

Let number of ₹ $50$ notes $= x$
and number of ₹ $100$ notes $=y$
$\text { Here } x+y=25 \text {------(i) }$
$50 x+100 y=2000 \text { or } x+2 y=40 \text {------(ii) }$
Solving eq.$(i)$ and eq.$(ii)$, we get
$x=10$ and $y=15$
Therefore $10$ notes of ₹ $50$ and $15$ notes of ₹ $100$ are received.

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Question 63 Marks

A box contains 90 discs which are numbered 1 to 90 . If one disc is drawn at random from the box, find the probability that it bears a :
(i) 2-digit number less than 40 .
(ii) number divisible by 5 and greater than 50 .
(iii) a perfect square number.

Answer

Total outcomes $=90$
(i) $P (2$ digit number less than 40$)=\frac{30}{90}$ or $\frac{1}{3}$
(ii) $P ( a$ number divisible by 5 and greater than 50$)=\frac{8}{90}$ or $\frac{4}{45}$
(iii) $P ($ a perfect square number $)=\frac{9}{90}$ or $\frac{1}{10}$

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Question 73 Marks

Three coins are tossed simultaneously. What is the probability of getting
(i) at least one head ?
(ii) exactly two tails ?
(iii) at most one tail?

Answer

Total number of outcomes $=8$
(i) $P ($ at least one head $)=\frac{7}{8}$
(ii) $P ($ exactly 2 tails $)=\frac{3}{8}$
(iii) $P ($ at most one tail $)=\frac{4}{8}$ or $\frac{1}{2}$

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Question 83 Marks

Prove that $\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$

Answer

$\text { L.H.S }=\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}$
Divide Numerator and Denominator by $\cos \theta$.
$=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta}$
$=\frac{\tan \theta-1+\sec \theta}{(\tan \theta-\sec \theta)+\left(\sec ^2 \theta-\tan ^2 \theta\right)}$
$=\frac{\tan \theta-1+\sec \theta}{(\sec \theta-\tan \theta)(\tan \theta+\sec \theta-1)}$
$=\frac{1}{\sec \theta-\tan \theta}=\text { R.H.S }$

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Maths 3 Marks Question

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