lear CALCULUS 13 f: R- (0) R given by f(x) = 8. EXERCISE - can be...

 

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lear CALCULUS 13 f: R- (0) R given by f(x) = 8. EXERCISE - can be made continuous at x = 0 by delining u log(3+x)- log (3-x)-k, the value f0) is If x0 2) 2 3) -1 4) 0 1) ! - JI- px of kis if -Isr<0 2) 3) 4)-3 1) 0 f(x) = 2x-1 if OsxsI is X-2 continuous on 1-1, 1] then p is 2) 3) e 4) e 12 1) (1-cos x)sin 5x 1) 0 2) -1 3) 1/2 4) -12 sin 3x 3 2) 10 10. Observe the following lists: List - I 10 4) I) 3) then f(x) is A) fx) = x + [x] at x = Ji- cos 2x Ex B) fx) = x #0. f(0) = 1 X-40 C) fx)= Tan(x+2) ++2] 3) 0 1) 1 4) doesnot exist 2) -1 x*-2 and RSI Statement I: VA+4x+1-x D) f(x)= xCos **0.f(0)=1 X-3 Statement II: Li loga = log, 6 List - II Va +6-3 -cos( Sin Statement III: x-1/2 1-Tan Sin I. Continuous II. right continuous then f(x) ut x = 0 is III. Left continuous f(-2) = -1, then at x = - 2, f(x) is Which of the above is/are correct IV. Discontinuous then f(x) at x=0 is 1) only 1, II 3) 1, II, II 2) only II, II 4) only 1. II The value of f at x = 0 so that the function The correct match is : A В С D A B II IV 2)I C D III IV I) I II II 2-2 (x0) is continuous 3) IV III II I 4) IV II III 1 at x=0 is 11. f(x)=COSX- sin x cos 2x is continuous af 4 I) log 4 2) e4 3) log 2 4) 0 72-94-8+1 Let f(x)= 2-+cosx then 4. %3D klog 2 log 3; If f is continuous function at x = 0, then k = x=0 E 3) J2 4)-VE 2) 24 12. If f: R -R is continuous such that f(r+y 3)185 4) 242 f(x).f(y) V X, y e R and f(1) = 2, then f(100) 1) 100 104 2) 101 31 200 lear CALCULUS 13 f: R- (0) R given by f(x) = 8. EXERCISE - can be made continuous at x = 0 by delining u log(3+x)- log (3-x)-k, the value f0) is If x0 2) 2 3) -1 4) 0 1) ! - JI- px of kis if -Isr<0 2) 3) 4)-3 1) 0 f(x) = 2x-1 if OsxsI is X-2 continuous on 1-1, 1] then p is 2) 3) e 4) e 12 1) (1-cos x)sin 5x 1) 0 2) -1 3) 1/2 4) -12 sin 3x 3 2) 10 10. Observe the following lists: List - I 10 4) I) 3) then f(x) is A) fx) = x + [x] at x = Ji- cos 2x Ex B) fx) = x #0. f(0) = 1 X-40 C) fx)= Tan(x+2) ++2] 3) 0 1) 1 4) doesnot exist 2) -1 x*-2 and RSI Statement I: VA+4x+1-x D) f(x)= xCos **0.f(0)=1 X-3 Statement II: Li loga = log, 6 List - II Va +6-3 -cos( Sin Statement III: x-1/2 1-Tan Sin I. Continuous II. right continuous then f(x) ut x = 0 is III. Left continuous f(-2) = -1, then at x = - 2, f(x) is Which of the above is/are correct IV. Discontinuous then f(x) at x=0 is 1) only 1, II 3) 1, II, II 2) only II, II 4) only 1. II The value of f at x = 0 so that the function The correct match is : A В С D A B II IV 2)I C D III IV I) I II II 2-2 (x0) is continuous 3) IV III II I 4) IV II III 1 at x=0 is 11. f(x)=COSX- sin x cos 2x is continuous af 4 I) log 4 2) e4 3) log 2 4) 0 72-94-8+1 Let f(x)= 2-+cosx then 4. %3D klog 2 log 3; If f is continuous function at x = 0, then k = x=0 E 3) J2 4)-VE 2) 24 12. If f: R -R is continuous such that f(r+y 3)185 4) 242 f(x).f(y) V X, y e R and f(1) = 2, then f(100) 1) 100 104 2) 101 31 200