LFM Stats And Pure

jee-main 2016 Q80 Finding Parameters for Continuity View

Let $a , b \in R , ( a \neq 0 )$. If the function $f$, defined as $$f ( x ) = \left\{ \begin{array} { c } \frac { 2 x ^ { 2 } } { a } , 0 \leq x < 1 \\ a , 1 \leq x < \sqrt { 2 } \\ \frac { 2 b ^ { 2 } - 4 b } { x ^ { 3 } } , \sqrt { 2 } \leq x < 8 \end{array} \right.$$ is continuous in the interval $[ 0 , \infty )$, then an ordered pair $( a , b )$ can be
(1) $( - \sqrt { 2 } , 1 - \sqrt { 3 } )$
(2) $( \sqrt { 2 } , - 1 + \sqrt { 3 } )$
(3) $( \sqrt { 2 } , 1 - \sqrt { 3 } )$
(4) $( - \sqrt { 2 } , 1 + \sqrt { 3 } )$

jee-main 2017 Q70 Injectivity, Surjectivity, or Bijectivity Classification View

The function $f : \mathbb { R } \to \left[ - \frac { 1 } { 2 } , \frac { 1 } { 2 } \right]$ defined as $f ( x ) = \frac { x } { 1 + x ^ { 2 } }$, is:
(1) Surjective but not injective
(2) Neither injective nor surjective
(3) Invertible
(4) Injective but not surjective

jee-main 2017 Q77 Injectivity, Surjectivity, or Bijectivity Classification View

The function $f : \mathbb{R} \rightarrow \left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$ defined as $f(x) = \dfrac{x}{1 + x^2}$, is:
(1) Invertible
(2) Injective but not surjective
(3) Surjective but not injective
(4) Neither injective nor surjective

jee-main 2017 Q78 Inverse trigonometric equation View

The value of $\tan ^ { - 1 } \left[ \frac { \sqrt { 1 + x ^ { 2 } } + \sqrt { 1 - x ^ { 2 } } } { \sqrt { 1 + x ^ { 2 } } - \sqrt { 1 - x ^ { 2 } } } \right] , | x | < \frac { 1 } { 2 } , x \neq 0$, is equal to:
(1) $\frac { \pi } { 4 } + \frac { 1 } { 2 } \cos ^ { - 1 } x ^ { 2 }$
(2) $\frac { \pi } { 4 } - \cos ^ { - 1 } x ^ { 2 }$
(3) $\frac { \pi } { 4 } - \frac { 1 } { 2 } \cos ^ { - 1 } x ^ { 2 }$
(4) $\frac { \pi } { 4 } + \cos ^ { - 1 } x ^ { 2 }$

jee-main 2017 Q79 Recover a Function from a Composition or Functional Equation View

Let $f ( x ) = 2 ^ { 10 } x + 1$ and $g ( x ) = 3 ^ { 10 } x - 1$. If $( f \circ g ) ( x ) = x$, then $x$ is equal to:
(1) $\frac { 2 ^ { 10 } - 1 } { 2 ^ { 10 } - 3 ^ { - 10 } }$
(2) $\frac { 1 - 2 ^ { - 10 } } { 3 ^ { 10 } - 2 ^ { - 10 } }$
(3) $\frac { 3 ^ { 10 } - 1 } { 3 ^ { 10 } - 2 ^ { - 10 } }$
(4) $\frac { 1 - 3 ^ { - 10 } } { 2 ^ { 10 } - 3 ^ { - 10 } }$

jee-main 2019 Q61 Determine Domain or Range of a Composite Function View

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x) = \frac{x}{1+x^2}$, $x \in \mathbb{R}$. Then the range of $f$ is:
(1) $\mathbb{R} - [-1, 1]$
(2) $(-1, 1) - \{0\}$
(3) $\left[-\frac{1}{2}, \frac{1}{2}\right]$
(4) $\left(-\frac{1}{2}, \frac{1}{2}\right)$

jee-main 2019 Q73 Continuity and Discontinuity Analysis of Piecewise Functions View

If $f ( x ) = [ x ] - \left[ \frac { x } { 4 } \right] , x \in R$, where $[ x ]$ denotes the greatest integer function, then:
(1) $\lim _ { x \rightarrow 4 + } f ( x )$ exists but $\lim _ { x \rightarrow 4 - } f ( x )$ does not exist
(2) $f$ is continuous at $x = 4$
(3) $\lim _ { x \rightarrow 4 - } f ( x )$ exists but $\lim _ { x \rightarrow 4 + } f ( x )$ does not exist
(4) Both $\lim _ { x \rightarrow 4 - } f ( x )$ and $\lim _ { x \rightarrow 4 + } f ( x )$ exist but are not equal

jee-main 2019 Q76 Injectivity, Surjectivity, or Bijectivity Classification View

If the function $f : R - \{ 1 , - 1 \} \rightarrow A$ defined by $f ( x ) = \frac { x ^ { 2 } } { 1 - x ^ { 2 } }$, is surjective, then $A$ is equal to
(1) $[ 0 , \infty )$
(2) $R - \{ - 1 \}$
(3) $R - [ - 1,0 )$
(4) $R - ( - 1,0 )$

jee-main 2019 Q77 Addition/Subtraction Formula Evaluation View

If $\alpha = \cos^{-1}\frac{3}{5}$, $\beta = \tan^{-1}\frac{1}{3}$, where $0 < \alpha, \beta < \frac{\pi}{2}$, then $\alpha - \beta$ is equal to
(1) $\tan^{-1}\frac{9}{14}$
(2) $\cos^{-1}\frac{9}{5\sqrt{10}}$
(3) $\sin^{-1}\frac{9}{5\sqrt{10}}$
(4) $\tan^{-1}\frac{9}{5\sqrt{10}}$

jee-main 2019 Q78 Evaluate Composition from Algebraic Definitions View

If $f(x) = \log_e\frac{1-x}{1+x}$, $|x| < 1$, then $f\left(\frac{2x}{1+x^2}\right)$ is equal to
(1) $f(x^2)$
(2) $2f(x^2)$
(3) $-2f(x)$
(4) $2f(x)$

jee-main 2019 Q78 Find or Apply an Inverse Function Formula View

If the function $f$ defined on $\left( \frac { \pi } { 6 } , \frac { \pi } { 3 } \right)$ by $f ( x ) = \left\{ \begin{array} { l l } \frac { \sqrt { 2 } \cos x - 1 } { \cot x - 1 } , & x \neq \frac { \pi } { 4 } \\ k , & x = \frac { \pi } { 4 } \end{array} \right.$ is continuous, then $k$ is equal to
(1) $\frac { 1 } { 2 }$
(2) 1
(3) 2
(4) $\frac { 1 } { \sqrt { 2 } }$

jee-main 2019 Q79 Determine Domain or Range of a Composite Function View

The domain of the definition of the function $f ( x ) = \frac { 1 } { 4 - x ^ { 2 } } + \log _ { 10 } \left( x ^ { 3 } - x \right)$ is:
(1) $( - 1,0 ) \cup ( 1,2 ) \cup ( 2 , \infty )$
(2) $( 1,2 ) \cup ( 2 , \infty )$
(3) $( - 2 , - 1 ) \cup ( - 1,0 ) \cup ( 2 , \infty )$
(4) $( - 1,0 ) \cup ( 1,2 ) \cup ( 3 , \infty )$

jee-main 2019 Q80 Finding Parameters for Continuity View

If the function $f ( x ) = \left\{ \begin{array} { l } a | \pi - x | + 1 , x \leq 5 \\ b | x - \pi | + 3 , x > 5 \end{array} \right.$ is continuous at $x = 5$, then the value of $a - b$ is:
(1) $\frac { 2 } { 5 - \pi }$
(2) $\frac { - 2 } { \pi + 5 }$
(3) $\frac { 2 } { \pi + 5 }$
(4) $\frac { 2 } { \pi - 5 }$

jee-main 2019 Q81 Injectivity, Surjectivity, or Bijectivity Classification View

Let $A = \{x \in R : x$ is not a positive integer$\}$. Define a function $f: A \rightarrow R$ as $f(x) = \frac{2x}{x-1}$, then $f$ is:
(1) Injective but not surjective
(2) Not injective
(3) Surjective but not injective
(4) Neither injective nor surjective

jee-main 2020 Q61 Recover a Function from a Composition or Functional Equation View

If $g(x) = x ^ { 2 } + x - 1$ and $(g \circ f)(x) = 4x ^ { 2 } - 10x + 5$, then $f \left( \frac { 5 } { 4 } \right)$ is equal to
(1) $\frac { 3 } { 2 }$
(2) $- \frac { 1 } { 2 }$
(3) $\frac { 1 } { 2 }$
(4) $- \frac { 3 } { 2 }$

jee-main 2020 Q61 Determine Domain or Range of a Composite Function View

Let $f:(1,3) \rightarrow R$ be a function defined by $f(x) = \frac{x[x]}{1 + x^{2}}$, where $[x]$ denotes the greatest integer $\leq x$. Then the range of $f$ is
(1) $\left(\frac{2}{5}, \frac{3}{5}\right] \cup \left(\frac{3}{4}, \frac{4}{5}\right)$
(2) $\left(\frac{2}{5}, \frac{1}{2}\right) \cup \left(\frac{3}{5}, \frac{4}{5}\right]$
(3) $\left(\frac{2}{5}, \frac{4}{5}\right]$
(4) $\left(\frac{3}{5}, \frac{4}{5}\right)$

jee-main 2020 Q62 Finding Parameters for Continuity View

$f ( x ) = \left\{ \begin{array} { c l } \frac { \sin ( a + 2 ) x + \sin x } { x } & ; x < 0 \\ b & ; x = 0 \\ \frac { \left( x + 3 x ^ { 2 } \right) ^ { 1 / 3 } - x ^ { 1 / 3 } } { x ^ { 1 / 3 } } & ; x > 0 \end{array} \right.$ is continuous at $x = 0$, then $a + 2 b$ is equal to:
(1) 1
(2) - 1
(3) 0
(4) - 2

jee-main 2020 Q64 Finding parameter values from differentiability or equation constraints View

The function $f ( x ) = \left\{ \begin{array} { l l } \frac { \pi } { 4 } + \tan ^ { - 1 } x , & | x | \leq 1 \\ \frac { 1 } { 2 } ( | x | - 1 ) , & | x | > 1 \end{array} \right.$ is:
(1) continuous on $R - \{ 1 \}$ and differentiable on $R - \{ - 1,1 \}$.
(2) both continuous and differentiable on $R - \{ 1 \}$
(3) continuous on $R - \{ - 1 \}$ and differentiable on $R - \{ - 1,1 \}$
(4) both continuous and differentiable on $R - \{ - 1 \}$

jee-main 2021 Q70 Injectivity, Surjectivity, or Bijectivity Classification View

Let $f : R \rightarrow R$ be defined as $f(x) = 2x - 1$ and $g : R - \{1\} \rightarrow R$ be defined as $g(x) = \frac { x - \frac { 1 } { 2 } } { x - 1 }$. Then the composition function $f(g(x))$ is:
(1) neither one-one nor onto
(2) onto but not one-one
(3) both one-one and onto
(4) one-one but not onto

jee-main 2021 Q70 Determine Domain or Range of a Composite Function View

Let $f ( x ) = \sin ^ { - 1 } x$ and $g ( x ) = \frac { x ^ { 2 } - x - 2 } { 2 x ^ { 2 } - x - 6 }$. If $g ( 2 ) = \lim _ { x \rightarrow 2 } g ( x )$, then the domain of the function $f o g$ is
(1) $( - \infty , - 1 ] \cup [ 2 , \infty )$
(2) $( - \infty , - 2 ] \cup \left[ - \frac { 3 } { 2 } , \infty \right)$
(3) $( - \infty , - 2 ] \cup \left[ - \frac { 4 } { 3 } , \infty \right)$
(4) $( - \infty , - 2 ] \cup [ - 1 , \infty )$

jee-main 2021 Q70 Determine Domain or Range of a Composite Function View

Let $\alpha \in R$ be such that the function $f ( x ) = \left\{ \begin{array} { l l } \frac { \cos ^ { - 1 } \left( 1 - \{ x \} ^ { 2 } \right) \sin ^ { - 1 } ( 1 - \{ x \} ) } { \{ x \} - \{ x \} ^ { 3 } } , & x \neq 0 \\ \alpha , & x = 0 \end{array} \right.$ is continuous at $x = 0$, where $\{ x \} = x - [ x ] , [ x ]$ is the greatest integer less than or equal to $x$. Then :
(1) $\alpha = \frac { \pi } { \sqrt { 2 } }$
(2) $\alpha = 0$
(3) no such $\alpha$ exists
(4) $\alpha = \frac { \pi } { 4 }$

jee-main 2021 Q71 Recover a Function from a Composition or Functional Equation View

Let $f : R - \left\{ \frac { \alpha } { 6 } \right\} \rightarrow R$ be defined by $f ( x ) = \left( \frac { 5 x + 3 } { 6 x - \alpha } \right)$. Then the value of $\alpha$ for which $( f \circ f ) ( x ) = x$, for all $x \in R - \left\{ \frac { \alpha } { 6 } \right\}$, is
(1) No such $\alpha$ exists
(2) 5
(3) 8
(4) 6

jee-main 2021 Q72 Determine Domain or Range of a Composite Function View

Let a function $f : R \rightarrow R$ be defined as, $f ( x ) = \begin{cases} \sin x - e ^ { x } & \text { if } x \leq 0 \\ a + [ - x ] & \text { if } 0 < x < 1 \\ 2 x - b & \text { if } x \geq 1 \end{cases}$
Where $[ x ]$ is the greatest integer less than or equal to $x$. If $f$ is continuous on $R$, then ( $a + b$ ) is equal to:
(1) 4
(2) 3
(3) 2
(4) 5

jee-main 2021 Q72 Injectivity, Surjectivity, or Bijectivity Classification View

Let $f , g : N \rightarrow N$ such that $f ( n + 1 ) = f ( n ) + f ( 1 ) \forall n \in N$ and $g$ be any arbitrary function. Which of the following statements is NOT true?
(1) If $f$ is onto, then $f ( n ) = n \forall n \in N$
(2) If $g$ is onto, then $f o g$ is one-one
(3) $f$ is one-one
(4) If $f \circ g$ is one-one, then $g$ is one-one

jee-main 2021 Q73 Determine Domain or Range of a Composite Function View

Let the functions $f : R \rightarrow R$ and $g : R \rightarrow R$ be defined as : $f ( x ) = \left\{ \begin{array} { l l } x + 2 , & x < 0 \\ x ^ { 2 } , & x \geq 0 \end{array} \right.$ and $g ( x ) = \begin{cases} x ^ { 3 } , & x < 1 \\ 3 x - 2 , & x \geq 1 \end{cases}$ Then, the number of points in $R$ where $( f \circ g ) ( x )$ is NOT differentiable is equal to :
(1) 3
(2) 1
(3) 0
(4) 2

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LFM Stats And Pure

Completing the square and sketching 80

Inequalities 215

Discriminant and conditions for roots 106

Solving quadratics and applications 106

Curve Sketching 295

Factor & Remainder Theorem 90

Polynomial Division & Manipulation 44

Binomial Theorem (positive integer n) 244

Function Transformations 65

Composite & Inverse Functions 249

Partial Fractions 22

Generalised Binomial Theorem 16

Generalised Binomial Theorem and Partial Fractions 3

Complex Numbers Arithmetic 283

Complex Numbers Argand & Loci 135

Permutations & Arrangements 276

Combinations & Selection 268

Measures of Location and Spread 233

Probability Definitions 413

Tree Diagrams 5

Principle of Inclusion/Exclusion 23

Independent Events 51

Conditional Probability 127

Discrete Probability Distributions 210

Binomial Distribution 107

Geometric Distribution 11

Hypergeometric Distribution 10

Negative Binomial Distribution 2

Modelling and Hypothesis Testing 38

Hypothesis test of binomial distributions 12

Data representation 29

Continuous Probability Distributions and Random Variables 30

Continuous Uniform Random Variables 15

Geometric Probability 28

Normal Distribution 96

Approximating Binomial to Normal Distribution 7

Sign Change & Interval Methods 74

Fixed Point Iteration 12