LFM Pure

jee-main 2019 Q81 Chain Rule with Composition of Explicit Functions View

If $f ( 1 ) = 1 , f ^ { \prime } ( 1 ) = 3$, then the derivative of $f ( f ( f ( x ) ) ) + ( f ( x ) ) ^ { 2 }$ at $x = 1$ is:
(1) 9
(2) 12
(3) 15
(4) 33

jee-main 2020 Q58 Limit Evaluation Involving Composition or Substitution View

$\lim _ { x \rightarrow a } \frac { ( a + 2 x ) ^ { \frac { 1 } { 3 } } - ( 3 x ) ^ { \frac { 1 } { 3 } } } { ( 3 a + x ) ^ { \frac { 1 } { 3 } } - ( 4 x ) ^ { \frac { 1 } { 3 } } } ( a \neq 0 )$ is equal to:
(1) $\left( \frac { 2 } { 9 } \right) \left( \frac { 2 } { 3 } \right) ^ { \frac { 1 } { 3 } }$
(2) $\left( \frac { 2 } { 3 } \right) ^ { \frac { 4 } { 3 } }$
(3) $\left( \frac { 2 } { 9 } \right) ^ { \frac { 4 } { 3 } }$
(4) $\left( \frac { 2 } { 3 } \right) \left( \frac { 2 } { 9 } \right) ^ { \frac { 1 } { 3 } }$

jee-main 2020 Q58 Limit Evaluation Involving Composition or Substitution View

If $\alpha$ is the positive root of the equation, $p ( x ) = x ^ { 2 } - x - 2 = 0$, then $\lim _ { x \rightarrow \alpha ^ { + } } \frac { \sqrt { 1 - \cos p ( x ) } } { x + \alpha - 4 }$ is equal to
(1) $\frac { 3 } { 2 }$
(2) $\frac { 3 } { \sqrt { 2 } }$
(3) $\frac { 1 } { \sqrt { 2 } }$
(4) $\frac { 1 } { 2 }$

jee-main 2020 Q59 Limit Evaluation Involving Composition or Substitution View

$\lim_{x\rightarrow 0} \frac{x\left(e^{\left(\sqrt{1+x^2+x^4}-1\right)/x}-1\right)}{\sqrt{1+x^2+x^4}-1}$
(1) is equal to $\sqrt{e}$
(2) is equal to 1
(3) is equal to 0
(4) does not exist

jee-main 2020 Q62 Compute slope at a point via implicit differentiation (single-step) View

Let $y = y ( x )$ be a function of $x$ satisfying $y \sqrt { 1 - x ^ { 2 } } = k - x \sqrt { 1 - y ^ { 2 } }$ where $k$ is a constant and $y \left( \frac { 1 } { 2 } \right) = - \frac { 1 } { 4 }$. Then $\frac { d y } { d x }$ at $x = \frac { 1 } { 2 }$, is equal to
(1) $- \frac { \sqrt { 5 } } { 4 }$
(2) $- \frac { \sqrt { 5 } } { 2 }$
(3) $\frac { 2 } { \sqrt { 5 } }$
(4) $\frac { \sqrt { 5 } } { 2 }$

jee-main 2020 Q63 Determine parameters from function or curve conditions View

If a function $f(x)$ defined by $$f(x) = \begin{cases} ae^{x} + be^{-x}, & -1 \leq x < 1 \\ cx^{2}, & 1 \leq x \leq 3 \\ ax^{2} + 2cx, & 3 < x \leq 4 \end{cases}$$ be continuous for some $a, b, c \in R$ and $f'(0) + f'(2) = e$, then the value of $a$ is
(1) $\frac{1}{e^{2} - 3e + 13}$
(2) $\frac{e}{e^{2} - 3e - 13}$
(3) $\frac{e}{e^{2} + 3e + 13}$
(4) $\frac{e}{e^{2} - 3e + 13}$

jee-main 2020 Q64 Derivative of an Inverse Function View

Let $f$ and $g$ be differentiable functions on $R$ such that $f \circ g$ is the identity function. If for some $a , b \in R , g ^ { \prime } ( a ) = 5$ and $g ( a ) = b$, then $f ^ { \prime } ( b )$ is equal to:
(1) $\frac { 1 } { 5 }$
(2) 1
(3) 5
(4) $\frac { 2 } { 5 }$

jee-main 2020 Q64 Evaluate derivative at a point or find tangent slope View

The derivative of $\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $\tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$ at $x = \frac{1}{2}$ is:
(1) $\frac{2\sqrt{3}}{5}$
(2) $\frac{\sqrt{3}}{12}$
(3) $\frac{2\sqrt{3}}{3}$
(4) $\frac{\sqrt{3}}{10}$

jee-main 2021 Q67 Limit Evaluation Involving Composition or Substitution View

$\lim _ { x \rightarrow 0 } \frac { \sin ^ { 2 } \left( \pi \cos ^ { 4 } x \right) } { x ^ { 4 } }$ is equal to :
(1) $2 \pi ^ { 2 }$
(2) $\pi ^ { 2 }$
(3) $4 \pi ^ { 2 }$
(4) $4 \pi$

jee-main 2021 Q68 Limit evaluation involving derivatives or asymptotic analysis View

The value of $\lim _ { x \rightarrow 0 ^ { + } } \frac { \cos ^ { - 1 } \left( x - [ x ] ^ { 2 } \right) \cdot \sin ^ { - 1 } \left( x - [ x ] ^ { 2 } \right) } { x - x ^ { 3 } }$, where $[ x ]$ denotes the greatest integer $\leq x$ is:
(1) $\pi$
(2) 0
(3) $\frac { \pi } { 4 }$
(4) $\frac { \pi } { 2 }$

jee-main 2021 Q84 Limit Evaluation Involving Composition or Substitution View

If $\lim _ { x \rightarrow 0 } \frac { a x - \left( e ^ { 4 x } - 1 \right) } { a x \left( e ^ { 4 x } - 1 \right) }$ exists and is equal to $b$, then the value of $a - 2 b$ is $\underline{\hspace{1cm}}$.

jee-main 2021 Q84 Limit Evaluation Involving Composition or Substitution View

If $\lim _ { x \rightarrow 0 } \frac { a e ^ { x } - b \cos x + c e ^ { - x } } { x \sin x } = 2$, then $a + b + c$ is equal to $\_\_\_\_$.

jee-main 2021 Q86 Limit Evaluation Involving Composition or Substitution View

If $\lim _ { x \rightarrow 0 } \left[ \frac { \alpha x e ^ { x } - \beta \log _ { e } ( 1 + x ) + \gamma x ^ { 2 } e ^ { - x } } { x \sin ^ { 2 } x } \right] = 10 , \alpha , \beta , \gamma \in R$, then the value of $\alpha + \beta + \gamma$ is $\underline{\hspace{1cm}}$.

jee-main 2021 Q87 Continuity Conditions via Composition View

If $\lim _ { x \rightarrow 0 } \frac { a e ^ { x } - b \cos x + c e ^ { - x } } { x \sin x } = 2$, then $a + b + c$ is equal to

jee-main 2022 Q3 Chain Rule with Composition of Explicit Functions View

If $t = \sqrt { x } + 4$, then $\left( \frac { \mathrm { d } x } { \mathrm {~d} t } \right) _ { t = 4 }$ is:
(1) 4
(2) Zero
(3) 8
(4) 16

jee-main 2022 Q66 Limit Evaluation Involving Composition or Substitution View

$\lim _ { x \rightarrow 0 } \frac { \cos ( \sin x ) - \cos x } { x ^ { 4 } }$ is equal to
(1) $\frac { 1 } { 3 }$
(2) $\frac { 1 } { 6 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 1 } { 12 }$

jee-main 2022 Q66 Inverse trigonometric equation View

$\lim _ { x \rightarrow \frac { 1 } { \sqrt { 2 } } } \frac { \sin \left( \cos ^ { - 1 } x \right) - x } { 1 - \tan \left( \cos ^ { - 1 } x \right) }$ is equal to
(1) $\frac { 1 } { \sqrt { 2 } }$
(2) $\frac { - 1 } { \sqrt { 2 } }$
(3) $\sqrt { 2 }$
(4) $- 1$

jee-main 2022 Q69 Limit Evaluation Involving Composition or Substitution View

$\lim_{x \rightarrow \frac{\pi}{4}} \frac{8\sqrt{2} - (\cos x + \sin x)^7}{\sqrt{2} - \sqrt{2}\sin 2x}$ is equal to
(1) 14
(2) 7
(3) $14\sqrt{2}$
(4) $7\sqrt{2}$

jee-main 2022 Q74 Evaluate derivative at a point or find tangent slope View

If $a = \lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } \frac { 2 n } { n ^ { 2 } + k ^ { 2 } }$ and $f ( x ) = \sqrt { \frac { 1 - \cos x } { 1 + \cos x } } , x \in ( 0,1 )$, then:
(1) $2 \sqrt { 2 } f \left( \frac { a } { 2 } \right) = f ^ { \prime } \left( \frac { a } { 2 } \right)$
(2) $f \left( \frac { a } { 2 } \right) f ^ { \prime } \left( \frac { a } { 2 } \right) = \sqrt { 2 }$
(3) $\sqrt { 2 } f \left( \frac { a } { 2 } \right) = f ^ { \prime } \left( \frac { a } { 2 } \right)$
(4) $f \left( \frac { a } { 2 } \right) = \sqrt { 2 } f ^ { \prime } \left( \frac { a } { 2 } \right)$

jee-main 2022 Q87 Higher-order or nth derivative computation View

If $y ( x ) = \left( x ^ { x } \right) ^ { x } , x > 0$ then $\frac { d ^ { 2 } x } { d y ^ { 2 } } + 20$ at $x = 1$ is equal to

jee-main 2023 Q65 Convergence proof and limit determination View

$\lim _ { t \rightarrow 0 } 1 ^ { \frac { 1 } { \sin ^ { 2 } t } } + 2 ^ { \frac { 1 } { \sin ^ { 2 } t } } + 3 ^ { \frac { 1 } { \sin ^ { 2 } t } } \ldots \ldots n ^ { \frac { 1 } { \sin ^ { 2 } t } } \sin ^ { 2 } t$ is equal to
(1) $n ^ { 2 } + n$
(2) $n$
(3) $\frac { n n + 1 } { 2 }$
(4) $n ^ { 2 }$

jee-main 2023 Q68 Limit evaluation using series expansion or exponential asymptotics View

If $\lim _ { x \rightarrow 0 } \frac { e ^ { ax } - \cos ( bx ) - \frac { cx e ^ { cx } } { 2 } } { 1 - \cos ( 2 x ) } = 17$, then $5 a ^ { 2 } + b ^ { 2 }$ is equal to
(1) 64
(2) 72
(3) 68
(4) 76

jee-main 2023 Q72 Limit Evaluation Involving Composition or Substitution View

Let $x = 2$ be a root of the equation $x ^ { 2 } + p x + q = 0$ and $f ( x ) = \left\{ \begin{array} { c l } \frac { 1 - \cos \left( x ^ { 2 } - 4 p x + q ^ { 2 } + 8 q + 16 \right) } { ( x - 2 p ) ^ { 4 } } , & x \neq 2 p \\ 0 , & x = 2 p \end{array} \right.$. Then $\lim _ { x \rightarrow 2 p ^ { + } } [ f ( x ) ]$ where $[ \cdot ]$ denotes greatest integer function, is
(1) 2
(2) 1
(3) 0
(4) $- 1$

jee-main 2023 Q79 Higher-Order Derivatives of Products/Compositions View

Let $y ( x ) = ( 1 + x ) \left( 1 + x ^ { 2 } \right) \left( 1 + x ^ { 4 } \right) \left( 1 + x ^ { 8 } \right) \left( 1 + x ^ { 16 } \right)$. Then $y ^ { \prime } - y ^ { \prime \prime }$ at $x = - 1$ is equal to
(1) 976
(2) 464
(3) 496
(4) 944

jee-main 2023 Q85 Higher-Order Derivatives of Products/Compositions View

If $f(x) = x^2 + g'(1)x + g''(2)$ and $g(x) = f(1)x^2 + xf'(x) + f''(x)$, then the value of $f(4) - g(4)$ is equal to $\_\_\_\_$.

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LFM Pure

Straight Lines & Coordinate Geometry 242

Circles 702

Simultaneous equations 134

Proof 868

Proof by induction 36

Sine and Cosine Rules 185

Radians, Arc Length and Sector Area 22

Trig Graphs & Exact Values 95

Standard trigonometric equations 142

Trig Proofs 92

Quadratic trigonometric equations 8

Trigonometric equations in context 11

Modulus function 37

Matrices 1085

Linear transformations 72

Invariant lines and eigenvalues and vectors 339

Reciprocal Trig & Identities 63

Addition & Double Angle Formulae 88

Harmonic Form 14

Small angle approximation 14

Chain Rule 281

Product & Quotient Rules 13

Differentiating Transcendental Functions 160

Connected Rates of Change 48

Standard Integrals and Reverse Chain Rule 82

Integration by Substitution 213

Integration by Parts 44

Implicit equations and differentiation 27

Differential equations 308

3x3 Matrices 164