LFM Pure and Mechanics

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csat-suneung 2024 Q5 3 marks Recovering Function Values from Derivative Information View
A polynomial function $f(x)$ satisfies $$f'(x) = 3x(x-2), \quad f(1) = 6$$ Find the value of $f(2)$. [3 points]
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5
csat-suneung 2024 Q8 3 marks Definite Integral Evaluation (Computational) View
A cubic function $f(x)$ satisfies $$xf(x) - f(x) = 3x^4 - 3x$$ for all real numbers $x$. Find the value of $\int_{-2}^{2} f(x)\,dx$. [3 points]
(1) 12
(2) 16
(3) 20
(4) 24
(5) 28
csat-suneung 2025 Q7 3 marks Finding a Function from an Integral Equation View
A polynomial function $f(x)$ satisfies $$\int_{0}^{x} f(t)\, dt = 3x^{3} + 2x$$ for all real numbers $x$. What is the value of $f(1)$? [3 points]
(1) 7
(2) 9
(3) 11
(4) 13
(5) 15
csat-suneung 2025 Q9 4 marks Definite Integral Evaluation (Computational) View
For the function $f(x) = 3x^{2} - 16x - 20$, $$\int_{-2}^{a} f(x)\, dx = \int_{-2}^{0} f(x)\, dx$$ When this condition is satisfied, what is the value of the positive number $a$? [4 points]
(1) 16
(2) 14
(3) 12
(4) 10
(5) 8
csat-suneung 2025 Q17 3 marks Recovering Function Values from Derivative Information View
For a polynomial function $f(x)$, $f'(x) = 9x^{2} + 4x$ and $f(1) = 6$. What is the value of $f(2)$? [3 points]
csat-suneung 2025 Q26C 3 marks Definite Integral Evaluation (Computational) View
As shown in the figure, a solid figure has as its base the region enclosed by the curve $y = \sqrt{\frac{x+1}{x(x + \ln x)}}$, the $x$-axis, and the two lines $x = 1$ and $x = e$. When the cross-section of this solid figure cut by a plane perpendicular to the $x$-axis is a square, what is the volume of this solid figure? [3 points]
(1) $\ln(e+1)$
(2) $\ln(e+2)$
(3) $\ln(e+3)$
(4) $\ln(2e+1)$
(5) $\ln(2e+2)$
csat-suneung 2026 Q15 4 marks Accumulation Function Analysis View
The function $f ( x )$ is $$f ( x ) = \begin{cases} - x ^ { 2 } & ( x < 0 ) \\ x ^ { 2 } - x & ( x \geq 0 ) \end{cases}$$ and for a positive number $a$, the function $g ( x )$ is $$g ( x ) = \left\{ \begin{array} { c l } a x + a & ( x < - 1 ) \\ 0 & ( - 1 \leq x < 1 ) \\ a x - a & ( x \geq 1 ) \end{array} \right.$$ Let $k$ be the maximum value of $a$ such that the function $h ( x ) = \int _ { 0 } ^ { x } ( g ( t ) - f ( t ) ) d t$ has exactly one extremum. When $a = k$, what is the value of $k + h ( 3 )$? [4 points]
(1) $\frac { 9 } { 2 }$
(2) $\frac { 11 } { 2 }$
(3) $\frac { 13 } { 2 }$
(4) $\frac { 15 } { 2 }$
(5) $\frac { 17 } { 2 }$
csat-suneung 2026 Q17 3 marks Antiderivative Verification and Construction View
For the function $f ( x ) = 4 x ^ { 3 } - 2 x$, let $F ( x )$ be an antiderivative with $F ( 0 ) = 4$. Find the value of $F ( 2 )$. [3 points]
gaokao 2015 Q11 Definite Integral Evaluation (Computational) View
11. $\int _ { 0 } ^ { 2 } ( x - 1 ) d x = $ $\_\_\_\_$.
grandes-ecoles 2010 QI.C.1 Convergence and Evaluation of Improper Integrals View
We denote by $E$ the vector space of polynomial functions on $\mathbb{R}$ and, for all $n \in \mathbb{N}$, we denote by $E_n$ the vector subspace of $E$ formed by polynomial functions of degree at most $n$.
Show that, for every pair $(f, g)$ of $E \times E$, the function: $$x \mapsto f(x) g(x) \frac{1}{\sqrt{1 - x^2}}$$ is integrable on the interval $]-1,1[$.
grandes-ecoles 2011 QI.A.1 Integral Inequalities and Limit of Integral Sequences View
Let $f$ be a real function, defined, continuous and decreasing on $[ a , + \infty [$, where $a \in \mathbb { R }$. Show that for every integer $k \in \left[ a + 1 , + \infty \left[ \right. \right.$, we have $\int _ { k } ^ { k + 1 } f ( x ) d x \leqslant f ( k ) \leqslant \int _ { k - 1 } ^ { k } f ( x ) d x$.
grandes-ecoles 2011 QVI.C Integral Inequalities and Limit of Integral Sequences View
For all $i \in \{1,2,3,4\}$, we set $$\theta(N_{i}) = \frac{1}{2N_{i}} + \int_{0}^{+\infty} \frac{h(u)}{(u+N_{i})^{2}} du$$
VI.C.1) Show that for all $i \in \{1,2,3,4\}$ $$0 < \theta(N_{i}) < \frac{1}{N_{i}}$$
VI.C.2) Show the existence of a strictly positive real $K$ such that for all $i \in \{1,2,3,4\}$ $$N_{i} = K e^{\mu \varepsilon_{i}} e^{-\theta(N_{i})}$$
grandes-ecoles 2011 QII.A Definite Integral Evaluation (Computational) View
We define the sequence of polynomials $\left(P_n\right)_{n \in \mathbb{N}}$ by: $$\left\{\begin{array}{l} P_0 = 1 \\ \forall n \in \mathbb{N}^*, \quad P_n = [X(X-1)]^n \end{array}\right.$$ Moreover, we set: $$\forall (P, Q) \in (\mathbb{R}[X])^2, \quad \langle P, Q \rangle = \int_0^1 P(t) Q(t) \, dt$$
Show that the map $(P, Q) \mapsto \langle P, Q \rangle$ is an inner product on $\mathbb{R}[X]$.
grandes-ecoles 2011 QII.D Definite Integral Evaluation (Computational) View
We define the sequence of polynomials $\left(P_n\right)_{n \in \mathbb{N}}$ by: $$\left\{\begin{array}{l} P_0 = 1 \\ \forall n \in \mathbb{N}^*, \quad P_n = [X(X-1)]^n \end{array}\right.$$ We define the sequence of polynomials $\left(L_n\right)_{n \in \mathbb{N}}$ by: $$\left\{\begin{array}{l} L_0 = 1 \\ \forall n \in \mathbb{N}^*, \quad L_n = \frac{1}{P_n^{(n)}(1)} P_n^{(n)} \end{array}\right.$$
II.D.1) For all $n \in \mathbb{N}$, we set $I_n = \int_0^1 P_n(u) \, du$.
Calculate, for all $n \in \mathbb{N}$, the value of $I_n$.
II.D.2) Deduce for all $n \in \mathbb{N}$ the relation: $\langle L_n, L_n \rangle = \frac{1}{2n+1}$.
grandes-ecoles 2011 QII.F Definite Integral Evaluation (Computational) View
We define the sequence of polynomials $\left(P_n\right)_{n \in \mathbb{N}}$ by: $$\left\{\begin{array}{l} P_0 = 1 \\ \forall n \in \mathbb{N}^*, \quad P_n = [X(X-1)]^n \end{array}\right.$$ Moreover, we set: $$\forall (P, Q) \in (\mathbb{R}[X])^2, \quad \langle P, Q \rangle = \int_0^1 P(t) Q(t) \, dt$$ The family $\left(K_n\right)_{n \in \mathbb{N}}$ is the unique family of polynomials such that for all $n \in \mathbb{N}$, the degree of $K_n$ equals $n$ with strictly positive leading coefficient, and for all $N \in \mathbb{N}$, $\left(K_n\right)_{0 \leqslant n \leqslant N}$ is an orthonormal basis of $\mathbb{R}_N[X]$ for $\langle \cdot, \cdot \rangle$.
Calculate $K_0$, $K_1$ and $K_2$.
grandes-ecoles 2011 QIII.B.1 Definite Integral Evaluation (Computational) View
For all $n \in \mathbb{N}^*$, we define the matrix $H_n$ by: $$\forall (i,j) \in \llbracket 1; n \rrbracket^2, \quad (H_n)_{i,j} = \frac{1}{i+j-1}$$ We extend to $C^0([0;1], \mathbb{R})$ the inner product $\langle \cdot, \cdot \rangle$ by setting $$\forall f, g \in C^0([0;1], \mathbb{R}), \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, dt$$ and we denote by $\|\cdot\|$ the associated norm.
Let $n \in \mathbb{N}$. Show that there exists a unique polynomial $\Pi_n \in \mathbb{R}_n[X]$ such that $$\left\|\Pi_n - f\right\| = \min_{Q \in \mathbb{R}_n[X]} \|Q - f\|$$
grandes-ecoles 2011 QIII.B.5 Definite Integral Evaluation (Computational) View
For all $n \in \mathbb{N}^*$, we define the matrix $H_n$ by: $$\forall (i,j) \in \llbracket 1; n \rrbracket^2, \quad (H_n)_{i,j} = \frac{1}{i+j-1}$$ We extend to $C^0([0;1], \mathbb{R})$ the inner product $\langle \cdot, \cdot \rangle$ by setting $$\forall f, g \in C^0([0;1], \mathbb{R}), \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, dt$$ For each $n \in \mathbb{N}$, $\Pi_n$ denotes the unique polynomial in $\mathbb{R}_n[X]$ minimizing $\|Q - f\|$ over $\mathbb{R}_n[X]$.
Determine explicitly $\Pi_2$ when $f$ is the function defined for all $t \in [0,1]$ by $f(t) = \frac{1}{1+t^2}$.
grandes-ecoles 2011 QIV.A.3 Definite Integral Evaluation (Computational) View
For $n \in \mathbb{N}^*$ and $(i,j) \in \llbracket 1, n \rrbracket^2$, we denote by $h_{i,j}^{(-1,n)}$ the coefficient at position $(i,j)$ of the matrix $H_n^{-1}$ and we denote by $s_n$ the sum of the coefficients of the matrix $H_n^{-1}$, that is: $$s_n = \sum_{1 \leqslant i,j \leqslant n} h_{i,j}^{(-1,n)}$$ We define, for all $n \in \mathbb{N}^*$, the polynomial $S_n$ by: $S_n = a_0^{(n)} + a_1^{(n)} X + \cdots + a_{n-1}^{(n)} X^{n-1}$, where $\left(a_p^{(n)}\right)_{0 \leqslant p \leqslant n-1}$ is the unique solution of the system in IV.A.2.
Show that $$\forall Q = \alpha_0 + \alpha_1 X + \cdots + \alpha_{n-1} X^{n-1} \in \mathbb{R}_{n-1}[X], \quad \langle S_n, Q \rangle = \sum_{p=0}^{n-1} \alpha_p$$
grandes-ecoles 2011 Q15 Definite Integral Evaluation (Computational) View
We consider a rectangle $]a,b[ \times ]c,d[$ of the plane $\mathbb{R}^{2}$, with $a < b$ and $c < d$. Calculate the real number $V(]a,b[ \times ]c,d[)$. What does it represent? (One may use functions of the type $$(x,y) \mapsto f(x,y) = \phi(x)\varphi(y)$$ where $\phi$ and $\varphi$ are well-chosen continuous and piecewise affine functions).
grandes-ecoles 2011 Q16 Integral Inequalities and Limit of Integral Sequences View
Let $\mathcal{A}$ and $\mathcal{B}$ be two open bounded non-empty subsets of $\mathbb{R}^{2}$ and $\lambda \in ]0,1[$. Verify that $\lambda\mathcal{A} + (1-\lambda)\mathcal{B}$ is an open bounded subset of $\mathbb{R}^{2}$. Then show that $$V(\lambda\mathcal{A} + (1-\lambda)\mathcal{B}) \geq V(\mathcal{A})^{\lambda} V(\mathcal{B})^{1-\lambda}$$ To prove this inequality, you will use the following admitted result. For all $f \in C(\mathcal{A})$ and $g \in C(\mathcal{B})$, the function $h$ determined by: $$\forall Z \in \mathbb{R}^{2},\quad h(Z) = \sup\left\{f(X)^{\lambda} g(Y)^{1-\lambda} \,/\, X, Y \in \mathbb{R}^{2},\, Z = \lambda X + (1-\lambda) Y\right\}$$ defines a continuous function on $\mathbb{R}^{2}$.
grandes-ecoles 2011 Q17 Integral Inequalities and Limit of Integral Sequences View
Let $u : \mathbb{R}^{2} \rightarrow ]0,+\infty[$ be a continuous and log-concave function in the sense of Part II. Prove that the preceding inequality remains true if we replace the application $V$ by the application $\gamma$ defined for all open bounded (non-empty) subsets $\mathcal{A}$ of $\mathbb{R}^{2}$ by $$\gamma(\mathcal{A}) = \sup_{f \in C(\mathcal{A})} \iint_{\mathbb{R}^{2}} f(x,y)\,u(x,y)\,dx\,dy$$
grandes-ecoles 2012 QII.E.1 Piecewise/Periodic Function Integration View
Let $g : [ 0,1 ] \rightarrow \mathbb { R }$ be the function such that $g ( x ) = 1 / x$ if $x \geqslant \mathrm { e } ^ { - 1 }$ and $g ( x ) = 0$ otherwise. We fix a real $\varepsilon \in ] 0 , \mathrm { e } ^ { - 1 } [$. We define two continuous applications $g ^ { + } , g ^ { - } : [ 0,1 ] \rightarrow \mathbb { R }$ as follows:
  • $g ^ { + }$ is affine on $\left[ \mathrm { e } ^ { - 1 } - \varepsilon , \mathrm { e } ^ { - 1 } \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } - \varepsilon \right] \cup \left[ \mathrm { e } ^ { - 1 } , 1 \right]$;
  • $g ^ { - }$ is affine on $\left[ \mathrm { e } ^ { - 1 } , \mathrm { e } ^ { - 1 } + \varepsilon \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } \left[ \cup \left[ \mathrm { e } ^ { - 1 } + \varepsilon , 1 \right] \right. \right.$.
Calculate $\int _ { 0 } ^ { 1 } g ^ { + } ( t ) \mathrm { d } t$ and $\int _ { 0 } ^ { 1 } g ^ { - } ( t ) \mathrm { d } t$.
grandes-ecoles 2013 QIII.A.2 Convergence and Evaluation of Improper Integrals View
Study the convergence of the improper integrals $$\int _ { 0 } ^ { \pi } \ln ( \sin \theta ) d \theta \quad \int _ { 0 } ^ { \pi } \ln ( 1 - \cos \theta ) d \theta \quad \int _ { 0 } ^ { \pi } \ln ( 1 + \cos \theta ) d \theta$$
Deduce that, for all $x \in \mathbb { R }$, the integral $\int _ { 0 } ^ { \pi } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) \mathrm { d } \theta$ converges.
grandes-ecoles 2013 QIII.A.3 Integral Inequalities and Limit of Integral Sequences View
Show that, as $x$ tends to $+ \infty$, $$2 \pi \ln ( x ) - \int _ { 0 } ^ { \pi } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) \mathrm { d } \theta$$ has a limit, which one will determine.
grandes-ecoles 2013 QIII.A.4 Integral Equation with Symmetry or Substitution View
Show that $x \mapsto \int _ { 0 } ^ { \pi } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) \mathrm { d } \theta$ is an even function of the variable $x \in \mathbb { R }$.