Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Differentiation question

2023 · 13 Apr · Shift 1 · Q35
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Differentiation
  5. /2023 · 13 Apr · Shift 1 · Q35

Differentiation question

2023 · 13 Apr · Shift 1 · Q35

JEE MainMathematicsDifferentiationMCQ+4 / −1
For the differentiable function f:R−{0}→Rf: \mathbb{R}-\{0\} \rightarrow \mathbb{R}f:R−{0}→R, let 3f(x)+2f(1x)=1x−103 f(x)+2 f\left(\frac{1}{x}\right)=\frac{1}{x}-103f(x)+2f(x1​)=x1​−10, then ∣f(3)+f′(14)∣\left|f(3)+f^{\prime}\left(\frac{1}{4}\right)\right|​f(3)+f′(41​)​ is equal to
  1. A
    13
  2. B
    295\frac{29}{5}529​
  3. C
    335\frac{33}{5}533​
  4. D
    7
View written solutionFree

Correct answer: A

  1. We are given 3f(x)+2f(1x)=1x−10(x≠0).3f(x)+2f\left(\frac{1}{x}\right)=\frac{1}{x}-10 \qquad (x\neq 0).3f(x)+2f(x1​)=x1​−10(x=0). We need to find ∣f(3)+f′(14)∣.\left|f(3)+f'\left(\frac14\right)\right|.​f(3)+f′(41​)​.

  2. First, replace xxx by 1x\dfrac{1}{x}x1​ in the given equation: 3f(1x)+2f(x)=x−10.3f\left(\frac{1}{x}\right)+2f(x)=x-10.3f(x1​)+2f(x)=x−10.

So we have the system: 3f(x)+2f(1x)=1x−10...(1)3f(x)+2f\left(\frac{1}{x}\right)=\frac{1}{x}-10 \qquad ...(1)3f(x)+2f(x1​)=x1​−10...(1) 2f(x)+3f(1x)=x−10...(2)2f(x)+3f\left(\frac{1}{x}\right)=x-10 \qquad ...(2)2f(x)+3f(x1​)=x−10...(2)

  1. Let a=f(x),b=f(1x).a=f(x), \qquad b=f\left(\frac{1}{x}\right).a=f(x),b=f(x1​). Then 3a+2b=1x−103a+2b=\frac{1}{x}-103a+2b=x1​−10 2a+3b=x−10.2a+3b=x-10.2a+3b=x−10.

Solve for aaa. Multiply the first by 333 and the second by 222: 9a+6b=3x−309a+6b=\frac{3}{x}-309a+6b=x3​−30 4a+6b=2x−20.4a+6b=2x-20.4a+6b=2x−20. Subtract: 5a=3x−2x−105a=\frac{3}{x}-2x-105a=x3​−2x−10 f(x)=a=15(3x−2x−10).f(x)=a=\frac{1}{5}\left(\frac{3}{x}-2x-10\right).f(x)=a=51​(x3​−2x−10).

Thus, f(x)=35x−2x5−2.f(x)=\frac{3}{5x}-\frac{2x}{5}-2.f(x)=5x3​−52x​−2.

  1. Now compute f(3)f(3)f(3): f(3)=315−65−2=15−65−2=−1−2=−3.f(3)=\frac{3}{15}-\frac{6}{5}-2=\frac{1}{5}-\frac{6}{5}-2=-1-2=-3.f(3)=153​−56​−2=51​−56​−2=−1−2=−3.

  2. Differentiate f(x)f(x)f(x): f(x)=35x−1−25x−2f(x)=\frac{3}{5}x^{-1}-\frac{2}{5}x-2f(x)=53​x−1−52​x−2 f′(x)=−35x−2−25=−35x2−25.f'(x)=-\frac{3}{5}x^{-2}-\frac{2}{5}=-\frac{3}{5x^2}-\frac{2}{5}.f′(x)=−53​x−2−52​=−5x23​−52​.

Now at x=14x=\frac14x=41​, f′(14)=−35⋅(1/16)−25=−485−25=−505=−10.f'\left(\frac14\right)=-\frac{3}{5\cdot (1/16)}-\frac{2}{5}=-\frac{48}{5}-\frac{2}{5}=-\frac{50}{5}=-10.f′(41​)=−5⋅(1/16)3​−52​=−548​−52​=−550​=−10.

  1. Therefore, f(3)+f′(14)=−3+(−10)=−13.f(3)+f'\left(\frac14\right)=-3+(-10)=-13.f(3)+f′(41​)=−3+(−10)=−13. Hence, ∣f(3)+f′(14)∣=∣−13∣=13.\left|f(3)+f'\left(\frac14\right)\right|=|-13|=13.​f(3)+f′(41​)​=∣−13∣=13.

  2. So the correct option is: A  (13)\boxed{A\; (13)}A(13)​

  3. Comparison with stored answer: Stored correct answer is A, which matches our result.

PreviousNext

More from Differentiation

  • Let f(x)=∑k=110​kxk,x∈R. If 2f(2)+f′(2)=119(2)n+1 then n is equal to ​2023 · Numerical
  • If f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3),x∈R, then2023 · MCQ
  • Let y(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y′−y′′ at x=−1 is equal to2023 · MCQ
  • Let f:R→R be a differentiable function that satisfies the relation f(x+y)=f(x)+f(y)−1,∀x,y∈R. If f′(0)=2, then ∣f(−2)∣ is equal to ​.2023 · Numerical
  • Let f and g be the twice differentiable functions on R such that f′′(x)=g′′(x)+6xf′(1)=4g′(1)−3=9f(2)=3g(2)=12. Then which of the following is NOT true?2023 · MCQ
  • Let f1(x)=2x+33x+2​,x∈R−{2−3​} For n≥2, define fn(x)=f1ofn−1(x). If f5(x)=bx+aax+b​,gcd(a,b)=1…2023 · Numerical
  • Let y=f(x)=sin3(3π​(cos(32​π​(−4x3+5x2+1)23​))). Then, at x = 1,2023 · MCQ
  • If y=tan−1(secx3−tanx3),2π​<x3<23π​, then2022 · MCQ