Sets, Relations & Functions PYQs Set 22 (PDF)

Set 22 of 25 · Questions 631-660 of 734

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Mere Teacher · Previous Year Questions JEE Main Mathematics · Sets, Relations & Functions · Set 22 (Q631-660)
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631. $The equation x^{2} - 4 x + [x] + 3 = x [x], where [x] denotes the greatest integer function, has:$ (1) $Exactly two solutions in (- \infty\infty)$ (2) $No Solution$ (3) $A unique solution in (- \infty, 1)$ (4) $A unique solution in (- \infty, \infty)$ 632. $If a > 0, [\cdot] denotes greatest integer function and lim_{x \to a^{-}} (\frac{∣ x ∣ ^{3}}{a} - [\frac{x}{a}]^{3}) = k, lim_{x \to a^{+}} (\frac{∣ x ∣ ^{3}}{a} - [\frac{x}{a}]^{3}) = l, then$ (1) $k = l$ (2) $k - l = 1$ (3) $l - k = 1$ (4) $l = a^{2}, k does not exist$ 633. $If f (x) = {x^{2} - 1, 2 x + 3, 0 < x < 2 2 \leq x < 3, the quadratic equation whose roots are lim_{x \to 2^{-}} f (x) and lim_{x \to 2^{+}} f (x) is$ (1) $x^{2} - 14 x + 49 = 0$ (2) $x^{2} - 6 x + 9 = 0$ (3) $x^{2} - 10 x + 21 = 0$ (4) $x^{2} - 7 x + 8 = 0$ 634. $If f (x) = \frac{2 ^{x}}{2 ^{x} + 2}, x \in R, then \sum_{k = 1}^{81} f (\frac{k}{82}) is equal to$ (1) $1.81 2$ (2) $41$ (3) $82$ (4) $\frac{81}{2}$ 635. $lim_{x \to - 9} \frac{( 2.5 ) ^{81 - x^{2}} - ( 0.4 ) ^{x + 9}}{x + 9} =$ (1) $18 log (2.5) + log (0.4)$ (2) $log (2.5) - log (0.4)$ (3) $18 (log (2.5) + log (0.4))$ (4) $- 19 log (0.4)$ 636. $f (x) = \frac{1}{2} - tan (\frac{π x}{2}), - 1 < x < 1 and g (x) = 3 + 4 x - 4 x^{2} . Find the domain of (f + g) .$ (1) $(\frac{- 1}{2}, 1]$ (2) $(- 1, 1)$ (3) $[\frac{- 1}{2}, 1)$ (4) $[- \frac{1}{2}, \frac{3}{2}]$ 637. $Let f : R \to R be a function defined by f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b, a \neq = 1 . If f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y, then the value of 28 \sum_{i = 1}^{5} ∣ f (i) ∣ is$ (1) $545$ (2) $715$ (3) $735$ (4) $675$ 638. $For x \in R if f (x) = lo g_{10} (\frac{3 - x}{x}), then the domain of f is$ (1) $[0, \frac{3}{2}]$ (2) $(0, \frac{3}{2}]$ (3) $[0, 1]$ (4) $(0, 1]$ 639. $If [\cdot] denotes greatest integer function, then lim_{x \to - \frac{3}{5}} \frac{1}{x} [\frac{- 1}{x}] =$ (1) $\frac{- 5}{3}$ (2) $\frac{5}{3}$ (3) $\frac{10}{3}$ (4) $\frac{- 10}{3}$ 640. $Let [\cdot] denote the greatest integer function. Assertion (A) : lim_{x \to \infty} \frac{[ x ]}{x} = 1 . Reason (R): f (x) = x - 1, g (x) = [x], h (x) = x and lim_{x \to \infty} \frac{f [ x ]}{x} = lim_{x \to \infty} \frac{h ( x )}{x} = 1$ (1) $A is true, R is true: R is correct explanation of A$ (2) $A, R are true; R is not the correct explanation of A$ (3) $A is true, R is false$ (4) $A is false, R is true$ 641. $If f (x) = f^{’} (x) + f^{’ ’} (x) + f^{’ ’ ’} (x) + and f (0) = 1, then f (x) =$ (1) $e^{x /2}$ (2) $e^{x}$ (3) $e^{2 x}$ (4) $e^{4 x}$ 642. $If y = \frac{a x + b}{c x + d}, then \frac{d x}{d y} =$ (1) $\frac{a d - b c}{( a x + b ) ^{2}}$ (2) $\frac{a d - b c}{( a - cy ) ^{2}}$ (3) $\frac{a d + b c}{( c x + d ) ^{2}}$ (4) $\frac{a d - b c}{( y c - a ) ^{2}}$ 643. $Let f and g be functions satisfying f (x + y) = f (x) f (y), f (1) = 7 and g (x + y) = g (x y), g (1) = 1, for all x, y \in N . If \sum_{x = 1}^{n} \frac{f ( x )}{g ( x )} = 19607, then n is equal to:$ (1) $7$ (2) $6$ (3) $4$ (4) $5$ 644. $Let R_{1} and R_{2} be two relations defined as follows: R_{1} = {(a, b) \in R^{2} : a^{2} + b^{2} \in Q} and R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \in / Q}, where Q is the set of all rational numbers. Then:$ (1) $Neither R_{1} nor R_{2} is transitive.$ (2) $R2is transitive but R_{1} is not transitive.$ (3) $R_{1} and R_{2} are both transitive.$ (4) $R_{1} is transitive but R_{2} is not transitive.$ 645. $lim_{x \to - a} \frac{x ^{7} + a ^{7}}{x + a} = 7 \Rightarrow a =$ (1) $\pm 7$ (2) $\pm 6$ (3) $\pm 1$ (4) $\pm 2$ 646. $If a and b are fixed non-zero constants, then the derivative of \frac{a}{x ^{4}} - \frac{b}{x ^{2}} + cos x is ma + nb - p$ (1) $m = 4 x^{3}; n = \frac{- 2}{x ^{3}}; p = sin x$ (2) $m = \frac{- 4}{x ^{5}}; n = \frac{2}{x ^{3}}; p = sin x$ (3) $m = \frac{- 4}{x ^{5}}; n = \frac{- 2}{x ^{3}}; p = - sin x$ (4) $m = 4 x^{3}; n = \frac{2}{x ^{3}}; p = - sin x$ 647. $If f (x) is a function such that f^{''} (x) + f (x) = 0 and g (x) = (f (x))^{2} + (f^{'} (x))^{2} and g (3) = 3 then g (8) =$ (1) $5$ (2) $0$ (3) $3$ (4) $8$ 648. $If f (x) = ⎩ ⎨ ⎧ \frac{1}{2} (b^{2} - a^{2}), \frac{1}{2} b^{2} - \frac{x ^{2}}{6} - \frac{a ^{3}}{3 x}, \frac{1}{3} (\frac{b ^{3} - a ^{3}}{x}), 0 \leq x \leq a a < x \leq b x > b then$ (1) $f^{’ ’} (a) = 2 b$ (2) $f^{’ ’} (a) = 1$ (3) $f^{’ ’} (a) = b^{2} - a^{2}$ (4) $f^{’} (x) is not differentiable at x = a$ 649. $Suppose f : R \to (0, \infty) be a differentiable function such that 5 f (x + y) = f (x) \cdot f (y), \forall x, y \in R . If f (3) = 320, then \sum_{n = 0}^{5} f (n) is equal to$ (1) $6875$ (2) $6575$ (3) $6825$ (4) $6528$ 650. $lim_{n \to \infty} (\frac{1 ^{k} + 2 ^{k} + 3 ^{k} + \dots + n ^{k}}{n ^{k + 1}}) =$ (1) $\frac{1}{k}$ (2) $\frac{2}{k + 1}$ (3) $\frac{1}{k + 1}$ (4) $\frac{2}{k}$ 651. $Define f (x) = ⎩ ⎨ ⎧ \frac{1 + p x - 1 - p x}{x}, \frac{2 x + 1}{x - 2}, - 1 \leq x < 0 0 \leq x \leq 1 If lim_{x \to 0} f (x) exists, then p =$ (1) $-1$ (2) $- \frac{1}{2}$ (3) $\frac{1}{2}$ (4) $1$ 652. $lim x \to \infty (\frac{6 x ^{2} - cos 3 x}{x ^{2} + 5} - \frac{5 x ^{3} + 3}{x ^{6} + 2}) =$ (1) $11$ (2) $0$ (3) $-1$ (4) $1$ 653. $Let f (x) = sin^{- 1} x and g (x) = \frac{x ^{2} - x - 2}{2 x ^{2} - x - 6} . If g (2) = lim_{x \to 2} g (x), then the domain of the function f \circ g is:$ (1) $(- \infty, - 2] \cup [- \frac{4}{3}, \infty)$ (2) $(- \infty, - 1] \cup [2, \infty)$ (3) $(- \infty, - 2] \cup [- 1, \infty)$ (4) $(- \infty, - 2] \cup [- \frac{3}{2}, \infty)$ 654. $Let A = {1, 2, 3, \dots, 10} and f : A \to A be defined as f (k) = {k + 1 k if k is odd if k is even . Then the number of possible functions g : A \to A such that g \circ f = f is :$ (1) $10^{5}$ (2) $^{10} C_{5}$ (3) $5^{5}$ (4) $5!$ 655. $If A = {x \in R : ∣ x ∣ < 2} and B = {x \in R : ∣ x - 2∣ \geq 3}; then:$ (1) $A - B = [- 1, 2)$ (2) $A \cup B = R - (2, 5)$ (3) $B - A = R - (- 2, 5)$ (4) $A \cap B = (- 2, - 1)$ 656. $If g (x) = x^{2} + x - 1 and (g \circ f) (x) = 4 x^{2} - 10 x + 5, then f (\frac{5}{4}) is equal to.$ (1) $\frac{1}{2}$ (2) $- \frac{1}{2}$ (3) $- \frac{1}{3}$ (4) $\frac{1}{3}$ 657. $lim x \to 0 \frac{tan ^{4} x - sin ^{4} x}{x ^{6}} =$ (1) $\frac{1}{2}$ (2) $-$ (3) $2$ (4) $4$ 658. $Let f : R \to R , g : R \to R and h : R \to R be differentiable functions such that f(x) = x^{3} + 3x + 2, g(f(x)) = x and h(g(g(x))) = x for all x \in R. Then$ (1) $g^{'} (2) = \frac{1}{15}$ (2) $h^{'} (1) = 666$ (3) $h (0) = 16$ (4) $h (g (3)) = 36$ 659. $Let f be a function such that 3 f (x) + 2 f (\frac{m}{19 x}) = 5 x, x \neq = 0, where m = \sum_{i = 1}^{9} (i)^{2} . Then f (5) - f (2) is equal to$ (1) $18$ (2) $36$ (3) $9$ (4) $-9$ 660. $Suppose that a function f : R \to R satisfies f (x + y) = f (x) f (y) for all x, y \in R and f (1) = 3 . If \sum_{i = 1}^{n} f (i) = 363, then n is equal to$