Title Logarithm & Modulus Function( A and R).pdf ✅

Sol.[D] The reason R, being cancellation law, is true but in the assertion A there is a mistake of cancelling log 2 1 . Since log 2 1 < 0 that is why, an absurd result 5 < 4 has appeared. Thus assertion is false. Q.10 Assertion (A) : The least value of the expression 2 log10x – logx(0.01) for x > 0 is 4. Reason (R) : Sum of positive number and its reciprocal is always greater than or equal to 2. Sol.[D] A is false, R is true 2 log10x + 2 logx10  4 if x > 1 So A is false. Q.11 If 0 <  < 1 then Assertion (A) : log2 > log3 > loge > log10 Reason (R) : logxy > logx z  y > z if x > 1  y < z if 0 < x < 1 Sol.[D] A is false, R is true log2 > log3 > loge > log10  log2  < log3  < loge  < log10 Q.12 Assertion : log ( 13 12) log ( 14 13) 10 −  0.1 − Reason : (i) If a > 1, then x > 1  logax > 0 and 0 < x < 1  loga x < 0 (ii) If 0 < a < 1, then x > 1  logax < 0 and 0 < x < 1  logax > 0 Sol. [A] Q.13 Assertion (A) : log10x < log3x < logex < log2x, Reason (R) : If 0 < x < 1 then logxa > logxb  a < b Sol. [D] Questions Add (23-10-09) Q.14 Assertion (A) : log2 (2 17 – 2x ) = 1 – log1/2(x–1) has a solution. Reason (R) : change of base in log is possible. Sol.[B] (A) and (R) are true but (R) is not correct explanation of (A) Q.15 Assertion (A) : logex > logey x>y where x,y > 0 Reason (R) :If logax >logay, then x> y for a > 0, a  1, x, y > 0 Sol.[C]  loga x > logay  x > y if a > 1 Q.16 Statement-1 : log(2 + sin 2x) (x4 + x2 + 1)esin x > 0. Statement-2 : Product of non-negative expressions can not assume negative value. Sol.[D] Q.17 Assertion (A) : The equation log3(5 + 4 log3(x – 1)) = 2 has only one solution. Reason (R) : 2 x ;x 0 log3 x log3 2 = = . Sol.[B] Q.18 Assertion (A) : logab > 0, if a > 1 and 0 < b < 1. Reason (R) : The number of digits in 2100 is 31. Sol.[D] Q.19 Assertion (A) : If N = 20 0.4 1       then N contains 7 digits before decimal. Reason (R): Characteristic of the logarithm of N to the base 10 is 7. Sol.[D] Q.20 Consider the following statements Assertion (A) : The equation 5 – x 1 log(x 1) 2 3 = + has two distinct real solution. Reason (R) : a N loga N = when a > 0, a  1 and N > 0 Sol.[B] Q.21 Assertion (A): The equation log3(5 + 4 log3) (x –1) = 2 has only one solution. Reason (R) : log3 x 2 = log3 2 x ; x = 0. Sol.[B] Q.22 Assertion (A): log57 > log83. Reason (R) : 1 < log57 < 2 and 0 < log83 < 1. Sol.[A] Q.23 Assertion (A): The number of solution of log | x | = ex is two. Reason (R) : If log303 = a, log305 = b then log308 = 3(1 – a – b) Sol.[D] Q.24 Assertion (A): 9 x + 6x = 2.4x has no solution. Reason (R) : log2(9 –2 x ) = 10log(3–x) has only one solution. Sol.[C]
Q.