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JEE Advanced Mathematics Application of Derivatives 2025 JEE Advanced 2025 (Paper 1)

JEE Advanced Mathematics Question (2025) — Solution

Question

Let R denote the set of all real numbers. For a real number x, let [x] denote the greatest integer less than or equal to x. Let n denote a natural number. Match each entry in List-I to the correct entry in List-II and choose the correct option. LIST - I LIST - II (P) The minimum value of n for which the function f(x)= [ 10 x^3-45 x^2+60 x+35 n ] is continuous on the interval [1,2],is (1) 8 (Q) The minimum value of n for which g(x)= (2 n^2-13 n-15 ) (x^3+3 x ),x R ,is an increasing function on R ,is (2) 9 (R) The smallest natural number n which is greater than 5,such that x=3 is a point of local minima of h(x)= (x^2-9 )^ n (x^2+2 x+3 ),is (3) 5 (S) Number of x_0 R such that l(x)= _ k=0 ^4 ( |x-k|+ |x-k+ 1 2 | ),x R ,is NOT differentiable at x_0,is (4) 6 (5) 10

Options

  1. A. ( P ) (1),( Q ) (3),( R ) (2),( S ) (5)
  2. B. ( P ) (2),( Q ) (1),( R ) (4),( S ) (3)
  3. C. ( P ) (5),( Q ) (1),( R ) (4),( S ) (3)
  4. D. ( P ) (2),( Q ) (3),( R ) (1),( S ) (5)

Answer

B. ( P ) (2),( Q ) (1),( R ) (4),( S ) (3)

Step-by-step solution

aligned & P(x)=10 x^3-45 x^2+60 x+35 \\ & P^ (x)=30(x-1)(x-2) aligned P ( x ) decreases in [1,2] Range of P ( x )=[55,60] f( x )= [ P ( x ) n ] value of n =9 (Q) For g(x) to be increasing 2 n^2-13 n-15 0 (R) aligned & h(x)= (x^2-9 )^n (x^2+2 x+3 ) \\ & h^ (x)= (x^2-9 )^n(2 x+2)+ (x^2+2 x+3 ) n (x^2-9 )^ n-1 2 x \\ & = (x^2-9 )^ n-1 [2 (x^2-9 )(x+1)+2 n x (x^2+2 x+3 ) ] \\ & =(x+3)^ n-1 (x-3)^ n-1 q(x) aligned Derivative must change sign at x=3 aligned & n -1= odd \\ & n = even \\ & n =6 aligned (S) | x - k + 1 2 | is differentiable everywhere | x - k | is NOT diff. at k =0,1,2,3,4

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