Let f: R to (0, infty) be a twice differentiable function such that f(3) = 18 , f'(3) = 0 and f''(3) = 4 . Then lim_x to 1 left( log_e left( fracf(2 + x)f(3) right)^frac18(x-1)^2 right) is equal to:

Solution & Explanation

### Related Formula For a limit of 1^infty form, lim_x to a [g(x)]^h(x) equals: e^lim_x to a h(x)[g(x) - 1] ### Core Logic Let T = lim_x to 1 left( fracf(x + 2)f(3) right)^frac18(x - 1)^2. As x to 1, fracf(x+2)f(3) to fracf(3)f(3) = 1. The exponent goes to infty. This is a standard 1^infty form. T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)f(3) right) ### Step 1: Simplify Exponent Limit Given f(3) = 18: T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)18 right) T = e^lim_x to 1 fracf(x + 2) - f(3)(x - 1)^2 This is a frac00 form limit. ### Step 2: Apply L'Hopital's Rule Differentiate numerator and denominator w.r.t x: T = e^lim_x to 1 fracf'(x + 2)2(x - 1) This is still a frac00 form since f'(3) = 0. Apply L'Hopital's Rule again: T = e^lim_x to 1 fracf''(x + 2)2 Substitute x = 1: T = e^fracf''(3)2 ### Step 3: Final Calculation Given f''(3) = 4: T = e^frac42 = e^2 The question asks for log_e(T): log_e(T) = log_e(e^2) = 2 ### Pattern Recognition When expanding f(x) around an extrema (f'(a)=0) inside a 1^infty limit form, double L'Hopital isolates the second derivative divided by the Taylor factorial (2! = 2). The limit elegantly shrinks directly to f''(a)/2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Limits, Continuity and Differentiability

Reference Study Guides

Limits Continuity And Differentiability

More Limits, Continuity and Differentiability Previous-Year Questions — Page 7

Q18 jee_main_2024_27_jan_morning Standard Limits
If a=lim_xrightarrow0fracsqrt1+sqrt1+x^4-sqrt2x^4 and b=lim_xrightarrow0fracsin^2xsqrt2-sqrt1+cos x, then the value of ab^3 is :
  • A. 36
  • B. 32
  • C. 25
  • D. 30

Solution

### Related Formula lim_x to 0 fracsin xx = 1 Rationalization: (u-v)(u+v) = u^2 - v^2 ### Core Logic Evaluate limit a by rationalizing the numerator: a = lim_x to 0 fracsqrt1+sqrt1+x^4-sqrt2x^4 Multiply by conjugate: a = lim_x to 0 frac(1+sqrt1+x^4) - 2x^4 (sqrt1+sqrt1+x^4 + sqrt2) a = lim_x to 0 fracsqrt1+x^4 - 1x^4 (sqrt1+sqrt1+x^4 + sqrt2) Rationalize again: a = lim_x to 0 frac(1+x^4) - 1x^4 (sqrt1+sqrt1+x^4 + sqrt2) (sqrt1+x^4 + 1) Cancel x^4: a = lim_x to 0 frac1(sqrt1+sqrt1+0 + sqrt2) (sqrt1+0 + 1) a = frac1(sqrt2 + sqrt2)(1 + 1) = frac14sqrt2 ### Step 1: Evaluating Limit b Evaluate limit b by rationalizing the denominator: b = lim_x to 0 fracsin^2 xsqrt2-sqrt1+cos x Multiply by conjugate: b = lim_x to 0 fracsin^2 x (sqrt2 + sqrt1+cos x)2 - (1+cos x) b = lim_x to 0 frac(1-cos^2 x)(sqrt2 + sqrt1+cos x)1 - cos x Using 1-cos^2 x = (1-cos x)(1+cos x): b = lim_x to 0 (1+cos x)(sqrt2 + sqrt1+cos x) Apply limit x to 0 (so cos 0 = 1): b = (1+1)(sqrt2 + sqrt1+1) = 2(2sqrt2) = 4sqrt2 ### Step 2: Final Output Calculate the value of ab^3: ab^3 = left(frac14sqrt2right) times (4sqrt2)^3 ab^3 = frac(4sqrt2)^34sqrt2 = (4sqrt2)^2 ab^3 = 16 times 2 = 32 ### Pattern Recognition Double square-root structures require double rationalization. Do not rush to L'Hopital's rule when roots are stacked; iterative conjugation resolves x^n terms naturally. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 11 Maths: Trigonometric Functions
Q6 jee_main_2024_29_jan_morning L'Hopital's Rule with Integration
lim_xrightarrowfracpi2 fracint_x^3^(fracpi2)^3 cos(t^1/3) dt(x-fracpi2)^2 is equal to
  • A. frac3pi8
  • B. frac3pi^24
  • C. frac3pi^28
  • D. frac3pi4

Solution

### Related Formula textNewton-Leibniz Formula: fracddx int_h(x)^g(x) f(t)dt = f(g(x)) cdot g'(x) - f(h(x)) cdot h'(x) lim_x to a fracf(x)g(x) = lim_x to a fracf'(x)g'(x) quad text(L'Hopital's Rule for frac00 text forms) ### Core Logic Evaluate the limit L = lim_xrightarrowfracpi2 fracint_x^3^(pi/2)^3 cos(t^1/3) dt(x-fracpi2)^2. When x to fracpi2, the integral limits become from (fracpi2)^3 to (fracpi2)^3, so the numerator is 0. The denominator evaluates to 0^2 = 0. This is a frac00 form, meaning L'Hopital's rule must be applied. Differentiate the numerator using Newton-Leibniz theorem: N'(x) = fracddx left[ int_x^3^(pi/2)^3 cos(t^1/3) dt right] = cosleft(left((pi/2)^3right)^1/3right) cdot 0 - cosleft((x^3)^1/3right) cdot fracddx(x^3) = 0 - cos(x) cdot 3x^2 = -3x^2 cos(x) Differentiate the denominator: D'(x) = fracddxleft[ (x-fracpi2)^2 right] = 2(x-fracpi2) ### Step 1: Simplify and Re-evaluate Limit Substitute the derivatives back into the limit expression: L = lim_xrightarrowfracpi2 frac-3x^2 cos x2(x-fracpi2) Notice that cos(x) = sin(fracpi2 - x) = -sin(x - fracpi2). Substituting this equivalence: L = lim_xrightarrowfracpi2 frac-3x^2 cdot (-sin(x - fracpi2))2(x-fracpi2) L = lim_xrightarrowfracpi2 left[ fracsin(x-fracpi2)x-fracpi2 right] times left[ frac3x^22 right] ### Step 2: Apply Standard Limit Since lim_theta to 0 fracsin thetatheta = 1, where theta = x - fracpi2: L = 1 times frac3(pi/2)^22 L = frac3 cdot fracpi^242 = frac3pi^28 ### Pattern Recognition Integral over a variable boundary over a 0-yielding polynomial denominator is the classic signal for the Newton-Leibniz differentiation combined with L'Hopital's rule. Watch out for shifting cos x to -sin(x - fracpi2) to match the denominator structure for standard trigonometric limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limit and Continuity Class 12 Mathematics: Integral Calculus
Q19 jee_main_2024_29_jan_morning First Principle of Differentiation
Suppose f(x)=frac(2^x+2^-x)tan xsqrttan^-1(x^2-x+1)(7x^2+3x+1)^3. Then the value of f'(0) is equal to
  • A. pi
  • B. 0
  • C. sqrtpi
  • D. fracpi2

Solution

### Related Formula f'(0) = lim_h to 0 fracf(h) - f(0)h Standard Limits: lim_h to 0 fractan hh = 1 ### Core Logic First, evaluate f(0) to ensure the first principle approach simplifies: f(0) = frac(2^0 + 2^-0)tan(0)sqrttan^-1(0-0+1)(0+0+1)^3 Since tan(0) = 0, the entire numerator collapses, giving f(0) = 0. Set up the limit definition of the derivative at x = 0: f'(0) = lim_h to 0 fracf(h) - 0h f'(0) = lim_h to 0 frac1h left( frac(2^h + 2^-h)tan hsqrttan^-1(h^2-h+1)(7h^2+3h+1)^3 right) ### Step 1: Group Standard Limit Forms Regroup the expression to isolate the known limit forms: f'(0) = lim_h to 0 left( fractan hh right) times left( 2^h + 2^-h right) times fracsqrttan^-1(h^2-h+1)(7h^2+3h+1)^3 Now, evaluate the limit of each independent non-zero segment as h to 0: 1. lim_h to 0 fractan hh = 1 2. lim_h to 0 (2^h + 2^-h) = 2^0 + 2^-0 = 1 + 1 = 2 3. lim_h to 0 sqrttan^-1(h^2-h+1) = sqrttan^-1(1) = sqrtfracpi4 = fracsqrtpi2 4. lim_h to 0 (7h^2+3h+1)^3 = (0+0+1)^3 = 1 ### Step 2: Combine Limits Multiply the evaluated continuous components together: f'(0) = 1 times 2 times fracfracsqrtpi21 f'(0) = sqrtpi ### Pattern Recognition If you are asked to find f'(0) for a massive, horrifying fraction where f(0)=0 (usually due to a rogue sin x, tan x, or x term), completely ignore the quotient rule. Use the first principle formula lim_hto0 f(h)/h to instantly isolate standard limit identities and plug 0 into everything else. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limit and Continuity Class 11 Mathematics: Derivatives
Q14 jee_main_2024_30_january_evening Differentiability
Let a and b be real constants such that the function f defined by f(x) = begincases x^2 + 3x + a, & x le 1 \\ bx + 2, & x gt 1 endcases be differentiable on mathbbR. Then, the value of int_-2^2f(x)dx equals
  • A. frac156
  • B. frac196
  • C. 21
  • D. 17

Solution

### Related Formula textFor differentiability at x=c:\\ lim_x to c^- f(x) = lim_x to c^+ f(x) quad text(Continuity)\\ lim_x to c^- f'(x) = lim_x to c^+ f'(x) quad text(Differentiability) ### Core Logic Function f(x) is continuous at x=1: lim_x to 1^- (x^2 + 3x + a) = lim_x to 1^+ (bx + 2) 1 + 3 + a = b + 2 Rightarrow 4 + a = b + 2 Rightarrow a = b - 2 quad dots(i) Function f(x) is differentiable at x=1: f'(x) = begincases 2x + 3, & x lt 1 \\ b, & x gt 1 endcases Equating left-hand and right-hand derivatives at x=1: 2(1) + 3 = b Rightarrow b = 5 Substitute b = 5 into (i): a = 5 - 2 = 3 ### Step 1: Setting up the Integral Now we have the full function: f(x) = begincases x^2 + 3x + 3, & x le 1 \\ 5x + 2, & x gt 1 endcases We need to evaluate int_-2^2 f(x) dx: I = int_-2^1 (x^2 + 3x + 3) dx + int_1^2 (5x + 2) dx ### Step 2: Evaluating the Integrals First integral: int_-2^1 (x^2 + 3x + 3) dx = left[ fracx^33 + frac3x^22 + 3x right]_-2^1 = left( frac13 + frac32 + 3 right) - left( frac-83 + frac122 - 6 right) = left( frac13 + frac32 + 3 right) - left( frac-83 + 0 right) = frac93 + frac32 + 3 = 3 + frac32 + 3 = frac152 Second integral: int_1^2 (5x + 2) dx = left[ frac5x^22 + 2x right]_1^2 = left( frac202 + 4 right) - left( frac52 + 2 right) = 14 - frac92 = frac192 Total sum: I = frac152 + frac192 = frac342 = 17 ### Pattern Recognition Piecewise unknown parameters are locked by continuity first, then differentiability. Splitting the integral limit at the critical node correctly processes the integration paths. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Continuity and Differentiability Class 12 Maths: Integral Calculus
Q20 jee_main_2024_30_jan_morning Limits
Let f:left[-fracpi2,fracpi2right] to mathbbR be a differentiable function such that f(0) = frac12. If the lim_x to 0 fracx int_0^x f(t) dte^x^2 - 1 = alpha, then 8alpha^2 is equal to:
  • A. 16
  • B. 2
  • C. 1
  • D. 4

Solution

### Related Formula lim_y to 0 frace^y - 1y = 1 Leibniz Integral Rule: fracddx int_0^x f(t) dt = f(x) ### Core Logic Given limit is: alpha = lim_x to 0 fracx int_0^x f(t) dte^x^2 - 1 Multiply and divide the denominator by x^2 to use standard exponential limit: alpha = lim_x to 0 fracx int_0^x f(t) dtleft(frace^x^2 - 1x^2right) cdot x^2 Since lim_xto 0 frace^x^2 - 1x^2 = 1, the expression simplifies to: alpha = lim_x to 0 fracx int_0^x f(t) dt1 cdot x^2 = lim_x to 0 fracint_0^x f(t) dtx ### Step 1: Applying L'Hôpital's Rule This is a 0/0 form. Apply L'Hôpital's Rule by differentiating numerator and denominator w.r.t x: alpha = lim_x to 0 fracfracddx int_0^x f(t) dtfracddx(x) = lim_x to 0 fracf(x)1 By continuity of differentiable function f at 0: alpha = f(0) ### Step 2: Final Calculation We are given f(0) = frac12, so alpha = frac12. We need to find 8alpha^2: 8alpha^2 = 8 left(frac12right)^2 = 8 left(frac14right) = 2 ### Pattern Recognition Standard expansion/limits on isolated terms in denominators immediately reduce the power of x, setting up a trivial Leibniz derivative application. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Integrals

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