Lakshya JEE 2024 Math Lecture Notes PDF

## LAKSHYA JEE 2024 ### Lecture - 14 **Mathematics** LIMIT, CONTINUITY, DIFFERENTIABILITY By- ASHISH SIR ### Today's Targets 1. Practice Problems 2. Determination of Functions ### Question Let $f(x) = \begin{cases} x^2cos(\frac{1}{x}), & x < 0 \\ 0, & x = 0\\ x^2sin(\frac{1}{x}), & x > 0 \en...

## LAKSHYA JEE 2024 ### Lecture - 14 **Mathematics** LIMIT, CONTINUITY, DIFFERENTIABILITY By- ASHISH SIR ### Today's Targets 1. Practice Problems 2. Determination of Functions ### Question Let $f(x) = \begin{cases} x^2cos(\frac{1}{x}), & x < 0 \\ 0, & x = 0\\ x^2sin(\frac{1}{x}), & x > 0 \end{cases}$ then which of the following is(are) correct? A. f(x) is continuous but not differentiable at x = 0. B. f(x) is continuous and differentiable at x = 0. C. f'(x) is continuous but not differentiable at x = 0. D. f'(x) is discontinuous at x = 0. ### Solution **RHL at x=0** $\lim_{x \to 0^+} \frac{x^2sin(\frac{1}{x})}{x} = 0$ **LHL** $\lim_{x\to0^-} \frac{x^2cos(\frac{1}{x})}{x} = 0$ LHL = RHL = 0 = f(0) f is continuous at x=0 **RHD at x=0** $\lim_{h \to 0} \frac{h^2sin(\frac{1}{h})}{h} = 0$ **LHD at x=0** $\lim_{h \to 0} \frac{(-h)^2cos(-\frac{1}{h}) - 0}{h} = \lim_{h \to 0} -hcos\frac{1}{h} = 0$ LHD = RHD = 0 = f'(0) f is derivable at x=0 $f'(X) = \begin{cases} \frac{d(x^2cos(\frac{1}{x}))}{dx}, & x < 0 \\ 0, & x = 0\\ \frac{d(x^2sin(\frac{1}{x}))}{dx}, & x > 0 \end{cases}$ $f'(x)= \begin{cases} sinx + 2xcosx, & x < 0 \\ 0, & x = 0\\ x^2cosx(-\frac{1}{x^2}) + 2xsinx, & x > 0 \end{cases}$ $f'(x)= \begin{cases} sinx + 2xcosx, & x < 0 \\ 0, & x = 0\\ -cosx + 2xsinx, & x > 0 \end{cases}$ ### For f'(x) RHL, LHL at x=0 D.N.E f'(x) is discontinuous at x=0 f'(x) is not differentiable at x=0. ### Question Let f: [-1,2] -> R and g: [-1,2] -> R be functions defined by f(x) = [x^2 - 3] and g(x) = |x|f(x) + 4x - 7|f(x), where [y] denotes the greatest integer less than or equal to y for y ∈ R. Then [JEE Advanced 2016] A. f is discontinuous exactly at three points in [-1,2] B. f is discontinuous exactly at four points in [-1,2] C. g is NOT differentiable exactly at four points in (-1,2) D. g is NOT differentiable exactly at five points in (-1,2) ### Solution <start_of_image> f(x) = [x^2 - 3] f is discontinuous exactly at four points in [-1,2] $g(x) = |x|f(x) + 4x - 7|f(x) = (|x|+ 4x-7)f(x) = (|x|+ 4x-7)([x^2] - 3)$ $g(x)= (|x|+ 4x-7)([x^2] - 3)$ g is discontinuous at x = 1, √2, √3 g is also non differentiable at x = 1, √2, √3. Again $g(x)= (|x|+ 4x-7)([x^2] - 3)$ g is non diff at x = 0, 7/4. But continuous at x = 0, 7/4. g is not derivable at x = 0, 1, √2, √3. ### Question Let f₁: R → R, f2: [0, ∞) → R, f3: R → R and f4: R → [0,∞) be defined by $f_1(x) = \begin{cases} |x|, & x < 0 \\ e^x, & x ≥ 0 \end{cases}$ and $f_2(x) = x^2$ $f_3(x) = \begin{cases} sinx, & x < 0 \\ x, & x ≥ 0 \end{cases}$ and $f_4(x) = \begin{cases} f_2(f_1(x)), & x < 0 \\ f_2(f_1(x))-1, & x ≥ 0 \end{cases}$ [JEE Advanced 2014] | List-I | List-II | |:---|---| | (P) f4 is | (1) onto but not one-one | | (Q) f3 is | (2) neither continuous nor one-one | | (R) f2 o f₁ is | (3) differentiable but not one-one | | (S) f₂ is | (4) continuous and one-one | A. P → 3; Q → 1; R → 4; S → 2 B. P → 1; Q → 3; R → 4; S → 2 C. P → 3; Q → 1; R → 2; S → 4 D. P → 1; Q → 3; R → 2; S → 4 ### Solution $f_1(x) = \begin{cases} |x|, & x < 0 \\ e^x, & x ≥ 0 \end{cases}$ $f_2(x) = x^2$ $f_3(x) = \begin{cases} sinx, & x < 0 \\ x, & x ≥ 0 \end{cases}$ $f_4(x) = \begin{cases} f_2(f_1(x)), & x < 0 \\ f_2(f_1(x))-1, & x ≥ 0 \end{cases}$ $f_4(x) = \begin{cases} (-x)^2 & x < 0 \\ e^{2x} - 1 & x ≥ 0 \end{cases}$ f4 is onto but not one-one $f_3(x) = \begin{cases} sinx, & x < 0 \\ x, & x ≥ 0 \end{cases}$ f3 is neither continuous nor one-one $f_2 o f_1 (x) = f_2(f_1(x)) =\begin{cases} (-x)^2 & x < 0 \\ (e^x)^2 & x ≥ 0 \end{cases}$ $f_2 o f_1 (x) = f_2(f_1(x)) =\begin{cases} x^2 & x < 0 \\ e^{2x} & x ≥ 0 \end{cases}$ f2 o f1 is not continuous, not one-one and not differentiable $f_2(x) = x^2$ f2 is continuous and one-one P → 1; Q → 3; R → 2; S → 4 ### Question Let a, b ∈ R and f: R → R be defined by f(x) = a cos(|x³ – x|) + b|x|sin(|x³ + x|). Then f is [JEE Advanced 2016] A. differentiable at x = 0 if a = 0 and b = 1 B. differentiable at x = 1 if a = 1 and b = 0 C. NOT differentiable at x = 0 if a = 1 and b = 0 D. NOT differentiable at x = 1 if a = 1 and b = 1 ### Solution f(x) = a cos(|x³ – x|) + b|x|sin(|x³ + x|) $f(x) = \begin{cases} a cos(x^3-x) + bxsin(x(x+1)) & x ≥ 0 \\ a cos(x^3-x) + b(-x)sin(-x(x+1)) & x < 0 \end{cases}$ $f(x) = a cos(x^3-x) + bxsin(x(x+1))$ f(x) is continuous and derivable ### Question Number of points at which the function $f(x) = \begin{cases} min · (|x|, x^2) & if − ∞ < x < 1 \\ min. (2x – 1, x^2) & if x ≥ 1 \end{cases}$ is not derivable is ### Solution <start_of_image> solve y = x^2 and y = -x x = x^2 -> x = 0, 1 solve y = x^2 and y = 2x -1 We can see from the image that there is one point where y = x^2 and y = 2x -1 cross. solve y = x and y = x^2 x = x^2 -> x = 0, 1 solve y = x and y = 2x -1 dy/dx = 2x|x=1 = 2 Number of points of non diff is one only i.e. x=-1 ### Question Consider the function f(x) = Min(tan x, cotx) in (0,2π). Number of points where f fails to be derivable is 'm' and number of points where it is discontinuous in 'n'. Find (m, n) ### Solution From the image we can see that there are 7 points where f fails to be derivable and 3 points where it is discontinuous. Therefore (m, n) = (7, 3) ### Determination of function which are differentiable and satisfying the given functional rule **Basic Steps** 1. Write down the expression for f'(x) as $f'(x) = \frac{f(x+h)-f(x)}{h}$ 2. Manipulate f(x + h) – f(x) in such a way that the given functional rule is applicable. Now apply the functional rule and simplify the RHS to get f'(x) as a function of x along with constants if any. 3. Integrate f'(x) to get f(x) as a function of x and a constant of integration. In some cases a Differential Equation in formed which can be solved to get f(x). 4. Apply the boundary value conditions to determine the value of this constant. ### Question Suppose f is a derivable function that satisfies the equation f(x + y) = f(x) + f(y) + x²y + xy² for all real numbers x and y. Suppose that $lim_{x\to 0} \frac{f(x)}{x} = 1$, find [Ans. a) 0, b) 1, c) x² + 1, d) 12] (a) f(0) (b) f'(0) (c) f'(x) (d) f(3) ### Solution f(0) = 2 f(0) + 0 + 0 f(0) = 0 $f'(x) = lim_{h\to 0} \frac{f(x+h)-f(x)}{h} - 1$ Now f(x+y) = f(x) + f(y) + x^2y + xy^2 y = h f(x+h) = f(x) + f(h) + x^2h + xh^2 $f'(x) = lim_{h\to 0} \frac{f(x) + f(h) + x^2h + xh^2 - f(x)}{h}$ $f'(x)= lim_{h\to 0} \frac{f(h)}{h} + x^2 + xh = 1 + x^2$ $f'(x) = 1 + x^2$ $\int f'(x) dx =\int (1+x^2) dx$ f(x) = x + x^3/3 + C from f(0) = 0 0 = 0 +0 + c -> c = 0 f(x) = x + x^3/3 f(3) = 3 + 27/3 = 12 ### Question A differentiable function satisfies the relation f(x + y) = f(x) + f(y) + 2xy − 1 ∀ x, y ∈ R . If '(0) = √3 + a – a² find f(x) and prove that f(x) > 0 ∀ x ∈ R [Ans. ?] ### Solution $f'(x)= lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ $f'(x)= lim_{h\to 0} \frac{f(x)+f(h) + 2xh - 1 - f(x)}{h}$ $f'(x)= lim_{h\to 0} \frac{f(h) + 2xh - 1}{h}$ $f'(x)= lim_{h \to 0} \frac{f(h) - f(0) + 2xh}{h}$ $f'(x)= lim_{h \to 0} \frac{f(h) - f(0)}{h} + 2x = f'(0) + 2x$ f'(x) = f'(0) + 2x f'(x) = √3 + a - a^2 + 2x f(x) = x√3 + a - a^2 + x^2 + c Now f(0) = 1 f(0) = 0 + 0 + c c = 1 f(x) = x√3 + a - a^2 + x^2 + 1 f(x) = x^2 + x√3 + a - a^2 + 1 $D= (3 + a - a^2) - 4 .1 .1 = -a^2 + a + 3 - 4 = -a^2 + a - 1$ $D= -(a^2 - a + 1) = -(a - 1/2)^2 + 3/4 < 0$ Therefore, f(x) > 0 ∀ x ∈ R ### Question Let f: R → R be a differentiable function with f(0) = 1 and satisfying the equation f(x + y) = f(x)f' (y) + f'(x)f(y) for all x, y ∈ R. Then, the value of loge (f(4)) is [JEE Advanced 2018] ### Solution put y = 0 f(x) = f(x)f'(0) + f'(x)f(0) -> f(x) = f(x)f'(0) + f'(x) put x = 0, y = 0 f(0 + 0) = f(0)f'(0) + f'(0)f(0) 1 = 1. f'(0) + f'(0).1 2f'(0) = 1 f'(0) =1/2. $f'(x) = \frac{f'(x)}{f(x)} = \frac{1}{2}$ $\int \frac{f'(x) dx}{f(x)} = \int \frac{1}{2} dx$ $lnf(x) = \frac{1}{2}x + c$ put x = 0, f(0) = 1 lnf(0) = 1/2.0 + c c = enf(0) = ln1 = 0 lnf(x) = 1/2x ln(f(4)) = 4/2 = 2.