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In the section each question has some statements (A, B,C,D,…) given in Column – I and some statements (p,q,r,s,t,…) in column – II. Any given statement is column – I can have correct matching with ONE OR MORE statement(s) in column – II for example, if for a given question, statement B matches with the statements given in q and r, then for that particular question against statement B, darken the bubbles corresponding to q and r in the ORS. i.e., answer r will be q and r

1) A → p; B → r 2) A → r; B → q 3) A → p; B → q 4) A → q; B → r Solution: (1) A → p; B → r A → p; B → r (A) f (x) = x + sin x... View Article

In the section, each question has some statements (A, B, C, D,…) given in Column – I and some statements (p,q,r,s,t,…) in column – II. Any given statement is column – I can have correct matching with ONE OR MORE statement(s) in the column – II for example, if for a given questions, statement B matches with the statements given in q and r, then for that particular question against statement B, darken the bubbles corresponding to q and r in the ORS. i.e., answer r will be q and r. In the following [x] denotes the greatest integer less than or equal to x. Match the functions in Column – I with the properties in the column – II

In the following [x] denotes the greatest integer less than or equal to x. Match the functions in Column – I with the properties in the column –... View Article

In the section, each question has some statements (A, B, C, D,…) given in Column– I and some statements (p,q,r,s,t,…) in the column – II. Any given statement is column – I can have correct matching with ONE OR MORE statement(s) in the column – II for example, if for a given questions, statement B matches with the statements given in q and r, then for that particular question against statement B, darken the bubbles corresponding to q and r in the ORS. i.e., answer r will be q and r. Match that statements are given in Column – I with the intervals/union of intervals given in Column – II

Match that statements are given in Column – I with the intervals/union of intervals given in Column – II 1) A→r; B→q, s; C→r, s; D→p, r... View Article

Let f (x) = (1 – x)2 + sin2 x + x2 for all x ∈ IR, and g (x) = integral from 1 to x [2 (t – 1)] / [t + 1] – ln t] f (t) dt for all x ∈ (1, ∞]. Consider the statements. P: There exists some x ∈ IR such that f (x) + 2x = 2 (1 + x2). Q: There exists some x ∈ IR such that 2 f (x) + 1 = 2x (1 + x). Then,

Consider the statements P: There exists some x ∈ IR such that f (x) + 2x = 2 (1 + x2) Q: There exists some x ∈ IR such that 2 f (x) + 1 = 2x (1... View Article

It is a continuous function f defined on the real line R, assume positive and negative values in R then the equation f(x) = 0 has root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum value is negative then the equation f(x) = 0 has a root in R. Consider f(x) = ke^x – x for all real x where k is a real constant. For k > 0, the set of all values of k for which ke^x – x = 0 has two distinct roots is

1) (0, 1 / e) 2) (1 / e, 1) 3) (1 / e, ∞) 4) (0, 1) Solution: (1) (0, 1 / e) If k > 0, then for 2 distinct roots local minima... View Article

It is a continuous function f defined on the real line R, assume positive and negative values in R then the equation f(x) = 0 has root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum value is negative then the equation f(x) = 0 has a root in R. Consider f(x) = kex – x for all real x where k is a real constant. The positive value of k for which kex – x = 0 has only one root is

The positive value of k for which kex – x = 0 has only one root is 1) 1 / e 2) 1 3) e 4) loge 2 Solution: (1) 1 / e If k > 0,... View Article

It is a continuous function f defined on the real line R, assume positive and negative values in R then the equation f(x) = 0 has root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum value is negative then the equation f(x) = 0 has a root in R. Consider f (x) = kex – x for all real x where k is a real constant.The line y = x meets y = kex for k ≤ 0 at

1) No point 2) One point 3) Two points 4) More than two points Solution: (2) One point f (x) = kex – x f’ (x) = kex - 1 < 0 if... View Article