Assertion (A) & Reason (B) MCQ

MCQ 11 Mark

Assertion $( A )$ : The maximum value of the function $f(x)$ $=x^5, x \in[-1,1]$, is attained at its critical point, $x=0$. Reason (R): The maximum of a function can only occur at points where derivative is zero.

A
Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation for (A).
B
Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation for (A).
C
(A) is false but $(R)$ is true.
✓
Both $(A)$ and $(R)$ are false.

Answer

Correct option: D.

Both $(A)$ and $(R)$ are false.

View full question & answer→

MCQ 21 Mark

Consider the function $f(x)=x^{1 / 3}, x \in R$.
Assertion (A) : $f$ has a point of inflexion at $x=0$.
Reason $( R ): f^{\prime \prime}(0)=0$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
✓
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: C.

(A) is true but (R) is false.

(c) : Given $f(x)=x^{1 / 3}, x \in R$
$\Rightarrow f^{\prime}(x)=\frac{1}{3} x^{-2 / 3}$
and $f^{\prime \prime}(x)=\frac{1}{3}\left(-\frac{2}{3}\right) x^{-5 / 3}$
$
=\frac{-2}{9 x^{5 / 3}}
$

Both $f^{\prime}(x)$ and $f^{\prime \prime}(x)$ exist at all points except at $x=0$
So, $f^{\prime \prime}(0)$ does not exist.
Thus, reason is wrong.
However, we note that $f^{\prime \prime}(x)>0$ for $x<0$ and $f^{\prime \prime}(x)<0$ for $x>0$
$\Rightarrow f^{\prime \prime}(x)$ changes sign from positive to negative as we move from left to right through 0 .
So, $f$ has a point of inflexion at $x=0$.

View full question & answer→

MCQ 31 Mark

Assertion (A) : Let $f: R \rightarrow R$ be a function such that $f(x)=x^3+x^2+3 x+\sin x$. Then, $f$ is an increasing function.
Reason (R) : If $f^{\prime}(x)<0$, then $f(x)$ is a decreasing function.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

$
\begin{array}{l}
\text { (b) : } f^{\prime}(x)=3 x^2+2 x+3+\cos x \\
=3\left\{x^2+\frac{2 x}{3}+1\right\}+\cos x=3\left\{\left(x+\frac{1}{3}\right)^2-\frac{1}{9}+1\right\}+\cos x \\
=3\left(x+\frac{1}{3}\right)^2+\frac{8}{3}+\cos x>0 \quad\left[\because 3\left(x+\frac{1}{3}\right)^2 \geq 0,-1 \leq \cos x \leq 1\right]
\end{array}
$
$\therefore f(x)$ is an increasing function.
If $f^{\prime}(x)<0$, then $f(x)$ is a decreasing function.

View full question & answer→

MCQ 41 Mark

Assertion (A) : Both $\sin x$ and $\cos x$ are decreasing functions in $\left(\frac{\pi}{2}, \pi\right)$.
Reason (R): If a differentiable function decreases in $(a, b)$, then its derivative also decreases in $(a, b)$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Reason is wrong which is clear from the graph that $f(x)$ is differentiable in $(a, b)$.
Also, $a < x_1 < x_2 < b$
But $f^{\prime}\left(x_1\right)=\tan \phi_1<\tan \phi_2=f^{\prime}\left(x_2\right)$
$\Rightarrow$ Derivative is increasing.

View full question & answer→

MCQ 51 Mark

Assertion (A): If the function $f(x)=\frac{a e^x+b e^{-x}}{c e^x+d e^{-x}}$ is increasing function of $x$, then $b c>a d$.
Reason (R): A function $f(x)$ is increasing if $f^{\prime}(x)>0$ for all $x$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
✓
(A) is false but (R) is true.

Answer

Correct option: D.

(A) is false but (R) is true.

(d) : $f^{\prime}(x)=\frac{2(a d-b c)}{\left(c e^x+d e^{-x}\right)^2}$and $f(x)$ is an increasing function.
$
\begin{array}{l}
\therefore \quad f^{\prime}(x)>0 \Rightarrow \frac{2(a d-b c)}{\left(c e^x+d e^{-x}\right)^2}>0 \\
\Rightarrow \quad 2(a d-b c)>0 \Rightarrow a d>b c \Rightarrow b c < ad
\end{array}
$

View full question & answer→

MCQ 61 Mark

Assertion (A) : A window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is $12 m$, then length $1.782 m$ and breadth $2.812 m$ of the rectangle will produce the largest area of the window.
Reason $( R )$ : For maximum or minimum, $f^{\prime}(x)=0$.

✓
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: A.

Both (A) and (R) are true and (R) is the correct explanation of (A).

(a) : Let each side of the equilateral triangle be $x m$ and $l m$ be the height (length) of the rectangular part of the window, then
$
\begin{aligned}
& x+x+x+l+l=12 \\
\Rightarrow & 3 x+2 l=12 \Rightarrow l=6-\frac{3}{2} x ...(i)
\end{aligned}
$
Let $A m ^2$ be the area of the window corresponding to the above dimensions, then
$
\begin{aligned}
A & =l x+\frac{\sqrt{3}}{4} x^2 ; x>0, l>0 \\
& =\left(6-\frac{3}{2} x\right) x+\frac{\sqrt{3}}{4} x^2 (Using (i))
\end{aligned}
$
Now, $\frac{d A}{d x}=6-3 x+\frac{\sqrt{3}}{2} x$
$
\begin{array}{l}
\frac{d A}{d x}=0 \Rightarrow 6-3 x+\frac{\sqrt{3}}{2} x=0 \\
\Rightarrow 12-6 x+\sqrt{3} x=0 \Rightarrow x=\frac{12}{6-\sqrt{3}} \approx 2.812
\end{array}
$
Now, $\frac{d^2 A}{d x^2}=-3+\frac{\sqrt{3}}{2}<0$
$\therefore \quad A$ has local maxima at $x=2.812$
For $x=2.812, l=6-\frac{3}{2}(2.812)=1.782$
$\therefore$ Height of rectangular part $=1.782 m$ and breadth $=2.812 m$

View full question & answer→

MCQ 71 Mark

Assertion (A) : If $f^{\prime}(x)=(x-1)^3(x-2)^8$, then $f(x)$ has neither maximum nor minimum at $x=2$.
Reason $( R ): f^{\prime}(x)$ changes sign from negative to positive at $x=2$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
✓
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: C.

(A) is true but (R) is false.

(c) : It is clear from figure that $f^{\prime}(x)$ has no sign change at $x=2$. Hence, $f(x)$ is neither maximum nor minimum at $x=2$.

View full question & answer→

MCQ 81 Mark

Assertion (A): The graph $y=x^3+a x^2+b x+c$ has extremum, if $a^2<3 b$.
Reason (R): A function, $y=f(x)$ has an extremum, if $\frac{d y}{d x}>0$ or $\frac{d y}{d x}<0$ for all $x \in R$.

✓
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: A.

Both (A) and (R) are true and (R) is the correct explanation of (A).

(a) : For an extremum
$
\begin{aligned}
& \frac{d y}{d x}>0 \text { or } \frac{d y}{d x}<0 \text { for all } x \in R \\
\therefore & \frac{d y}{d x}=3 x^2+2 a x+b>0 \\
\Rightarrow & 3 x^2+2 a x+b>0 \Rightarrow D<0 \\
\Rightarrow & 4 a^2-4 \cdot 3 \cdot b<0 \Rightarrow a^2<3 b
\end{aligned}
$

View full question & answer→

MCQ 91 Mark

Assertion (A) : The absolute maximum and minimum values of $f(x)=x^2 \sqrt{1+x}$ in $\left[-1, \frac{1}{2}\right]$ are $\frac{\sqrt{6}}{8}$ and 0 respectively.
Reason (R) : Let $f$ be a differentiable function on $I$ and $x_0$ be any interior point of $I$. If $f$ attains its absolute maximum or minimum value at $x_0$, then $f^{\prime}\left(x_0\right)=0$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
✓
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: C.

(A) is true but (R) is false.

(c) : Given, $f(x)=x^2 \sqrt{1+x}$ ....(i)
The given function is differentiable for all $x$ in $\left[-1, \frac{1}{2}\right]$.
Differentiating (i) w.r.t. $x$, we get
$
\begin{array}{l}
f^{\prime}(x)=x^2 \cdot \frac{1}{2}(1+x)^{-1 / 2}+\sqrt{1+x} \cdot 2 x \\
=\frac{x^2}{2 \sqrt{1+x}}+2 x \sqrt{1+x}=\frac{5 x^2+4 x}{2 \sqrt{1+x}}
\end{array}
$
Now $f^{\prime}(x)=0 \Rightarrow \frac{5 x^2+4 x}{2 \sqrt{1+x}}=0 \Rightarrow 5 x^2+4 x=0$
$
\Rightarrow x(5 x+4)=0 \Rightarrow x=0,-\frac{4}{5}
$
Also $0,-\frac{4}{5}$ both lie in $\left[-1, \frac{1}{2}\right]$, therefore, 0 and $-\frac{4}{5}$ both are stationary points.
Further, $f(0)=0, f\left(-\frac{4}{5}\right)=\frac{16}{25} \cdot \frac{1}{\sqrt{5}}=\frac{16}{25 \sqrt{5}}$
$
f(-1)=0, f\left(\frac{1}{2}\right)=\frac{1}{4} \cdot \sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{8}
$
Therefore, the absolute maximum value is $\frac{\sqrt{6}}{8}$ and the absolute minimum value is 0 .
The point of maxima is $\frac{1}{2}$ and points of minima are $\{-1,0\}$.

View full question & answer→

MCQ 101 Mark

Assertion (A) : $f(x)=\frac{1}{x-7}$ is decreasing $\forall x \in R-\{7\}$.
Reason (R) : $f^{\prime}(x)<0 \forall x \neq 7$.

✓
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: A.

Both (A) and (R) are true and (R) is the correct explanation of (A).

(a) : $f(x)=\frac{1}{x-7}$
$
\Rightarrow \quad f^{\prime}(x)=\frac{-1}{(x-7)^2}<0 \forall x \in R-\{7\}
$

View full question & answer→

MCQ 111 Mark

Assertion (A): Let $f(x)=x^3+a x^2+b x+5 \sin ^2 x$, then the condition that $f(x)$ is always one-one function is $a^2-3 b+15<0$.
Reason (R) : $f(x)$ to be one one either $f$ is strictly increasing or strictly decreasing.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

$
\begin{array}{l}
\text { (a): } f(x)=x^3+a x^2+b x+5 \sin ^2 x \\
\therefore \quad f^{\prime}(x)=3 x^2+2 a x+b+5 \sin 2 x
\end{array}
$
For one-one function, $f^{\prime}(x)>0$ for $x \in R$
$
\begin{array}{l}
\Rightarrow \quad 3 x^2+2 a x+b+5 \sin 2 x>0 \\
\Rightarrow \quad 3 x^2+2 a x+(b-5)>0 \Rightarrow(2 a)^2-4 \cdot 3(b-5)<0 \\
\Rightarrow \quad a^2-3 b+15<0
\end{array}
$

View full question & answer→

Maths Assertion (A) & Reason (B) MCQ

Test yourself on this topic