Let $[x]$ denote the greatest integer less than or equal to $x$. The number of positive integer solu...

$$ \left\lfloor\frac{x}{19}\right\rfloor=n \quad \text { and } \quad\left\lfloor\frac{x}{20}\right\rfloor=n $$where $n$ is an integer. This implies:$$ \begin{gathered} n \leq \frac{x}{19} < n+1 \text { and } n \leq \frac{x}{20} < n+1 \\ 19 n \leq x < 19(n+1) \\ 20 n \leq x < 20(n+1) \end{gathered} $$Finding the Intersection:$$ \begin{gathered} \max (19 n, 20 n) \leq x < \min (19(n+1), 20(n+1)) \\ 20 n \leq x < 19(n+1) \end{gathered} $$The number of integers in this range is:$$ 19(n+1)-20 n=19 n+19-20 n=19-n $$We also need to determine the range of $n$ for which there are valid solutions. We need:$$ \begin{gathered} 20 n \leq 19(n+1) \\ 20 n \leq 19 n+19 \\ n \leq 19 \end{gathered} $$Thus, $n$ ranges from 0 to 18 (since $n=19$ would result in $x \geq 380$, which falls outside our desired range as $\frac{x}{20}=19$ ).For each integer $n$ from 0 to 18 , the number of valid $x$ values is $19-n$. We sum these values:$$ \sum\limits_{n = 0}^{18} {}(19-n)=19+18+17 \ldots+1 $$Which is the sum of AP with $a=1, n=19, d=1$ :$$ \frac{19}{2}(2+18)=190 $$Since $x=0$ is not a valid positive integer, we need to subtract 1 from the total count:$$ 190-1=189 $$Therefore, the number of positive integer solutions is:$$ 189 $$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be two functio...

The above is the plot for:$$ g \circ f(x)= \begin{cases}x^2, & \text { if } x \geq 0 \\ e^x-1, & \text { if } x<0\end{cases} $$It can be verified in general that $g \circ f$ is one-one implies that $f$ is one-one. Similarly, $g \circ f$ is onto implies that g is onto. (NCERT Old Edition Class 12 page 14)Note: There is an argument to be made about this question as it is based on a "general" case, but for now we will work with the NCERT statement.So, since $g \circ f(x)$ is one-one, this implies that $f(x)$ will also be one-one in nature.

For what value of $t$ does $f(x)=x^3+t x^2+(2-t) x-3$ have only real roots?...

$$ f(x)=x^3+t x^2+(2-t) x-3 $$Testing $f(1)$ :$$ f(1)=1+t+(2-t)-3=0 $$So, $x=1$ is a root.Factoring $x-1$ out of the polynomial:$$ \frac{x^3+t x^2+(2-t) x-3}{x-1}=x^2+x(t+1)+3 $$For the polynomial to have all real roots, the quadratic $x^2+x(t-1)+3$ must have real roots. The discriminant of this quadratic must be non-negative:$$ \begin{gathered} D=(t+1)^2-4 \cdot 1 \cdot 3 \geq 0 \\ (t+1)^2-12 \geq 0 \\ (t+1)^2 \geq 12 \end{gathered} $$Solving for $t$ :$$ t+1 \geq \sqrt{12} \text { or } t+1 \leq-\sqrt{12} $$But we can ignore the negative solution as no relevant option is given$$ t \geq \sqrt{12}-1=2.464 $$From the given options $(0,4,2,-1)$, the value of $t$ that satisfies this condition is:4

Let $f: Q \rightarrow Q$ be a function such that $f(x+y)=f(x)+f(y)$ for all $x, y \in \mathbf{Q}$, a...

Given the function $ f: \mathbb{Q} \rightarrow \mathbb{Q} $ defined by the property $ f(x+y) = f(x) + f(y) $ for all $ x, y \in \mathbb{Q} $, and knowing that $ f(1) = 10 $, we can derive that for any rational number $ x $, the function is linear and can be expressed as $ f(x) = 10x $.To verify this, consider:For $ x = 1 $, we have $ f(1) = 10 $.For $ x = 2 $, using the functional equation:$ f(2) = f(1+1) = f(1) + f(1) = 10 + 10 = 20 $For $ x = 3 $, similarly:$ f(3) = f(1+1+1) = f(1) + f(1) + f(1) = 10 + 10 + 10 = 30 $Continuing this pattern, it follows that $ f(x) = 10x $ for any integer $ x $. Since $ \mathbb{Q} $ is closed under addition and scalar multiplication, $ f $ is defined for all rational numbers, and it can be shown that $ f(x) = 10x $ holds for all $ x \in \mathbb{Q} $.This function is linear with no restriction on $ x $, confirming it is not only injective but also surjective across $ \mathbb{Q} $:Injectivity: If $ f(a) = f(b) $, then $ 10a = 10b $ implies $ a = b $.Surjectivity: For any $ y \in \mathbb{Q} $, choose $ x = \frac{y}{10} $, then $ f(x) = 10 \left(\frac{y}{10}\right) = y $.Therefore, $ f(x)=10x $ demonstrates that $ f $ is bijective.

Functions - IAT-IISER Maths PYQs

Q: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be two functio...

The above is the plot for:$$ g \circ f(x)= \begin{cases}x^2, & \text { if } x \geq 0 \\ e^x-1, & \text { if } x<0\end{cases} $$It can be verified in general that $g \circ f$ is one-one implies that $f$ is one-one. Similarly, $g \circ f$ is onto implies that g is onto. (NCERT Old Edition Class 12 page 14)Note: There is an argument to be made about this question as it is based on a "general" case, but for now we will work with the NCERT statement.So, since $g \circ f(x)$ is one-one, this implies that $f(x)$ will also be one-one in nature.

Q: For what value of $t$ does $f(x)=x^3+t x^2+(2-t) x-3$ have only real roots?...

$$ f(x)=x^3+t x^2+(2-t) x-3 $$Testing $f(1)$ :$$ f(1)=1+t+(2-t)-3=0 $$So, $x=1$ is a root.Factoring $x-1$ out of the polynomial:$$ \frac{x^3+t x^2+(2-t) x-3}{x-1}=x^2+x(t+1)+3 $$For the polynomial to have all real roots, the quadratic $x^2+x(t-1)+3$ must have real roots. The discriminant of this quadratic must be non-negative:$$ \begin{gathered} D=(t+1)^2-4 \cdot 1 \cdot 3 \geq 0 \\ (t+1)^2-12 \geq 0 \\ (t+1)^2 \geq 12 \end{gathered} $$Solving for $t$ :$$ t+1 \geq \sqrt{12} \text { or } t+1 \leq-\sqrt{12} $$But we can ignore the negative solution as no relevant option is given$$ t \geq \sqrt{12}-1=2.464 $$From the given options $(0,4,2,-1)$, the value of $t$ that satisfies this condition is:4

Q: Let $f: Q \rightarrow Q$ be a function such that $f(x+y)=f(x)+f(y)$ for all $x, y \in \mathbf{Q}$, a...

Given the function $ f: \mathbb{Q} \rightarrow \mathbb{Q} $ defined by the property $ f(x+y) = f(x) + f(y) $ for all $ x, y \in \mathbb{Q} $, and knowing that $ f(1) = 10 $, we can derive that for any rational number $ x $, the function is linear and can be expressed as $ f(x) = 10x $.To verify this, consider:For $ x = 1 $, we have $ f(1) = 10 $.For $ x = 2 $, using the functional equation:$ f(2) = f(1+1) = f(1) + f(1) = 10 + 10 = 20 $For $ x = 3 $, similarly:$ f(3) = f(1+1+1) = f(1) + f(1) + f(1) = 10 + 10 + 10 = 30 $Continuing this pattern, it follows that $ f(x) = 10x $ for any integer $ x $. Since $ \mathbb{Q} $ is closed under addition and scalar multiplication, $ f $ is defined for all rational numbers, and it can be shown that $ f(x) = 10x $ holds for all $ x \in \mathbb{Q} $.This function is linear with no restriction on $ x $, confirming it is not only injective but also surjective across $ \mathbb{Q} $:Injectivity: If $ f(a) = f(b) $, then $ 10a = 10b $ implies $ a = b $.Surjectivity: For any $ y \in \mathbb{Q} $, choose $ x = \frac{y}{10} $, then $ f(x) = 10 \left(\frac{y}{10}\right) = y $.Therefore, $ f(x)=10x $ demonstrates that $ f $ is bijective.

%5Cleft%5Clfloor%5Cfrac%7Bx%7D%7B19%7D%5Cright%5Crfloor%3Dn%20%5Cquad%20%5Ctext%20%7B%20and%20%7D%20%5Cquad%5Cleft%5Clfloor%5Cfrac%7Bx%7D%7B20%7D%5Cright%5Crfloor%3Dn

where

is%20an%20integer.%20This%20implies%3A

%5Cbegin%7Bgathered%7D%20n%20%5Cleq%20%5Cfrac%7Bx%7D%7B19%7D%20%3C%20n%2B1%20%5Ctext%20%7B%20and%20%7D%20n%20%5Cleq%20%5Cfrac%7Bx%7D%7B20%7D%20%3C%20n%2B1%20%5C%5C%2019%20n%20%5Cleq%20x%20%3C%2019(n%2B1)%20%5C%5C%2020%20n%20%5Cleq%20x%20%3C%2020(n%2B1)%20%5Cend%7Bgathered%7D

Finding%20the%20Intersection%3A

%5Cbegin%7Bgathered%7D%20%5Cmax%20(19%20n%2C%2020%20n)%20%5Cleq%20x%20%3C%20%5Cmin%20(19(n%2B1)%2C%2020(n%2B1))%20%5C%5C%2020%20n%20%5Cleq%20x%20%3C%2019(n%2B1)%20%5Cend%7Bgathered%7D

The%20number%20of%20integers%20in%20this%20range%20is%3A

19(n%2B1)-20%20n%3D19%20n%2B19-20%20n%3D19-n

We%20also%20need%20to%20determine%20the%20range%20of

for%20which%20there%20are%20valid%20solutions.%20We%20need%3A

%5Cbegin%7Bgathered%7D%2020%20n%20%5Cleq%2019(n%2B1)%20%5C%5C%2020%20n%20%5Cleq%2019%20n%2B19%20%5C%5C%20n%20%5Cleq%2019%20%5Cend%7Bgathered%7D

Thus%2C

ranges%20from%200%20to%2018%20(since

n=19

would%20result%20in

x \geq 380

%2C%20which%20falls%20outside%20our%20desired%20range%20as

\frac{x}{20}=19

).For%20each%20integer

from%200%20to%2018%20%2C%20the%20number%20of%20valid

values%20is

19-n

.%20We%20sum%20these%20values%3A

%5Csum%5Climits_%7Bn%20%3D%200%7D%5E%7B18%7D%20%7B%7D(19-n)%3D19%2B18%2B17%20%5Cldots%2B1

Which%20is%20the%20sum%20of%20AP%20with

a=1, n=19, d=1

%3A

%5Cfrac%7B19%7D%7B2%7D(2%2B18)%3D190

Since

x=0

is%20not%20a%20valid%20positive%20integer%2C%20we%20need%20to%20subtract%201%20from%20the%20total%20count%3A

190-1%3D189

Therefore%2C%20the%20number%20of%20positive%20integer%20solutions%20is%3A

189

Given the function

f%3A%20%5Cmathbb%7BQ%7D%20%5Crightarrow%20%5Cmathbb%7BQ%7D

defined by the property

f(x%2By)%20%3D%20f(x)%20%2B%20f(y)

for all

x%2C%20y%20%5Cin%20%5Cmathbb%7BQ%7D

, and knowing that

f(1)%20%3D%2010

, we can derive that for any rational number

x

, the function is linear and can be expressed as

f(x)%20%3D%2010x

.To verify this, consider:For

x%20%3D%201

, we have

f(1)%20%3D%2010

.For

x%20%3D%202

, using the functional equation:

f(2)%20%3D%20f(1%2B1)%20%3D%20f(1)%20%2B%20f(1)%20%3D%2010%20%2B%2010%20%3D%2020

For

x%20%3D%203

, similarly:

f(3)%20%3D%20f(1%2B1%2B1)%20%3D%20f(1)%20%2B%20f(1)%20%2B%20f(1)%20%3D%2010%20%2B%2010%20%2B%2010%20%3D%2030

Continuing this pattern, it follows that

f(x)%20%3D%2010x

for any integer

x

. Since

%5Cmathbb%7BQ%7D

is closed under addition and scalar multiplication,

f

is defined for all rational numbers, and it can be shown that

f(x)%20%3D%2010x

holds for all

x%20%5Cin%20%5Cmathbb%7BQ%7D

.This function is linear with no restriction on

x

, confirming it is not only injective but also surjective across

%5Cmathbb%7BQ%7D

:Injectivity: If

f(a)%20%3D%20f(b)

, then

10a%20%3D%2010b

implies

a%20%3D%20b

.Surjectivity: For any

y%20%5Cin%20%5Cmathbb%7BQ%7D

, choose

x%20%3D%20%5Cfrac%7By%7D%7B10%7D

, then

f(x)%20%3D%2010%20%5Cleft(%5Cfrac%7By%7D%7B10%7D%5Cright)%20%3D%20y

.Therefore,

f(x)%3D10x

demonstrates that

f

is bijective.

Functions

High-Yield Trend

Questions
5 MCQs

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Official Solution

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Functions

High-Yield Trend

Questions 5 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Questions
5 MCQs