InterviewSolution

Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.	Does the fraph of the function ` f (x) = x^(2) -3` have y - axis symmetry?
Answer» We need to check if f(a) = f(-a). ` f(-a) = (-a)^(2) - 3 = a^(2) - 3 = f(a)` Therefore, the function ` f(x) = x^(2) - 3` must have y - axis symmetry.

2.	Which of the following pairs of graphs intersect? (i) ` y = x^(2) -x and y = 1` (ii) ` y = x^(2) - 2x + 3 and y = sin x ` (iii) ` y = x^(2) - x+1 and y = x-4`
Answer» (i) ` y = x^(2) - x and y = 1" intersect if "x^(2) - x = 1 or x^(2)-x-1 = 0`, which has real roots . Hence, the graphs intersect. (ii) ` y = x^(2) - 2x+3 and y = sin x` intersect if ` x^(2) - 2x+3 = sin x or (x-1)^(2) + 2=sin x`, which is not possible since L.H.S. has least value 2, while R.H.S. has maximum value 1. `(iii) y = x^(2)-x + 1 and y = x - 4" intersect if " x^(2) - x+1 = x-4 or x^(2) -2x + 5 = 0`, which has non-real roots. Hence, the graph do not intersect.

3.	Does the graph below represent a function or a relation?
Answer» Vertical line x = 2 meets the above graph at (2, 1) and (2, 2). Hence the graph does not represent a function.

4.	The graph of y = f(x) is shown, find the number of solution of f(f(x)) = 2.
Answer» Given equation if f(f(x)) = 2. Let f(x) = t ` :. F(t) = 2` `:. T = -3, 1/2, 3` Now, f(x) =- 3 has no solution ` f(x) = 1/2 ` has two solutions f(x) = 3 has one solution Hence total number of solutions is three.

5.	In how many points graph of `y=x^3-3x2+5x-3`interest the x-axis?
Answer» The number of points in which ` y = x^(3) - 3x^(2) + 5x - 3` intersects the x-axis is same as the number of real roots of the equation ` x ^(3) - 3x^(2) + 5x - 3 = 0`. Now we can see that x = 1 satisfies the equation, hence one root of the equation is x = 1. Now dividing ` x^(3) - 3x^(2) + 5x-3` by x - 1 , we have the quotient ` x^(2) - 2x + 3`. Hence the equation reduces to `(x-1)(x^(2)-2x+3)= 0`. Hence the only root of the equation is x = 1. Thus, the graph of ` y = x^(3) - 3x^(2) + 5x - 3` cuts the x - axis in one point only.

6.	Find the asymptote of the function ` y = (2x^(2) + 3x + 1)/ x` if any.
Answer» Vertical asymptote: Since denominator is x, vertical asymptote is x = 0. Oblique asymptote: Let the oblique asymptote by y = mx + n. Therefore, `underset(x to infty)"lim"[(2x^(2)+3x+1)/x -(mx + n)] = 0` ` rArrunderset(x to infty)"lim"[(2x^(2)+3x+1-(mx+n)x)/x ] =0` ` rArr underset(x to infty)"lim"[[(2-m)x^(2)+(3-n)x+1)/x]=0` Clearly, 2 - m = 0 ` rArr underset(x to infty)"lim" [((3-n)x+1)/x]=0` ` rArr underset(x to infty)"lim"[(3-n)+1/x]=0` ` rArr 3- n = 0` ` rArr n = 3` Thus, the oblique asymptote is y = 2x + 3. Alternative method: We have ` (2x^(2)+3x+1)/x = 2x+3+1/x` Here quotient is 2x+3, hence the oblique asymptote is y = 2x +3.

7.	In the following graph, state the absolute and local maximum and minimum values of the function.
Answer» x = 2 is the point of local minima, and the local minimum value is f(2) = 1. x = 4 is the point of absolute maxima, and the absolute maximum value is f(4) = 4. x = 5 is the point of local minima, and the local minimum value is f(5) = 2. x = 6 is the point of local maxima, and the local maximum value is f(6) = 3. x = 7 is the point of absolute minima, and the local minimum value is f (7) = 0.

8.	The graph of ` y = f(x)` is as shown in the following figure. Identify the points of discontinuity and give the reason for the same.
Answer» y = f(x) is discontinuous at x =- 3 as the value of the function is missing at this point. Assigning proper value to f(-3) [from the graph f(-3) = 2], f(x) can be made continuous at this point. y = f(x) is discontinuous at x = 0 as there is a jump in the value of the function at this point. Here ` underset(x to 0^(-))"lim" f(x) and underset(x to 0^(+))"lim" f(x) = 3`. At this point, the function cannot be made continuous. At x = 2, the function is discontinuous as the value of the value of the function is not same as the limiting value of function. That is, ` underset(x to 2) " lim" f (x) = 1` (limit of the function exists), but f(2) =- 1. At x = 6, the function is discontinuous as the value of the function is decreasing indefinitely when x is approached from either its left or its right.

9.	Which of the following functions has (have) y-symmetry or origin symmetry? (i) ` f(x) = x^(2) sin x" "(ii) f(x) = log (x+sqrt(1+x^(2)))` (iii)` f(x) = (e^(x)+e^(-x))/2" "(iv) f(x) = {{:(0", ""If x is rational "),(1", " "If x is irrational"):}`
Answer» (i) ` f(x) = (-x)^(2) sin (-x) =- x^(2) sin x =- f(x)`, hence function has origin symmetry (ii) `f(-x) = log (-x+sqrt(1+(-x)^(2)))` `= log(((-x+sqrt(1+x^(2)))(x+sqrt(1+x^(2))))/((x+sqrt(1+x^(2)))))` `= log(1/(x+sqrt(1+x^(2))))` ` = - log(x+sqrt(1+x^(2)))` =- f(x), hence function has origin symmetry. (iii) `f(-x) = (e^(-x)+e^(-(-x)))/2 =(e^(-x)+e^(x))/2 = f(x)`,hence function has y-symmetry. (iv) ` f(x) ={{:(0", ""If x is rational"),(1", ""If x is irrational"):}` `rArr f(-x) ={{:(0", ""If -x is rational"),(1", ""If -x is irrational"):}` `={{:(0", ""If x is rational"),(1", ""If x is irrational"):}` = f(x) Hence functions has y-symmetry.

10.	Let `f:R->R` and `g:R->R` be two one-one and onto functions such that they are mirror images of each other about the line `y=a`. If `h(x)=f(x)+g(x)`, then `h(x)` is (A) one-one onto (B) one-one into (D) many-one into (C) many-one onto
Answer» y = f(x) and y = g(x) are mirror image of each other about y = a. For some `x = b, g(b) - a - f(b)` ` rArr f(b) + g(b) = 2a` ` rArr h(b) = f(b) + g(b) = 2a` (constant) Hence h (x) is a constant function. Thus, it is neither one-one, nor onto.

11.	Does the graph of the function ` f(x) = 1//x^(3) ` have origin symmetry?
Answer» We must check if f(-a) =- f(a). ` f(-a)=1//(-a)^(3) =- 1//a^(3) =- f(a)` Since f(-a) =- f(a) for all values of a, the graph of the function ` f(x) = 1//x^(3)` must be symmetric around the origin.

12.	Find the horizontal, vertical and oblique asymptotes of each of the curves. `{:((a),y=x/(x+4),,(b),y=(x^(2)+4)/(x^(2)-1)),((c),y=x^(3)/(x^(2)+3x-10),,(d),y=(x^(3)+1)/(x^(3)+x)),((e),y=x/(root(4)(x^(4)+1)),,(f),y=(x-9)/(sqrt(4x^(2)+3x+2))),((g),y=1/(2^(x)-1),,(h),y=1/(log_(e) x)),((i),y= 1/(2^(x) - 1),,,):}`
Answer» (a) ` y= x/(x+4)` Since the degree of numerator and denominator is same, there is a horizontal asymptote y = 1 as `underset(x to infty )"lim"x/(x+4) = underset(x to infty)"lim" 1/(1+4/x) = 1`. Comparing the denominator to zero, we have x+4 = 0 or x =- 4, which is the vertical asymptote. Clearly, the curve has no oblique asymptote. (b) ` y = (x^(2)+4)/(x^(2)-1)` Since the degree of numerator and denominator is same, there is a horizontal asymptote y = 1 as ` underset(x to infty)"lim" (x^(2)+4)/(x^(2)-1) = underset( x to infty)"lim"(1+4/x^(2))/(1-1/x^(2)) = 1`. Comparing denominator to zero, we have ` x^(2) - 1 = 0`. So x = -1 , which are vertical asymptotes. Clearly, the curve has no obluque asymptote. (c) ` y = x^(3)/(x^(2)+3x-10)` Since the degree of numerator is higher than the degree of denominator, three is no horizontal asymptote. Comparing the denominator to zero, we have ` x^(2) + 3x - 10 = 0`. So x = 2 or x = -5, which are vertical asymptotes Also ` y = x- 3+(19x - 30)/(x^(2) + 3x -10)` , so y = x - 3 is an oblique asymptote. (d) ` y = (x^(3)+1)/(x^(3)+x)` Since the degree of numerator is same as the degree of denominator, there is a horizontal asymptote y = 1 as ` underset(x to infty)"lim" (x^(3)+1)/(x^(3)+x)=underset(x to infty)"lim" (1+1/x^(3))/(1-1/x^(2)) = 1`. Comparing the denominator to zero, we have ` x^(3) + x = 0`. So x = 1 , which is a vertical asymptote. Clearly, the curve has no oblique asymptote. (e) ` y = underset(x to infty)"lim" x/root(4)(x^(4)+1)` `underset(x to infty)"lim" 1/(1/xroot(4)(x^(4)+1))=underset(x to infty)"lim" 1/root(4)(1+1/x^(4))= 1` Hence x = 1 is a horizontal asymptote. Similarly, ` underset(x to - infty)"lim" 1/(1/x root(4)(x^(4)+1) )= underset(x to - infty)"lim"1/root(-4)(1+1/x^(4))=-1` So the curve has horizontal asymptotes ` y = pm 1/2`. Clearly, the curve has no vertical as well as oblique asymptote. (g) ` y = sqrt(4x^(2)+3)/(x-2)` ` underset(x to infty)"lim" sqrt(4x^(2)+3)/(x-2) = 2 and underset(x to - infty)"lim" sqrt(4x^(2)+3)/(x-2) =- 2` So the curve has horizontal asymptotes ` y = pm 2`. Comparing the denominator to zero, we have x - 2 = 0 or x = 2, which is a vertical asymptote. (h) ` y = 1/(log_(e)x)` ` underset(x to infty)"lim" 1/(log_(e)x) =0` So y = 0 is a horizontal asymptote. ` underset(x to 1^(+))"lim" 1/(log_(e) x) = infty and underset(x to 1^(-))"lim" 1/(log_(e) x) =- infty` So x = 0 is a vertical asymptote. (i) ` y = 1/(2^(x)-1)` ` underset(x to 0^(+))"lim" 1/(2^(x)-1)= infty underset(x to 0^(-))"lim" 1/(2^(x)-1) =- infty` So x = 0 is a vertical asymptote. ` underset(x to infty)"lim" 1/(2^(x)-1) = 0` So y = 0 is a horizontal asymptote. ` underset(x to - infty)"lim" 1/(2^(x)-1) =- 1` So y =- 1 is a horizontal asymptote.

13.	Which of the following functions has (have) y-symmetry or origin symmetry? (i) ` f(x) = x^(2) sin x" "(ii) f(x) = log ((1-x)/(1+x))` (iii)` f(x) = x/(e^(x)-1)+x/2 + 1`
Answer» (i) `f(-x) = (-x)^(2) sin (-x) =- x^(2) sin x =- f(x)`, hence the function has origin symmetry (ii) `f(-x)= log((1-(-x))/(1+(-x))) = log ((1+x)/(1-x))=- f(x)`hence the function has origin symmetry (iii) `f(x) = x/(e^(x)-1)+x/2 + 1` ` rArr f(-x) = (-x)/(e^(-x)-1)-x/2 + 1` ` = (xe^(x))/(e^(x)-1) - x/2 + 1` `=(xe^(x)-x+x)/(e^(x)-1) - x/2+1` ` = x+x/(e^(x)-1)-x/2 + 1` ` = x/(e^(x)-1) + x/2 + 1` = f(x) Hence the function has y-symmetry.